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CBSE Class 10 Maths Case Study Questions for Maths Chapter 6 (Published by CBSE)

Check case study questions released by cbse for class 10 maths chapter 6 - triangles. solve these questions to prepare the case study questions for the cbse class 10 maths exam 2021-22..

CBSE Class 10 Maths Case Study Questions for Chapter 6 - Triangles are available here. Students must practice with these questions to perform well in their Maths exam. All these case study questions have been published by the Central Board of Secondary Education (CBSE). For the convenience of students, all the questions are provided with answers.

Case Study Questions for Class 10 Maths Chapter 6 - Triangles

CASE STUDY 1:

Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles.The height of Vijay’s house if 20m when Vijay’s house casts a shadow 10m long on the ground. At the same time, the tower casts a shadow 50m long on the ground and the house of Ajay casts 20m shadow on the ground.

1. What is the height of the tower?

Answer: c) 100m

2. What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m?

Answer: d) 60m

3. What is the height of Ajay’s house?

Answer: b) 40m

4. When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house?

Answer: a) 16m

5. When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house?

Answer: d) 8m

CASE STUDY 2:

Rohan wants to measure the distance of a pond during the visit to his native. He marks points A and B on the opposite edges of a pond as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using rope connecting B with another point C are a distance of 12m, connecting C to point D at a distance of 40m from point C and the connecting D to the point A which is are a distance of 30m from D such the ∠ ADC=90 0 .

1. Which property of geometry will be used to find the distance AC?

a) Similarity of triangles

b) Thales Theorem

c) Pythagoras Theorem

d) Area of similar triangles

Answer: c)Pythagoras Theorem

2. What is the distance AC?

Answer: a) 50m

3. Which is the following does not form a Pythagoras triplet?

a) (7, 24, 25)

b) (15, 8, 17)

c) (5, 12, 13)

d) (21, 20, 28)

Answer: d) (21, 20, 28)

4. Find the length AB?

Answer: b) 38m

5. Find the length of the rope used.

Answer: c)82m

SCALE FACTOR

Case study:

A scale drawing of an object is the same shape at the object but a different size. The scale of a drawing is a comparison of the length used on a drawing to the length it represents. The scale is written as a ratio. The ratio of two corresponding sides in similar figures is called the scale factor

Scale factor= length in image / corresponding length in object

If one shape can become another using revising, then the shapes are similar. Hence, two shapes are similar when one can become the other after a resize, flip, slide or turn. In the photograph below showing the side view of a train engine. Scale factor is 1:200

This means that a length of 1 cm on the photograph above corresponds to a length of 200cm or 2 m, of the actual engine. The scale can also be written as the ratio of two lengths.

1. If the length of the model is 11cm, then the overall length of the engine in the photograph above, including the couplings (mechanism used to connect) is:

Answer: a)22m

2. What will affect the similarity of any two polygons?

a) They are flipped horizontally

b) They are dilated by a scale factor

c) They are translated down

d) They are not the mirror image of one another.

Answer: d)They are not the mirror image of one another

3. What is the actual width of the door if the width of the door in photograph is 0.35cm?

Answer: a)0.7m

4. If two similar triangles have a scale factor 5:3 which statement regarding the two triangles is true?

a) The ratio of their perimeters is 15:1

b) Their altitudes have a ratio 25:15

c) Their medians have a ratio 10:4

d) Their angle bisectors have a ratio 11:5

Answer: b)Their altitudes have a ratio 25:15

5. The length of AB in the given figure:

Answer: c)4cm

Also Check:

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Case Study Questions for Class 10 Maths Chapter 6 Triangles

• Last modified on: 9 months ago
• Reading Time: 4 Minutes

Case Study Questions:

Question 1:

Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles. The height of Vijay’s house if 20 m when Vijay’s house casts a shadow 10 m long on the ground. At the same time, the tower casts a shadow 50m long on the ground and the house of Ajay casts 20 m shadow on the ground.

(i) What is the height of the tower? (a) 20 m (b) 50 m (c) 100 m (d) 200 m

(ii) What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m? (a) 75 m (b) 50 m (c) 45 m (d) 60 m

(iii) What is the height of Ajay’s house? (a) 30 m (b) 40 m (c) 50 m (d) 20 m

(iv) When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house? (a) 16 m (b) 32 m (c) 20 m (d) 8 m

(v) When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house? (a) 15 m (b) 32 m (c) 16 m (d) 8 m

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CBSE Case Study Questions for Class 10 Maths Triangles Free PDF

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CBSE Case Study Questions for Class 10 Maths Triangles PDF

Mcq set 1 -, mcq set 2 -, checkout our case study questions for other chapters.

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Class 10th Maths - Triangles Case Study Questions and Answers 2022 - 2023

By QB365 on 09 Sep, 2022

QB365 provides a detailed and simple solution for every Possible Case Study Questions in Class 10th Maths Subject - Triangles, CBSE. It will help Students to get more practice questions, Students can Practice these question papers in addition to score best marks.

QB365 - Question Bank Software

Triangles case study questions with answer key.

10th Standard CBSE

Final Semester - June 2015

(ii) If m, n and r are the sides of right triangle ABJ, then which of the following can be correct?

(iii) If \(\Delta\) ABJ ~ \(\Delta\) ADH, then which similarity criterion is used here?

(iv) If  \(\angle\) ABJ = 90° and B, J are mid points of sides AD and AH respectively and BJ || DH, then which of the following option is false?

(ii) Distance travelled by aeroplane towards west after   \(1 \frac{1}{2}\)   hr is

(iii) In the given figure, \(\angle\) POQ is 

(iv) Distance between aeroplanes after  \(1 \frac{1}{2}\)   hr is

(v) Area of \(\Delta\) POQ is

(ii) Which criterion of similarity is applied here?

(iii) Height of the building is

(iv) In \(\Delta\) ABM, if LBAM = 30°, then LMCD is equal to 

(v) If \(\Delta\) ABM and \(\Delta\) CDMare similar where CD = 6 ern, MD = 8 cm and BM = 24 ern, then AB is equal to 

(ii) What will be the length of shadow of tower when Meenal's house casts a shadow of 15 m? 

(iii) Height of Aruns house is 

(iv) If tower casts a shadow of 40 rn, then find the length of shadow of Arun's house 

(v) If tower casts a shadow of 40 m, then what will be the length of shadow of Meenal's house? 

(ii) The length of CD is 

(iii) Area of whole empty land is 

(iv) Area of \(\Delta\) PAB is 

(v) Area of \(\Delta\) PCD is

(ii) The value of x is

(iii) The value of PR is 

(iv) The value of RQ is 

(v) How much distance will be saved in reaching city Q after the construction of highway? 

(ii) In if AB || CD, and DO = 3x - 19, OB = x - 5, OC = x - 3 and AO = 3, then the value of x can be 

(iii) In if OD = 3x - 1, OB = 5x - 3, OC = 2x + 1 and AO = 6x - 5, then the value of x is 

(iv) In \(\Delta\)  ABC, if PQ || BC and AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, then AB + PQ is equal to 

(v) In \(\Delta\) DEF, if RS || EF, DR = 4x - 3, DS = 8x - 7, ER = 3x - 1 and FS = 5x - 3, then the value of x is 

(ii) Length of BC =

(iii) Length of AD =

(iv) Length of ED = 

(v) Length of AE = 

(ii) If the distance between Aruna and the bus is twice as much as the height of the bus, then the height of the bus is 

(iii) If the distance of Aruna from the building is twelve times the height of the bus, then the ratio of the heights of bus and building is 

(iv) What is the ratio of the distance between Aruna and top of bus to the distance between the tops of bus and building? 

(v). What is the height of the building? 

(ii) The distance between the hotel A and hut-I is

(iii) If the horizontal distance between the hut -1 and hut -2 is 8 miles, then the distance between the two hotels is 

(iv) If the distance from mountain top to hut-1 is 5 miles more than that of distance from hotel B to mountain top, then what is the distance between hut-2 and mountain top?

(v) What is the ratio of areas of two parts formed in the complete figure?

(b) What is the length of string pulled in 12 seconds?

(c) What is the length of string after 12 seconds?

(d) What will be the horizontal distance of the fly from her after 12 seconds?

(e) The given problem is based on which concept?

(ii) What is the distance AC?

(iii) Which is the following does not form a Pythagoras triplet?

(iv)  Find the length AB.

(v)  Find the length of the rope used.

(ii) Which is the correct similarity criteria applicable for smaller triangles at the upper part of this kite?

(iii)    Sides of two similar triangles are in the ratio 4:9. Corresponding medians of these triangles are in the ratio,

(iv) In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. This theorem is called as,

(v) What is the area of the kite, formed by two perpendicular sticks of length 6 cm and 8 cm?

(ii) Find the correct similarity criteria applicable for triangles ABE and CDE.

(iii) Find the length of her shadow after 4 seconds.

(iv) Sides of two similar triangles are in the ratio 9:16. Find the ratio of Corresponding areas of these triangles.

(v) Find the ratio AC:CE.

(ii) State the criterion of similarity, that will be used in the above found triangles.

(iii) What is the height of the tree?

(iv) What is the distance between Rashmi and Gulmohar tree?

*****************************************

Triangles case study questions with answer key answer keys.

(i) (a): Speed = 1200 km/hr \(\text { Time }=1 \frac{1}{2} \mathrm{hr}=\frac{3}{2} \mathrm{hr}\) \(\therefore\)  Required distance = Speed x Time \(=1200 \times \frac{3}{2}=1800 \mathrm{~km}\) (ii) (c): Speed = 1500 km/hr Time =  \(\frac{3}{2}\)  hr. \(\therefore\)  Required distance = Speed x Time \(=1500 \times \frac{3}{2}=2250 \mathrm{~km}\) (iii) (b): Clearly, directions are always perpendicular to each other. \(\therefore \quad \angle P O Q=90^{\circ}\) (iv) (a): Distance between aeroplanes after  \(1\frac{1}{2}\)   hour  \(\begin{array}{l} =\sqrt{(1800)^{2}+(2250)^{2}}=\sqrt{3240000+5062500} \\ =\sqrt{8302500}=450 \sqrt{41} \mathrm{~km} \end{array}\) (v) (d): Area of  \(\Delta\) POQ= \(\frac{1}{2}\) x base x height \(=\frac{1}{2} \times 2250 \times 1800=2250 \times 900=2025000 \mathrm{~km}^{2}\)

(i) (c): Since, \(\angle\) B = \(\angle\) D = 90°, \(\angle\) AMB = \(\angle\) CMD ( \(\because\) Angle of incident = Angle of reflection) \(\therefore\) By similarity criterion, \(\Delta\) ABM \(\sim\) \(\Delta\) CDM (ii) (a) (iii) (c) :  \(\because\)   \(\Delta\) ABM \(\sim\) \(\Delta\) CDM \(\begin{array}{l} \therefore \quad \frac{A B}{C D}=\frac{B M}{D M} \Rightarrow \frac{A B}{1.8}=\frac{2.5}{1.5} \\ \Rightarrow \quad A B=\frac{2.5 \times 1.8}{1.5}=3 \mathrm{~m} \end{array}\) (iv) (b): Since    \(\Delta\) ABM \(\sim\) \(\Delta\) CDM \(\because\) \(\angle\) A= \(\angle\) C=30° [ \(\therefore\) Corresponding angles of similar triangles are also equal] (v) (b): Since    \(\Delta\) ABM \(\sim\) \(\Delta\) CDM \(\therefore \frac{A B}{C D}=\frac{B M}{M D} \Rightarrow \quad \frac{A B}{6}=\frac{24}{8} \Rightarrow A B=18 \mathrm{~cm}\)

(i) (a): In \(\Delta\) PAB and \(\Delta\) PQR, \(\angle\) P = \(\angle\) P (Common) \(\angle\) A = \(\angle\) Q (Corresponding angles) By AA similarity criterion, \(\Delta\) PAB \(\sim\) \(\Delta\) PQR \(\therefore \frac{A B}{Q R}=\frac{P A}{P Q} \Rightarrow \frac{A B}{12}=\frac{6}{24} \Rightarrow A B-3 \mathrm{~m}\) (ii) (d): Similarly, \(\Delta\) PCD and \(\Delta\) PQR are similar \(\therefore \frac{P C}{P Q}=\frac{C D}{Q R} \Rightarrow \frac{14}{24}=\frac{C D}{12} \Rightarrow C D=7 \mathrm{~m}\) (iii) (a): Area of whole empty land =  \(\frac{1}{2}\) x base x height =  \(\frac{1}{2}\)   x 1 x15= 90 m 2 . (iv) (b): Since,  \(\Delta\) PAB \(\sim\) \(\Delta\) PQR \(\therefore \frac{\operatorname{ar}(\Delta P A B)}{a r(\Delta P Q R)}=\left(\frac{P A}{P Q}\right)^{2}=\left(\frac{6}{24}\right)^{2}=\frac{1}{16}\) \(\Rightarrow \quad \operatorname{ar}(\Delta P A B)=\frac{1}{16} \times 90=\frac{45}{8} \mathrm{~m}^{2}\)   \(\left[\because \operatorname{ar}(\Delta P Q R)=90 \mathrm{~m}^{2}\right]\) (v) (d): Since,  \(\Delta\) PCD  \(\sim\) \(\Delta\) PQR \(\therefore \frac{\operatorname{ar}(\Delta P C D)}{\operatorname{ar}(\Delta P Q R)}=\left(\frac{P C}{P Q}\right)^{2}=\left(\frac{14}{24}\right)^{2}=\left(\frac{7}{12}\right)^{2}\) \(\Rightarrow \operatorname{ar}(\Delta P C D)=\frac{90 \times 49}{144}=\frac{245}{8} \mathrm{~m}^{2}\)

(i) (b) (ii) (c): Using Pythagoras theorem, we have PQ 2 = PR 2 + RQ 2 \(\Rightarrow(26)^{2}=(2 x)^{2}+(2(x+7))^{2} \Rightarrow 676=4 x^{2}+4(x+7)^{2} \) \(\Rightarrow 169=x^{2}+x^{2}+49+14 x \Rightarrow x^{2}+7 x-60=0\) \(\Rightarrow x^{2}+12 x-5 x-60=0 \) \(\Rightarrow x(x+12)-5(x+12)=0 \Rightarrow(x-5)(x+12)=0 \) \(\Rightarrow x=5, x=-12\) \(\therefore \quad x=5\)   [Since length can't be negative] (iii) (a) : PR = 2x = 2 x 5 = 10 km (iv) (b): RQ= 2(x + 7) = 2(5 + 7) = 24 km (v) (d): Since, PR + RQ = 10 + 24 = 34 km Saved distance = 34 - 26 = 8 km

(i) (c) (ii) (b): Since \(\Delta\) AOB ~ \(\Delta\) COD [ByAA similarity criterion] \(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{3}{x-3}=\frac{x-5}{3 x-19}\) \(\Rightarrow 3(3 x-19)=(x-5)(x-3) \) \(\Rightarrow 9 x-57=x^{2}-3 x-5 x+15 \Rightarrow x^{2}-17 x+72=0 \) \(\Rightarrow(x-8)(x-9)=0 \Rightarrow x=8 \text { or } 9\) (iii) (c) : Since, \(\Delta\) AOB ~ \(\Delta\) COD [ByAA similarity criterion] \(\therefore \frac{A O}{O C}=\frac{B O}{O D} \Rightarrow \frac{6 x-5}{2 x+1}=\frac{5 x-3}{3 x-1}\) \(\Rightarrow(6 x-5)(3 x=1)=(5 x-3)(2 x+1) \) \(\Rightarrow \quad 18 x^{2}-6 x-15 x+5=10 x^{2}+5 x-6 x-3 \) \(\Rightarrow \quad 8 x^{2}-20 x+8=0 \Rightarrow 2 x^{2}-5 x+2=0\) From options, x = 2 is the only value that satisfies this equation. (iv) (d): Since,  \(\Delta\) APQ ~ \(\Delta\) ABC [ByAA similarity criterion] \(\therefore \quad \frac{A P^{\circ}}{A B}=\frac{A Q}{A C}=\frac{P Q}{B C} \Rightarrow \frac{2.4}{A B}=\frac{2}{5}=\frac{P Q}{6} \) \(\therefore \quad A B=\frac{2.4 \times 5}{2}=6 \mathrm{~cm} \text { and } P Q=\frac{2 \times 6}{5}=2.4 \mathrm{~cm} \) \(\therefore \quad A B+P Q=6+2.4=8.4 \mathrm{~cm}\) (v) (a): Since,   \(\Delta\) DRS ~ \(\Delta\) DEF  [ByAA similarity criterion] \(\therefore \quad \frac{D E}{D R}=\frac{D F}{D S} \Rightarrow \frac{D E}{D R}-1=\frac{D F}{D S}-1 \) \(\Rightarrow \frac{D E-D R}{D R}=\frac{D F-D S}{D S} \Rightarrow \frac{E R}{D R}=\frac{F S}{D S} \) \(\Rightarrow \quad \frac{D R}{E R}=\frac{D S}{F S} \Rightarrow \frac{4 x-3}{3 x-1}=\frac{8 x-7}{5 x-3} \) \(\Rightarrow \quad 20 x^{2}-12 x-15 x+9=24 x^{2}-8 x-21 x+7 \) \(\Rightarrow \quad 4 x^{2}-2 x-2=0 \Rightarrow 2 x^{2}-x-1=0\) Only option (a) i.e., x = 1 satisfies this equation.

(i) (b): If \(\Delta\) AED and \(\Delta\) BEC, are similar by SAS similarity rule, then their corresponding proportional sides are  \(\frac{B E}{A E}=\frac{C E}{D E}\) (ii) (c): By Pythagoras theorem, we have \(\begin{array}{l} B C=\sqrt{C E^{2}+E B^{2}}=\sqrt{4^{2}+3^{2}}=\sqrt{16+9} \\ =\sqrt{25}=5 \mathrm{~cm} \end{array}\) (iii) (a): Since \(\Delta\) ADE and \(\Delta\) BCE are similar. \(\therefore \quad \frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{A D}{B C} \) \(\Rightarrow \frac{2}{3}=\frac{A D}{5} \Rightarrow A D=\frac{5 \times 2}{3}=\frac{10}{3} \mathrm{~cm}\) (iv) (b): \(\frac{\text { Perimeter of } \triangle A D E}{\text { Perimeter of } \Delta B C E}=\frac{E D}{C E} \) \(\Rightarrow \frac{2}{3}=\frac{E D}{4} \Rightarrow E D=\frac{4 \times 2}{3}=\frac{8}{3} \mathrm{~cm}\) (v) (d) :   \(\frac{\text { Perimeter of } \Delta A D E}{\text { Perimeter of } \Delta B C E}=\frac{A E}{B E} \Rightarrow \frac{2}{3} B E=A E\) \(\Rightarrow A E=\frac{2}{3} \sqrt{B C^{2}-C E^{2}} \) \(\text { Also, in } \triangle A E D, A E=\sqrt{A D^{2}-D E^{2}}\)

(i) (b):  AC 2 = AB 2 + BC 2     [ By Pythagoras theorem ] ⇒ AC 2 = (1.8)2 + (2.4) 2 ⇒ AC 2 = 3.24 + 5.76 ⇒ AC 2 = 9 ⇒ AC = 3m (ii) (c):  She pulls the string at the rate of 5cm/s ∴ String pulled in 12 second = 12 x 5 = 60cm = 0.6 (iii) (a):  Length of string out after 12 second is AP. ⇒ AP = AC – String pulled by Nazima in 12 seconds. ⇒ AP = (3 − 0.6)m = 2.4m (iv) (b):  In △ADB, AB 2 + BP 2 = AP 2     ⇒ (1.8)2 + BP 2 = (2.4) 2 ⇒ BP 2 = 5.76 − 3.24    ⇒ BP 2 = 5.76 − 3.24 ⇒ BP 2 = 2.52    ⇒ BP = 1.58 m Horizontal distance of fly = BP + 1.2m Horizontal distance of fly =1.58m + 1.2m ∴ Horizontal distance of fly = 2.78m (v) (a): Triangles

(i) (c): Pythagoras Theorem (ii) (a):  ADC is a right-angled triangle. By Pythagoras theorem we get AC 2 = (30) 2 + (40) 2 ⇒ AC 2 = 900 + 1600 ⇒ AC 2 = 2500 ⇒ AC = 50m (iii) (d):  (21, 20, 28) since, \((28)^{2} \neq(20)^{2}+(21)^{2}\) (iv) (b):  AC = 50m and BC = 12m ⇒ AB = AC - BC ⇒ AB = 50 - 12 = 38m (v) (c): Length of the rope used = 30 + 40 + 12 = 82m

(i) (c): In this kite shape, diagonals bisect each other at 90 o . (ii) (b): SAS (iii) (b):  Since ratio of sides of similar triangles = Ratio of their corresponding medians Therefore, the required ratio is 4 : 9. (iv) (d): Converse of Pythagoras theorem (v) (c): Since ,length sticks are 6 cm and 8 cm or length of diagonals Then, Area of kite  \(=\frac{1}{2} \times d_{1} \times d_{2}\) \(=\frac{6 \times 8}{2}\) = 24 cm 2

(i) (c):  Let AB denote the lamp-post and CD the girl after walking for 4 seconds away from the lamp-post. From the figure, DE is the shadow of the girl. Let DE be x metres. Now, her distance from the base of the lamp = BD = 1.2 m x 4 = 4.8 m. (ii) (a): In  \(\triangle\)   ABE and \(\triangle\) CDE, \(\angle\) B = \(\angle\) D ( Each is of 90 o ) and \(\angle\) E = \(\angle\) E (same angle) So,  \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\)  (AA similarity criterion) (iii) (d):   \(\Delta \mathrm{ABE} \sim \Delta \mathrm{CDE}\) \(\Rightarrow \frac{\mathrm{BE}}{\mathrm{DE}}=\frac{\mathrm{AB}}{\mathrm{CD}}\) \(\Rightarrow \frac{4.8+x}{x}=\frac{3.6}{0.9} \quad \Rightarrow 4.8+x=4 x\) \(\Rightarrow 3 x=4.8 \ \Rightarrow \ x=1.6\) So, the shadow of the girl after walking for 4 seconds is 1.6 m long. (iv) (b): Since ratio of the areas of two similar triangles is equal to the ratio of the square of their corresponding sides, Ratio of areas of similar triangles = \(\sqrt{9}\)  :  \(\sqrt{16}\) = 3 : 4 (v) (b):  \(\frac{A E}{C E}=\frac{B E}{D E}=\frac{4.8+1.6}{1.6}=\frac{6.4}{1.6}=4\) \(\Rightarrow\)   AE = 4 CE \(\Rightarrow\)  AC + CE = 4 CE \(\Rightarrow A C=3 C E \Rightarrow \frac{A C}{C E}=\frac{3}{1}\)

(i) (b):  (b)  \(\triangle\) s PQM and RSM (ii) (d): AA similarity As  \(\angle \mathrm{PQM}=\angle \mathrm{RSM}=90^{\circ}\) and angle of incidence is equal to angle of reflection. Then,  \(\angle \mathrm{PMQ}=\angle \mathrm{RMS}\) (iii) (a): Since,  \(\Delta \mathrm{PMQ} \sim \Delta \mathrm{RMS}\) Then,  \(\frac{\mathrm{PQ}}{\mathrm{RS}}=\frac{\mathrm{QM}}{\mathrm{MS}} \Rightarrow \mathrm{PQ}=\frac{2.8 \times 2}{1.4}\) = 4m (iv) (c): Distance between Rashmi and Gulmohar tree QS = QM + MS = 2.8 + 1.4 = 4.2 m

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Class 10 Maths Case Study Questions Chapter 6 Triangles PDF Download

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Case study Questions in the Class 10 Mathematics Chapter 6  are very important to solve for your exam. Class 10 Maths Chapter 6 Case Study Questions have been prepared for the latest exam pattern. You can check your knowledge by solving Class 10 Maths Case Study Questions Chapter 6  Triangles

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In CBSE Class 10 Maths Paper, Students will have to answer some questions based on Assertion and Reason. There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Triangles Case Study Questions With Answers

Here, we have provided case-based/passage-based questions for Class 10 Maths  Chapter 6 Triangles

Case Study/Passage-Based Questions

Case Study 1:

(i) What is the height of the tower?

Answer: (a) 100 m

(ii) What will be the length of the shadow of the tower when Meenal’s house casts a shadow of 15 m? 

Answer: (c) 75 m

(iii) Height of Aruns house is 

Answer: (d) 40 m

(iv) If the tower casts a shadow of 40 rn, then find the length of the shadow of Arun’s house 

Answer: (b) 17 m

(v) If the tower casts a shadow of 40 m, then what will be the length of the shadow of Meenal’s house? 

Answer: (d) 8 m

Case Study 2:

Rohan wants to measure the distance of a pond during the visit to his native. He marks points A and B on the opposite edges of a pond as shown in the figure below. To find the distance between the points, he makes a right-angled triangle using a rope connecting B with another point C is a distance of 12m, connecting C to point D at a distance of 40m from point C, and connecting D to the point A which is are a distance of 30m from D such the  ∠ ADC=90 0 .

1. Which property of geometry will be used to find the distance AC?

a) Similarity of triangles

b) Thales Theorem

c) Pythagoras Theorem

d) Area of similar triangles

Answer: c)Pythagoras Theorem

2. What is the distance AC?

Answer: a) 50m

3. Which is the following does not form a Pythagoras triplet?

a) (7, 24, 25)

b) (15, 8, 17)

c) (5, 12, 13)

d) (21, 20, 28)

Answer: d) (21, 20, 28)

4. Find the length AB?

Answer: b) 38m

5. Find the length of the rope used.

Answer: c)82m

Hope the information shed above regarding Case Study and Passage Based Questions for Class 10 Maths Chapter 6 Triangles with Answers Pdf free download has been useful to an extent. If you have any other queries about CBSE Class 10 Maths Triangles Case Study and Passage-Based Questions with Answers, feel free to comment below so that we can revert back to us at the earliest possible By Team Study Rate

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CBSE Class 10 Maths Syllabus

• Real Numbers
• Polynomials
• Pair of Linear Equations in Two Variables
• Quadratic Equations
• Arithmetic Progressions
• Coordinate Geometry
• Probability
• Introduction to Trigonometry
• Some Applications of Trigonometry
• Constructions
• Area Related to Circles
• Surface Areas and Volumes

UNIT I: NUMBER SYSTEMS

1. REAL NUMBER

Fundamental Theorem of Arithmetic - statements after reviewing work done earlier and after illustrating and motivating through examples, Proofs of irrationality.

UNIT II: ALGEBRA

• POLYNOMIALS Zeros of a polynomial. Relationship between zeros and coefficients of quadratic polynomials.
• PAIR OF LINEAR EQUATIONS IN TWO VARIABLES Pair of linear equations in two variables and graphical method of their solution, consistency/inconsistency. Algebraic conditions for number of solutions. Solution of a pair of linear equations in two variables algebraically - by substitution, by elimination. Simple situational problems.
• QUADRATIC EQUATIONS Standard form of a quadratic equation ax2 + bx + c = 0, (a ≠ 0). Solutions of quadratic equations (only real roots) by factorization, and by using quadratic formula. Relationship between discriminant and nature of roots. Situational problems based on quadratic equations related to day to day activities to be incorporated.
• ARITHMETIC PROGRESSIONS Motivation for studying Arithmetic Progression Derivation of the nth term and sum of the first n terms of A.P. and their application in solving daily life problems.

UNIT III: COORDINATE GEOMETRY Coordinate Geometry Review: Concepts of coordinate geometry, graphs of linear equations. Distance formula. Section formula (internal division).

UNIT IV: GEOMETRY

• TRIANGLES Definitions, examples, counter examples of similar triangles. 1. (Prove) If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. 2. (Motivate) If a line divides two sides of a triangle in the same ratio, the line is parallel to the third side. 3.(Motivate) If in two triangles, the corresponding angles are equal, their corresponding sides are proportional and the triangles are similar. 4.(Motivate) If the corresponding sides of two triangles are proportional, their corresponding angles are equal and the two triangles are similar. 5.(Motivate) If one angle of a triangle is equal to one angle of another triangle and the sides including these angles are proportional, the two triangles are similar.
• CIRCLES Tangent to a circle at, point of contact 1. (Prove) The tangent at any point of a circle is perpendicular to the radius through the point of contact. 2.(Prove) The lengths of tangents drawn from an external point to a circle are equal.

UNIT V: TRIGONOMETRY

• INTRODUCTION TO TRIGONOMETRY Trigonometric ratios of an acute angle of a right-angled triangle. Proof of their existence (well defined); motivate the ratios whichever are defined at 0° and 90°. Values of the trigonometric ratios of 30°, 45°, and 60°. Relationships between the ratios.
• TRIGONOMETRIC IDENTITIES Proof and applications of the identity sin2A + cos2A = 1 . Only simple identities to be given.
• HEIGHTS AND DISTANCES: Angle of elevation, Angle of Depression. (10)Periods Simple problems on heights and distances. Problems should not involve more than two right triangles. Angles of elevation / depression should be only 30°, 45°, and 60°.

UNIT VI: MENSURATION

• AREAS RELATED TO CIRCLES Area of sectors and segments of a circle. Problems based on areas and perimeter / circumference of the above said plane figures. (In calculating area of segment of a circle, problems should be restricted to central angle of 60°, 90° and 120° only.
• SURFACE AREAS AND VOLUMES Surface areas and volumes of combinations of any two of the following: cubes, cuboids, spheres, hemispheres and right circular cylinders/cones.

UNIT VII: STATISTICS AND PROBABILITY

• STATISTICS Mean, median and mode of grouped data (bimodal situation to be avoided).
• PROBABILITY (10) Periods Classical definition of probability. Simple problems on finding the probability of an event.

PRESCRIBED BOOKS:

• Mathematics - Textbook for class IX - NCERT Publication
• Mathematics - Textbook for class X - NCERT Publication
• Guidelines for Mathematics Laboratory in Schools, class IX - CBSE Publication
• Guidelines for Mathematics Laboratory in Schools, class X - CBSE Publication
• Laboratory Manual - Mathematics, secondary stage - NCERT Publication
• Mathematics exemplar problems for class IX, NCERT publication.
• Mathematics exemplar problems for class X, NCERT publication.

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There are times when students just run through any particular topic with casual awareness there by missing out on a few imperative “between the lines” concepts. Such things are the major causes of weak fundamental understandings of students. So in such cases Worksheets act as a boon and critical helpful tool which gauges the in-depth understanding of children highlighting doubts and misconceptions, if any.

Worksheets classifies the important aspects of any topic or chapter taught in the class in a very easy manner and increases the awareness amongst students.When students try to solve a worksheet they get to understand what are the key important factors which needs the main focus.Sometimes it happens that due to shortage of time all the major points of any particular topic gets skipped in the class or teacher rushes through , due to shortage of time. A worksheet thus provides a framework for the entire chapter and can help covering those important aspects which were rushed in the class and ensure that students record and understand all key items.

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Worksheets help students be focussed and attentive in the class because they know after the class is over they will be assigned a worksheet which they need to solve so if they miss or skip any point in the class they may not be able to solve the worksheet completely and thereby lose reputation in the class.

Often students revise the chapter at home reading their respective textbooks. Thus more often than not they do miss many important points. Worksheets thus can be used intentionally to help guide student’s to consult textbooks. Having students write out responses encourages their engagement with the textbooks, the questions chosen indicate areas on which to focus. Explicitly discussing the worksheets and why particular questions are asked helps students reflect on what is important.

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Class 10 Maths Chapter 6 Case Based Questions - Triangles

Case study - 1.

Case Study - 2

Vijay is trying to find the average height of a tower near his house. He is using the properties of similar triangles.The height of Vijay’s house if 20m when Vijay’s house casts a shadow 10m long on the ground. At the same time, the tower casts a shadow 50m long on the ground and the house of Ajay casts 20m shadow on the ground.

Q1: What is the height of the tower? (a) 20m (b) 50m (c) 100m (d) 200m Ans:  (c) Explanation: The properties of similar triangles state that the ratios of the corresponding sides of the triangles are equal. Hence the ratio of the height of Vijay's house to the length of its shadow equals to the ratio of the height of the tower to the length of its shadow. Therefore, the height of the tower can be calculated as follows: Height of the tower = (Height of Vijay's house / Length of Vijay's house's shadow) * Length of the tower's shadow = (20m / 10m) * 50m = 100m   Q2: What will be the length of the shadow of the tower when Vijay’s house casts a shadow of 12m? (a) 75m (b) 50m (c) 45m (d) 60m Ans:  (d) Explanation: Applying the same method, we can find the length of the shadow of the tower. Length of the tower's shadow = (Length of Vijay's house's shadow / Height of Vijay's house) * Height of the tower = (12m / 20m) * 100m = 60m   Q3: What is the height of Ajay’s house? (a) 30m (b) 40m (c) 50m (d) 20m Ans:  (b) Explanation: Similarly, the height of Ajay's house can be calculated as follows: Height of Ajay's house = (Height of Vijay's house / Length of Vijay's house's shadow) * Length of Ajay's house's shadow = (20m / 10m) * 20m = 40m   Q4: When the tower casts a shadow of 40m, same time what will be the length of the shadow of Ajay’s house? (a) 16m (b) 32m (c) 20m (d) 8m Ans: (a) Explanation: Length of Ajay's house's shadow = (Length of the tower's shadow / Height of the tower) * Height of Ajay's house = (40m / 100m) * 40m = 16m   Q5: When the tower casts a shadow of 40m, same time what will be the length of the shadow of Vijay’s house? (a) 15m (b) 32m (c) 16m (d) 8m Ans:  (d) Explanation: Length of Vijay's house's shadow = (Length of the tower's shadow / Height of the tower) * Height of Vijay's house = (40m / 100m) * 20m = 8m  

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CBSE Class 10 Maths Case Study Questions PDF

Download Case Study Questions for Class 10 Mathematics to prepare for the upcoming CBSE Class 10 Final Exam. These Case Study and Passage Based questions are published by the experts of CBSE Experts for the students of CBSE Class 10 so that they can score 100% on Boards.

CBSE Class 10 Mathematics Exam 2024  will have a set of questions based on case studies in the form of MCQs. The CBSE Class 10 Mathematics Question Bank on Case Studies, provided in this article, can be very helpful to understand the new format of questions. Share this link with your friends.

Table of Contents

Chapterwise Case Study Questions for Class 10 Mathematics

Inboard exams, students will find the questions based on assertion and reasoning. Also, there will be a few questions based on case studies. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

The above  Case studies for Class 10 Maths will help you to boost your scores as Case Study questions have been coming in your examinations. These CBSE Class 10 Mathematics Case Studies have been developed by experienced teachers of cbseexpert.com for the benefit of Class 10 students.

• Class 10th Science Case Study Questions
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Class 10 Maths Syllabus 2024

Chapter-1  real numbers.

Starting with an introduction to real numbers, properties of real numbers, Euclid’s division lemma, fundamentals of arithmetic, Euclid’s division algorithm, revisiting irrational numbers, revisiting rational numbers and their decimal expansions followed by a bunch of problems for a thorough and better understanding.

Chapter-2  Polynomials

This chapter is quite important and marks securing topics in the syllabus. As this chapter is repeated almost every year, students find this a very easy and simple subject to understand. Topics like the geometrical meaning of the zeroes of a polynomial, the relationship between zeroes and coefficients of a polynomial, division algorithm for polynomials followed with exercises and solved examples for thorough understanding.

Chapter-3  Pair of Linear Equations in Two Variables

This chapter is very intriguing and the topics covered here are explained very clearly and perfectly using examples and exercises for each topic. Starting with the introduction, pair of linear equations in two variables, graphical method of solution of a pair of linear equations, algebraic methods of solving a pair of linear equations, substitution method, elimination method, cross-multiplication method, equations reducible to a pair of linear equations in two variables, etc are a few topics that are discussed in this chapter.

Chapter-4  Quadratic Equations

The Quadratic Equations chapter is a very important and high priority subject in terms of examination, and securing as well as the problems are very simple and easy. Problems like finding the value of X from a given equation, comparing and solving two equations to find X, Y values, proving the given equation is quadratic or not by knowing the highest power, from the given statement deriving the required quadratic equation, etc are few topics covered in this chapter and also an ample set of problems are provided for better practice purposes.

Chapter-5  Arithmetic Progressions

This chapter is another interesting and simpler topic where the problems here are mostly based on a single formula and the rest are derivations of the original one. Beginning with a basic brief introduction, definitions of arithmetic progressions, nth term of an AP, the sum of first n terms of an AP are a few important and priority topics covered under this chapter. Apart from that, there are many problems and exercises followed with each topic for good understanding.

Chapter-6  Triangles

This chapter Triangle is an interesting and easy chapter and students often like this very much and a securing unit as well. Here beginning with the introduction to triangles followed by other topics like similar figures, the similarity of triangles, criteria for similarity of triangles, areas of similar triangles, Pythagoras theorem, along with a page summary for revision purposes are discussed in this chapter with examples and exercises for practice purposes.

Chapter-7  Coordinate Geometry

Here starting with a general introduction, distance formula, section formula, area of the triangle are a few topics covered in this chapter followed with examples and exercises for better and thorough practice purposes.

Chapter-8  Introduction to Trigonometry

As trigonometry is a very important and vast subject, this topic is divided into two parts where one chapter is Introduction to Trigonometry and another part is Applications of Trigonometry. This Introduction to Trigonometry chapter is started with a general introduction, trigonometric ratios, trigonometric ratios of some specific angles, trigonometric ratios of complementary angles, trigonometric identities, etc are a few important topics covered in this chapter.

Chapter-9  Applications of Trigonometry

This chapter is the continuation of the previous chapter, where the various modeled applications are discussed here with examples and exercises for better understanding. Topics like heights and distances are covered here and at the end, a summary is provided with all the important and frequently used formulas used in this chapter for solving the problems.

Chapter-10  Circle

Beginning with the introduction to circles, tangent to a circle, several tangents from a point on a circle are some of the important topics covered in this chapter. This chapter being practical, there are an ample number of problems and solved examples for better understanding and practice purposes.

Chapter-11  Constructions

This chapter has more practical problems than theory-based definitions. Beginning with a general introduction to constructions, tools used, etc, the topics like division of a line segment, construction of tangents to a circle, and followed with few solved examples that help in solving the exercises provided after each topic.

Chapter-12  Areas related to Circles

This chapter problem is exclusively formula based wherein topics like perimeter and area of a circle- A Review, areas of sector and segment of a circle, areas of combinations of plane figures, and a page summary is provided just as a revision of the topics and formulas covered in the entire chapter and also there are many exercises and solved examples for practice purposes.

Chapter-13  Surface Areas and Volumes

Starting with the introduction, the surface area of a combination of solids, the volume of a combination of solids, conversion of solid from one shape to another, frustum of a cone, etc are to name a few topics explained in detail provided with a set of examples for a better comprehension of the concepts.

Chapter-14  Statistics

In this chapter starting with an introduction, topics like mean of grouped data, mode of grouped data, a median of grouped, graphical representation of cumulative frequency distribution are explained in detail with exercises for practice purposes. This chapter being a simple and easy subject, securing the marks is not difficult for students.

Chapter-15  Probability

Probability is another simple and important chapter in examination point of view and as seeking knowledge purposes as well. Beginning with an introduction to probability, an important topic called A theoretical approach is explained here. Since this chapter is one of the smallest in the syllabus and problems are also quite easy, students often like this chapter

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Question 5 - Case Based Questions (MCQ) - Chapter 6 Class 10 Triangles

Last updated at April 16, 2024 by Teachoo

An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour.

This question is inspired from  Ex 6.5, 11 - Chapter 6 Class 10 - Triangles

What is the distance travelled by aeroplane towards north after 11/2 hours?

(a) 1000 , (c) 1500 .

What is the distance travelled by aeroplane towards west after 11/2 hours?

∠ AOB is

(a) 90°    , (b) 45°  , (c) 30°    , (d) 60°.

How far apart will the two planes be after 1 1/2 hours?

(a) √22,50,000  , (b) √32,40,000, (c) √54,90,000     , (d) none of these.

The given problem is based on which concept?

(a) triangles, (b) co-ordinate geometry, (c) height and distance.

Question An aeroplane leaves an airport and flies due north at a speed of 1000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1200 km per hour. We know that Speed = (𝐷𝑖𝑠𝑡𝑎𝑛𝑐𝑒 )/𝑇𝑖𝑚𝑒 Distance = Speed × Time Thus, OA = 1500 km & OB = 1800 km Question 1 What is the distance travelled by aeroplane towards north after 11/2 hours? (a) 1000 (b) 1200 (c) 1500 (d) 1800 Distance travelled towards north = OA = 1500 km So, the correct answer is (c) Question 2 What is the distance travelled by aeroplane towards west after 11/2 hours? (a) 1000 (b) 1200 (c) 1500 (d) 1800 Distance travelled towards north = OB = 1800 km So, the correct answer is (d) Question 3 ∠ AOB is (a) 90° (b) 45° (c) 30° (d) 60° Since North is always perpendicular to West Hence, ∠AOB = 90° So, the correct answer is (c) Question 4 How far apart will the two planes be after 1 1/2 hours? (a) √22,50,000 (b) √32,40,000 (c) √54,90,000 (d) none of these Distance between two planes after 1 1/2 hours = AB In right angled Δ AOB By Pythagoras Theorem AB2 = OA2 + OB2 AB2 = (1500)2 + (1800)2 AB2 = 22,50,000 + 32,40,000 AB2 = 54,90,000 + 32,40,000 AB = √𝟓𝟒𝟗𝟎𝟎𝟎𝟎 km So, the correct answer is (c) Question 5 The given problem is based on which concept? (a) Triangles (b) Co-ordinate geometry (c) Height and Distance (d) none of these Since this question involves Pythagoras Theorem Which is in Triangles Class 10 So, the correct answer is (a)

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• NCERT Solutions
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• Chapter 6: Triangles

NCERT Solutions for Class 10 Maths Chapter 6 Triangles

Ncert solutions for class 10 maths chapter 6 – download free pdf.

NCERT Solutions for Class 10 Maths Chapter 6 Triangles are provided here, which is considered to be one of the most important study materials for the students studying in CBSE Class 10. Chapter 6 of NCERT Solutions for Class 10 Maths is well structured in accordance with the CBSE Syllabus for 2023-24. It covers a vast topic, including a number of rules and theorems. Students often tend to get confused about which theorem to use while solving a variety of questions.

Download Exclusively Curated Chapter Notes for Class 10 Maths Chapter – 6 Triangles

Download most important questions for class 10 maths chapter – 6 triangles.

The solutions provided at BYJU’S are designed in such a way that every step is explained clearly and in detail. The Solutions for NCERT Class 10 Maths are prepared by the subject experts to help students prepare better for their board exams. These solutions will be helpful not only for exam preparations, but also in solving homework and assignments.

The CBSE Class 10 examination often asks questions, either directly or indirectly, from the NCERT textbooks. Thus, the NCERT Solutions for Chapter 6 Triangles of Class 10 Maths is one of the best resources to prepare, and equip oneself to solve any type of questions in the exam, from the chapter. It is highly recommended that the students practise these NCERT Solutions on a regular basis to excel in the Class 10 board examinations.

• Chapter 1 Real Numbers
• Chapter 2 Polynomials
• Chapter 3 Pair of Linear Equations in Two Variables
• Chapter 4 Quadratic Equations
• Chapter 5 Arithmetic Progressions
• Chapter 6 Triangles
• Chapter 7 Coordinate Geometry
• Chapter 8 Introduction to Trigonometry
• Chapter 9 Some Applications of Trigonometry
• Chapter 10 Circles
• Chapter 11 Constructions
• Chapter 12 Areas Related to Circles
• Chapter 13 Surface Areas and Volumes
• Chapter 14 Statistics
• Chapter 15 Probability

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Access Answers of NCERT Class 10 Maths Chapter 6 – Triangles

Exercise 6.1 page: 122.

1. Fill in the blanks using correct word given in the brackets:-

(i) All circles are __________. (congruent, similar)

Answer: Similar

(ii) All squares are __________. (similar, congruent)

(iii) All __________ triangles are similar. (isosceles, equilateral) Answer: Equilateral

(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)

Answer: (a) Equal

(b) Proportional

2. Give two different examples of pair of (i) Similar figures (ii) Non-similar figures

3. State whether the following quadrilaterals are similar or not:

From the given two figures, we can see their corresponding angles are different or unequal. Therefore, they are not similar.

Exercise 6.2 Page: 128

1. In figure. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).

(i) Given, in △ ABC, DE∥BC

∴ AD/DB = AE/EC [Using Basic proportionality theorem]

⇒1.5/3 = 1/EC

⇒EC = 3/1.5

EC = 3×10/15 = 2 cm

Hence, EC = 2 cm.

(ii) Given, in △ ABC, DE∥BC

⇒ AD/7.2 = 1.8 / 5.4

⇒ AD = 1.8 ×7.2/5.4 = (18/10)×(72/10)×(10/54) = 24/10

Hence, AD = 2.4 cm.

2. E and F are points on the sides PQ and PR, respectively of a ΔPQR. For each of the following cases, state whether EF || QR. (i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm

(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm (iii) PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm

Given, in ΔPQR, E and F are two points on side PQ and PR, respectively. See the figure below;

(i) Given, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2,4 cm

Therefore, by using Basic proportionality theorem, we get,

PE/EQ = 3.9/3 = 39/30 = 13/10 = 1.3

And PF/FR = 3.6/2.4 = 36/24 = 3/2 = 1.5

So, we get, PE/EQ ≠ PF/FR

Hence, EF is not parallel to QR.

(ii) Given, PE = 4 cm, QE = 4.5 cm, PF = 8cm and RF = 9cm

PE/QE = 4/4.5 = 40/45 = 8/9

And, PF/RF = 8/9

So, we get here,

PE/QE = PF/RF

Hence, EF is parallel to QR.

(iii) Given, PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm

From the figure,

EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm

And, FR = PR – PF = 2.56 – 0.36 = 2.20 cm

So, PE/EQ = 0.18/1.10 = 18/110 = 9/55 …………. (i)

And, PE/FR = 0.36/2.20 = 36/220 = 9/55 ………… (ii)

PE/EQ = PF/FR

3. In the figure, if LM || CB and LN || CD, prove that AM/AB = AN/AD

In the given figure, we can see, LM || CB,

By using basic proportionality theorem, we get,

AM/AB = AL/AC ……………………..(i)

Similarly, given, LN || CD and using basic proportionality theorem,

∴AN/AD = AL/AC ……………………………(ii)

From equation  (i)  and  (ii) , we get,

AM/AB = AN/AD

Hence, proved.

4. In the figure, DE||AC and DF||AE. Prove that BF/FE = BE/EC

In ΔABC, given as, DE || AC

Thus, by using Basic Proportionality Theorem, we get,

∴BD/DA = BE/EC ……………………………………………… (i)

In  ΔBAE, given as, DF || AE

∴BD/DA = BF/FE ……………………………………………… (ii)

From equation  (i)  and  (ii) , we get

BE/EC = BF/FE

5. In the figure, DE||OQ and DF||OR, show that EF||QR.

In ΔPQO, DE || OQ

So by using Basic Proportionality Theorem,

PD/DO = PE/EQ……………… ..(i)

Again given, in ΔPOR, DF || OR,

PD/DO = PF/FR………………… (ii)

Therefore, by converse of Basic Proportionality Theorem,

EF || QR, in ΔPQR.

6. In the figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.

Given here,

In ΔOPQ, AB || PQ

By using Basic Proportionality Theorem,

OA/AP = OB/BQ……………. (i)

Also given,

In ΔOPR, AC || PR

By using Basic Proportionality Theorem

∴ OA/AP = OC/CR……………(ii)

OB/BQ = OC/CR

In ΔOQR, BC || QR.

7. Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).

Given, in ΔABC, D is the midpoint of AB such that AD=DB.

A line parallel to BC intersects AC at E as shown in above figure such that DE || BC.

We have to prove that E is the mid point of AC.

Since, D is the mid-point of AB.

⇒AD/DB = 1 ………………………….  (i)

In ΔABC, DE || BC,

Therefore, AD/DB = AE/EC

From equation (i), we can write,

⇒ 1 = AE/EC

Hence, proved, E is the midpoint of AC.

8. Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).

Given, in ΔABC, D and E are the mid points of AB and AC, respectively, such that,

AD=BD and AE=EC.

We have to prove that: DE || BC.

Since, D is the midpoint of AB

⇒AD/BD = 1………………………………..  (i)

Also given, E is the mid-point of AC.

⇒ AE/EC = 1

AD/BD = AE/EC

By converse of Basic Proportionality Theorem,

9. ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that AO/BO = CO/DO.

Given, ABCD is a trapezium where AB || DC and diagonals AC and BD intersect each other at O.

We have to prove, AO/BO = CO/DO

From the point O, draw a line EO touching AD at E, in such a way that,

EO || DC || AB

In ΔADC, we have OE || DC

Therefore, by using Basic Proportionality Theorem

AE/ED = AO/CO ……………..(i)

Now, In ΔABD, OE || AB

DE/EA = DO/BO …………….(ii)

AO/CO = BO/DO

⇒AO/BO = CO/DO

10. The diagonals of a quadrilateral ABCD intersect each other at the point O such that AO/BO = CO/DO. Show that ABCD is a trapezium.

Given, Quadrilateral ABCD where AC and BD intersect each other at O such that,

AO/BO = CO/DO.

We have to prove here, ABCD is a trapezium

In ΔDAB, EO || AB

DE/EA = DO/OB ……………………(i)

Also, given,

AO/BO = CO/DO

⇒ AO/CO = BO/DO

⇒ CO/AO = DO/BO

⇒DO/OB = CO/AO …………………………..(ii)

DE/EA = CO/AO

Therefore, by using converse of Basic Proportionality Theorem,

EO || DC also EO || AB

⇒ AB || DC.

Hence, quadrilateral ABCD is a trapezium with AB || CD.

Exercise 6.3 Page: 138

1. State which pairs of triangles in the figure are similar. Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:

(i) Given, in ΔABC and ΔPQR,

∠A = ∠P = 60°

∠B = ∠Q = 80°

∠C = ∠R = 40°

Therefore, by AAA similarity criterion,

∴ ΔABC ~ ΔPQR

(ii) Given, in  ΔABC and ΔPQR,

AB/QR = 2/4 = 1/2,

BC/RP = 2.5/5 = 1/2,

CA/PA = 3/6 = 1/2

By SSS similarity criterion,

ΔABC ~ ΔQRP

(iii) Given, in ΔLMP and ΔDEF,

LM = 2.7, MP = 2, LP = 3, EF = 5, DE = 4, DF = 6

MP/DE = 2/4 = 1/2

PL/DF = 3/6 = 1/2

LM/EF = 2.7/5 = 27/50

Here , MP/DE = PL/DF ≠ LM/EF

Therefore, ΔLMP and ΔDEF are not similar.

(iv) In ΔMNL and ΔQPR, it is given,

MN/QP = ML/QR = 1/2

∠M = ∠Q = 70°

Therefore, by SAS similarity criterion

∴ ΔMNL ~ ΔQPR

(v) In ΔABC and ΔDEF, given that,

AB = 2.5, BC = 3, ∠A = 80°, EF = 6, DF = 5, ∠F = 80°

Here , AB/DF = 2.5/5 = 1/2

And, BC/EF = 3/6 = 1/2

Hence, ΔABC and ΔDEF are not similar.

(vi) In ΔDEF, by sum of angles of triangles, we know that,

∠D + ∠E + ∠F = 180°

⇒ 70° + 80° + ∠F = 180°

⇒ ∠F = 180° – 70° – 80°

Similarly, In ΔPQR,

∠P + ∠Q + ∠R = 180 (Sum of angles of Δ)

⇒ ∠P + 80° + 30° = 180°

⇒ ∠P = 180° – 80° -30°

Now, comparing both the triangles, ΔDEF and ΔPQR, we have

∠D = ∠P = 70°

∠F = ∠Q = 80°

∠F = ∠R = 30°

Hence, ΔDEF ~ ΔPQR

2.  In figure 6.35, ΔODC ~ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB.

As we can see from the figure, DOB is a straight line.

Therefore, ∠DOC + ∠ COB = 180°

⇒ ∠DOC = 180° – 125° (Given, ∠ BOC = 125°)

In ΔDOC, sum of the measures of the angles of a triangle is 180º

Therefore, ∠DCO + ∠ CDO + ∠ DOC = 180°

⇒ ∠DCO + 70º + 55º = 180°(Given, ∠ CDO = 70°)

⇒ ∠DCO = 55°

It is given that, ΔODC ~ ΔOBA,

Therefore, ΔODC ~ ΔOBA.

Hence, corresponding angles are equal in similar triangles

∠OAB = ∠OCD

⇒ ∠ OAB = 55°

⇒ ∠OAB = 55°

3. Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that AO/OC = OB/OD

In ΔDOC and ΔBOA,

AB || CD, thus alternate interior angles will be equal,

∴∠CDO = ∠ABO

∠DCO = ∠BAO

Also, for the two triangles ΔDOC and ΔBOA, vertically opposite angles will be equal;

∴∠DOC = ∠BOA

Hence, by AAA similarity criterion,

ΔDOC ~ ΔBOA

Thus, the corresponding sides are proportional.

DO/BO = OC/OA

⇒OA/OC = OB/OD

4. In the fig.6.36, QR/QS = QT/PR and ∠1 = ∠2. Show that ΔPQS ~ ΔTQR.

∠PQR = ∠PRQ

∴ PQ = PR ……………………… (i)

QR/QS = QT/PRUsing equation (i) , we get

QR/QS = QT/QP ……………….(ii)

In ΔPQS and ΔTQR, by equation (ii),

QR/QS = QT/QP

∴ ΔPQS ~ ΔTQR [By SAS similarity criterion]

5. S and T are point on sides PR and QR of ΔPQR such that ∠P = ∠RTS. Show that ΔRPQ ~ ΔRTS.

Given, S and T are point on sides PR and QR of ΔPQR

And ∠P = ∠RTS.

In ΔRPQ and ΔRTS,

∠RTS = ∠QPS (Given)

∠R = ∠R (Common angle)

∴ ΔRPQ ~ ΔRTS (AA similarity criterion)

6. In the figure, if ΔABE ≅ ΔACD, show that ΔADE ~ ΔABC.

Given, ΔABE ≅ ΔACD.

∴ AB = AC [By CPCT] ………………………………. (i)

And, AD = AE [By CPCT] …………………………… (ii)

In ΔADE and ΔABC, dividing eq.(ii) by eq(i),

AD/AB = AE/AC

∠A = ∠A [Common angle]

∴ ΔADE ~ ΔABC [SAS similarity criterion]

7. In the figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

(i) ΔAEP ~ ΔCDP (ii) ΔABD ~ ΔCBE (iii) ΔAEP ~ ΔADB (iv) ΔPDC ~ ΔBEC

Given, altitudes AD and CE of ΔABC intersect each other at the point P.

(i) In ΔAEP and ΔCDP,

∠AEP = ∠CDP (90° each)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by AA similarity criterion,

ΔAEP ~ ΔCDP

(ii) In ΔABD and ΔCBE,

∠ADB = ∠CEB ( 90° each)

∠ABD = ∠CBE (Common Angles)

ΔABD ~ ΔCBE

(iii) In ΔAEP and ΔADB,

∠AEP = ∠ADB (90° each)

∠PAE = ∠DAB (Common Angles)

ΔAEP ~ ΔADB

(iv) In ΔPDC and ΔBEC,

∠PDC = ∠BEC (90° each)

∠PCD = ∠BCE (Common angles)

ΔPDC ~ ΔBEC

8. E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ~ ΔCFB.

Given, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Consider the figure below,

In ΔABE and ΔCFB,

∠A = ∠C (Opposite angles of a parallelogram)

∠AEB = ∠CBF (Alternate interior angles as AE || BC)

∴ ΔABE ~ ΔCFB (AA similarity criterion)

9. In the figure, ABC and AMP are two right triangles, right angled at B and M, respectively, prove that:

(i) ΔABC ~ ΔAMP

(ii) CA/PA = BC/MP

Given, ABC and AMP are two right triangles, right angled at B and M, respectively.

(i) In ΔABC and ΔAMP, we have,

∠CAB = ∠MAP (common angles)

∠ABC = ∠AMP = 90° (each 90°)

∴ ΔABC ~ ΔAMP (AA similarity criterion)

(ii) As, ΔABC ~ ΔAMP (AA similarity criterion)

If two triangles are similar then the corresponding sides are always equal,

Hence, CA/PA = BC/MP

10. CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ~ ΔFEG, Show that:

(i) CD/GH = AC/FG (ii) ΔDCB ~ ΔHGE (iii) ΔDCA ~ ΔHGF

Given, CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG, respectively.

(i) From the given condition,

ΔABC ~ ΔFEG.

∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE

Since, ∠ACB = ∠FGE

∴ ∠ACD = ∠FGH (Angle bisector)

And, ∠DCB = ∠HGE (Angle bisector)

In ΔACD and ΔFGH,

∠ACD = ∠FGH

∴ ΔACD ~ ΔFGH (AA similarity criterion)

⇒CD/GH = AC/FG

(ii) In ΔDCB and ΔHGE,

∠DCB = ∠HGE (Already proved)

∠B = ∠E (Already proved)

∴ ΔDCB ~ ΔHGE (AA similarity criterion)

(iii) In ΔDCA and ΔHGF,

∠ACD = ∠FGH (Already proved)

∠A = ∠F (Already proved)

∴ ΔDCA ~ ΔHGF (AA similarity criterion)

11. In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ~ ΔECF.

Given, ABC is an isosceles triangle.

⇒ ∠ABD = ∠ECF

In ΔABD and ΔECF,

∠ADB = ∠EFC (Each 90°)

∠BAD = ∠CEF (Already proved)

∴ ΔABD ~ ΔECF (using AA similarity criterion)

12. Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see Fig 6.41). Show that ΔABC ~ ΔPQR.

Given, ΔABC and ΔPQR, AB, BC and median AD of ΔABC are proportional to sides PQ, QR and median PM of ΔPQR

i.e. AB/PQ = BC/QR = AD/PM

We have to prove: ΔABC ~ ΔPQR

As we know here,

AB/PQ = BC/QR = AD/PM

⇒AB/PQ = BC/QR = AD/PM (D is the midpoint of BC. M is the midpoint of QR)

⇒ ΔABD ~ ΔPQM [SSS similarity criterion]

∴ ∠ABD = ∠PQM [Corresponding angles of two similar triangles are equal]

⇒ ∠ABC = ∠PQR

In ΔABC and ΔPQR

AB/PQ = BC/QR …………………………. (i)

∠ABC = ∠PQR …………………………… (ii)

ΔABC ~ ΔPQR [SAS similarity criterion]

13. D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that CA 2  = CB.CD

Given, D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC.

In ΔADC and ΔBAC,

∠ADC = ∠BAC (Already given)

∠ACD = ∠BCA (Common angles)

∴ ΔADC ~ ΔBAC (AA similarity criterion)

We know that corresponding sides of similar triangles are in proportion.

∴ CA/CB = CD/CA

⇒ CA 2  = CB.CD.

14. Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that ΔABC ~ ΔPQR.

Given: Two triangles ΔABC and ΔPQR in which AD and PM are medians such that;

AB/PQ = AC/PR = AD/PM

We have to prove, ΔABC ~ ΔPQR

Let us construct first: Produce AD to E so that AD = DE. Join CE, Similarly produce PM to N such that PM = MN, also Join RN.

In ΔABD and ΔCDE, we have

AD = DE  [By Construction.]

BD = DC [Since, AP is the median]

and, ∠ADB = ∠CDE [Vertically opposite angles]

∴ ΔABD ≅ ΔCDE [SAS criterion of congruence]

⇒ AB = CE [By CPCT] ………………………….. (i)

Also, in ΔPQM and ΔMNR,

PM = MN [By Construction.]

QM = MR [Since, PM is the median]

and, ∠PMQ = ∠NMR [Vertically opposite angles]

∴ ΔPQM = ΔMNR [SAS criterion of congruence]

⇒ PQ = RN [CPCT] ……………………………… (ii)

Now, AB/PQ = AC/PR = AD/PM

From equation (i)  and  (ii) ,

⇒CE/RN = AC/PR = AD/PM

⇒ CE/RN = AC/PR = 2AD/2PM

⇒ CE/RN = AC/PR = AE/PN [Since 2AD = AE and 2PM = PN]

∴ ΔACE ~ ΔPRN [SSS similarity criterion]

Therefore, ∠2 = ∠4

Similarly, ∠1 = ∠3

∴ ∠1 + ∠2 = ∠3 + ∠4

⇒ ∠A = ∠P ……………………………………………. (iii)

Now, in ΔABC and ΔPQR, we have

AB/PQ = AC/PR (Already given)

From equation (iii),

∴ ΔABC ~ ΔPQR [ SAS similarity criterion]

15. A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

Given, Length of the vertical pole = 6m

Shadow of the pole = 4 m

Let Height of tower =  h  m

Length of shadow of the tower = 28 m

In ΔABC and ΔDEF,

∠C = ∠E (angular elevation of sum)

∠B = ∠F = 90°

∴ ΔABC ~ ΔDEF (AA similarity criterion)

∴ AB/DF = BC/EF (If two triangles are similar corresponding sides are proportional)

∴ 6/h = 4/28

⇒h = (6×28)/4

⇒  h  = 6 × 7

⇒  h  = 42 m

Hence, the height of the tower is 42 m.

16. If AD and PM are medians of triangles ABC and PQR, respectively where ΔABC ~ ΔPQR prove that AB/PQ = AD/PM.

Given, ΔABC ~ ΔPQR

We know that the corresponding sides of similar triangles are in proportion.

∴AB/PQ = AC/PR = BC/QR ……………………………(i )

Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R ………….….. (ii)

Since AD and PM are medians, they will divide their opposite sides.

∴ BD = BC/2 and QM = QR/2 ……………..…………. (iii)

From equations  (i)  and  (iii) , we get

AB/PQ = BD/QM ……………………….(iv)

In ΔABD and ΔPQM,

From equation (ii), we have

From equation  (iv), we have,

AB/PQ = BD/QM

∴ ΔABD ~ ΔPQM (SAS similarity criterion)

⇒AB/PQ = BD/QM = AD/PM

Exercise 6.4 Page: 143

1. Let ΔABC ~ ΔDEF and their areas be, respectively, 64 cm 2  and 121 cm 2 . If EF = 15.4 cm, find BC.

Solution: Given, ΔABC ~ ΔDEF,

Area of ΔABC = 64 cm 2

Area of ΔDEF = 121 cm 2

EF = 15.4 cm

As we know, if two triangles are similar, ratio of their areas are equal to the square of the ratio of their corresponding sides,

= AC 2 /DF 2  = BC 2 /EF 2

∴ 64/121 = BC 2 /EF 2

⇒ (8/11) 2  = (BC/15.4) 2

⇒ 8/11 = BC/15.4

⇒ BC = 8×15.4/11

⇒ BC = 8 × 1.4

⇒ BC = 11.2 cm

2. Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.

Given, ABCD is a trapezium with AB || DC. Diagonals AC and BD intersect each other at point O.

In ΔAOB and ΔCOD, we have

∠1 = ∠2 (Alternate angles)

∠3 = ∠4 (Alternate angles)

∠5 = ∠6 (Vertically opposite angle)

∴ ΔAOB ~ ΔCOD [AAA similarity criterion]

As we know, If two triangles are similar then the ratio of their areas are equal to the square of the ratio of their corresponding sides. Therefore,

Area of (ΔAOB)/Area of (ΔCOD) = AB 2 /CD 2

= (2CD) 2 /CD 2  [∴ AB = 2CD]

∴ Area of (ΔAOB)/Area of (ΔCOD)

= 4CD 2 /CD 2 = 4/1

Hence, the required ratio of the area of ΔAOB and ΔCOD = 4:1

3. In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.

Given, ABC and DBC are two triangles on the same base BC. AD intersects BC at O.

We have to prove: Area (ΔABC)/Area (ΔDBC) = AO/DO

Let us draw two perpendiculars AP and DM on line BC.

We know that area of a triangle = 1/2 × Base × Height

In ΔAPO and ΔDMO,

∠APO = ∠DMO (Each 90°)

∠AOP = ∠DOM (Vertically opposite angles)

∴ ΔAPO ~ ΔDMO (AA similarity criterion)

∴ AP/DM = AO/DO

⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.

4. If the areas of two similar triangles are equal, prove that they are congruent.

Say ΔABC and ΔPQR are two similar triangles and equal in area

Now let us prove ΔABC ≅ ΔPQR.

Since, ΔABC ~ ΔPQR

∴ Area of (ΔABC)/Area of (ΔPQR) = BC 2 /QR 2

⇒ BC 2 /QR 2  =1 [Since, Area(ΔABC) = (ΔPQR)

⇒ BC 2 /QR 2

Similarly, we can prove that

AB = PQ and AC = PR

Thus, ΔABC ≅ ΔPQR [SSS criterion of congruence]

5. D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.

6. Prove that the ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding medians.

Given: AM and DN are the medians of triangles ABC and DEF respectively and ΔABC ~ ΔDEF.

We have to prove: Area(ΔABC)/Area(ΔDEF) = AM 2 /DN 2

Since, ΔABC ~ ΔDEF (Given)

∴ Area(ΔABC)/Area(ΔDEF) = (AB 2 /DE 2 ) …………………………… (i)

and, AB/DE = BC/EF = CA/FD ……………………………………… (ii)

In ΔABM and ΔDEN,

Since ΔABC ~ ΔDEF

AB/DE = BM/EN [Already Proved in equation  (i) ]

∴ ΔABC ~ ΔDEF [SAS similarity criterion]

⇒ AB/DE = AM/DN ………………………………………………….. (iii)

∴ ΔABM ~ ΔDEN

As the areas of two similar triangles are proportional to the squares of the corresponding sides.

∴ area(ΔABC)/area(ΔDEF) = AB 2 /DE 2  = AM 2 /DN 2

7. Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.

Area(ΔBQC) = ½ Area(ΔAPC)

Since, ΔAPC and ΔBQC are both equilateral triangles, as per given,

∴ ΔAPC ~ ΔBQC [AAA similarity criterion]

∴ area(ΔAPC)/area(ΔBQC) = (AC 2 /BC 2 ) = AC 2 /BC 2

Since, Diagonal = √2 side = √2 BC = AC

⇒ area(ΔAPC) = 2 × area(ΔBQC)

⇒ area(ΔBQC) = 1/2area(ΔAPC)

Tick the correct answer and justify:

8. ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is (A) 2 : 1 (B) 1 : 2 (C) 4 : 1 (D) 1 : 4

Given , ΔABC and ΔBDE are two equilateral triangle. D is the midpoint of BC.

∴ BD = DC = 1/2BC

Let each side of triangle is 2 a .

As, ΔABC ~ ΔBDE

∴ Area(ΔABC)/Area(ΔBDE) = AB 2 /BD 2  = (2 a ) 2 /( a ) 2  = 4 a 2 / a 2  = 4/1 = 4:1

Hence, the correct answer is (C).

9. Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio (A) 2 : 3 (B) 4 : 9 (C) 81 : 16 (D) 16 : 81

Given, Sides of two similar triangles are in the ratio 4 : 9.

Let ABC and DEF are two similar triangles, such that,

ΔABC ~ ΔDEF

And AB/DE = AC/DF = BC/EF = 4/9

As, the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides,

∴ Area(ΔABC)/Area(ΔDEF) = AB 2 /DE 2 

∴ Area(ΔABC)/Area(ΔDEF) = (4/9) 2  = 16/81 = 16:81

Hence, the correct answer is (D).

Exercise 6.5 Page: 150

1.  Sides of triangles are given below. Determine which of them are right triangles. In case of a right triangle, write the length of its hypotenuse.

(i) 7 cm, 24 cm, 25 cm (ii) 3 cm, 8 cm, 6 cm (iii) 50 cm, 80 cm, 100 cm (iv) 13 cm, 12 cm, 5 cm

(i) Given, sides of the triangle are 7 cm, 24 cm, and 25 cm.

Squaring the lengths of the sides of the, we will get 49, 576, and 625.

49 + 576 = 625

(7) 2  + (24) 2  = (25) 2

Therefore, the above equation satisfies, Pythagoras theorem. Hence, it is right angled triangle.

Length of Hypotenuse = 25 cm

(ii) Given, sides of the triangle are 3 cm, 8 cm, and 6 cm.

Squaring the lengths of these sides, we will get 9, 64, and 36.

Clearly, 9 + 36 ≠ 64

Or, 3 2  + 6 2  ≠ 8 2

Therefore, the sum of the squares of the lengths of two sides is not equal to the square of the length of the hypotenuse.

Hence, the given triangle does not satisfies Pythagoras theorem.

(iii) Given, sides of triangle’s are 50 cm, 80 cm, and 100 cm.

Squaring the lengths of these sides, we will get 2500, 6400, and 10000.

However, 2500 + 6400 ≠ 10000

Or, 50 2  + 80 2  ≠ 100 2

As you can see, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.

Therefore, the given triangle does not satisfies Pythagoras theorem.

Hence, it is not a right triangle.

(iv) Given, sides are 13 cm, 12 cm, and 5 cm.

Squaring the lengths of these sides, we will get 169, 144, and 25.

Thus, 144 +25 = 169

Or, 12 2  + 5 2  = 13 2

The sides of the given triangle are satisfying Pythagoras theorem.

Therefore, it is a right triangle.

Hence, length of the hypotenuse of this triangle is 13 cm.

2. PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM 2  = QM × MR.

Given, ΔPQR is right angled at P is a point on QR such that PM ⊥QR

We have to prove, PM 2  = QM × MR

In ΔPQM, by Pythagoras theorem

PQ 2  = PM 2  + QM 2

Or, PM 2  = PQ 2  – QM 2  …………………………….. (i)

In ΔPMR, by Pythagoras theorem

PR 2  = PM 2  + MR 2

Or, PM 2  = PR 2  – MR 2  ……………………………………….. (ii)

Adding equation, (i)  and  (ii) , we get,

2PM 2  = (PQ 2  + PM 2 ) – (QM 2  + MR 2 )

= QR 2  – QM 2  – MR 2         [∴ QR 2  = PQ 2  + PR 2 ]

= (QM + MR) 2  – QM 2  – MR 2

∴ PM 2  = QM × MR

3. In Figure, ABD is a triangle right angled at A and AC ⊥ BD. Show that (i) AB 2  = BC × BD (ii) AC 2  = BC × DC (iii) AD 2  = BD × CD

(i) In ΔADB and ΔCAB,

∠DAB = ∠ACB (Each 90°)

∠ABD = ∠CBA (Common angles)

∴ ΔADB ~ ΔCAB [AA similarity criterion]

⇒ AB/CB = BD/AB

⇒ AB 2  = CB × BD

(ii) Let ∠CAB = x

∠CBA = 180° – 90° – x

∠CBA = 90° – x

Similarly, in ΔCAD

∠CAD = 90° – ∠CBA

= 90° –   x

∠CDA = 180° – 90° – (90° – x)

In ΔCBA and ΔCAD, we have

∠CBA = ∠CAD

∠CAB = ∠CDA

∠ACB = ∠DCA (Each 90°)

∴ ΔCBA ~ ΔCAD [AAA similarity criterion]

⇒ AC/DC = BC/AC

⇒ AC 2  =  DC × BC

(iii) In ΔDCA and ΔDAB,

∠DCA = ∠DAB (Each 90°)

∠CDA = ∠ADB (common angles)

∴ ΔDCA ~ ΔDAB [AA similarity criterion]

⇒ DC/DA = DA/DA

⇒ AD 2  = BD × CD

4. ABC is an isosceles triangle right angled at C. Prove that AB 2  = 2AC 2  .

Given, ΔABC is an isosceles triangle right angled at C.

In ΔACB, ∠C = 90°

AC = BC (By isosceles triangle property)

AB 2  = AC 2  + BC 2  [By Pythagoras theorem]

= AC 2  + AC 2  [Since, AC = BC]

AB 2  = 2AC 2

5. ABC is an isosceles triangle with AC = BC. If AB 2  = 2AC 2 , prove that ABC is a right triangle.

Given, ΔABC is an isosceles triangle having AC = BC and AB 2  = 2AC 2

AB 2  = AC 2  + AC 2

= AC 2  + BC 2  [Since, AC = BC]

Hence, by Pythagoras theorem ΔABC is right angle triangle.

6. ABC is an equilateral triangle of side 2a. Find each of its altitudes .

Given, ABC is an equilateral triangle of side 2a.

Draw, AD ⊥ BC

In ΔADB and ΔADC,

∠ADB = ∠ADC [Both are 90°]

Therefore, ΔADB ≅ ΔADC by RHS congruence.

Hence, BD = DC [by CPCT]

In right angled ΔADB,

AB 2  = AD 2  + BD 2

(2 a ) 2  = AD 2  +  a 2 

⇒ AD 2 = 4 a 2  –  a 2

⇒ AD 2 = 3 a 2

⇒ AD =  √3a

7. Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.

Given, ABCD is a rhombus whose diagonals AC and BD intersect at O.

We have to prove, as per the question,

AB 2  + BC 2  + CD 2  + AD 2  = AC 2  + BD 2

Since, the diagonals of a rhombus bisect each other at right angles.

Therefore, AO = CO and BO = DO

AB 2  = AO 2  + BO 2  ……………………..  (i)  [By Pythagoras theorem]

AD 2  = AO 2  + DO 2  ……………………..  (ii)

DC 2  = DO 2  + CO 2  ……………………..  (iii)

BC 2  = CO 2  + BO 2  ……………………..  (iv)

Adding equations  (i) + (ii) + (iii) + (iv) , we get,

AB 2  + AD 2  +   DC 2  +   BC 2  = 2(AO 2  + BO 2  + DO 2  + CO 2 )

= 4AO 2  + 4BO 2  [Since, AO = CO and BO =DO]

= (2AO) 2  + (2BO) 2  = AC 2  + BD 2

AB 2  + AD 2  +   DC 2  +   BC 2  = AC 2  + BD 2

8. In Fig. 6.54, O is a point in the interior of a triangle.

ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that: (i) OA 2  + OB 2  + OC 2  – OD 2  – OE 2  – OF 2  = AF 2  + BD 2  + CE 2  , (ii) AF 2  + BD 2  + CE 2  = AE 2  + CD 2  + BF 2 .

Given, in ΔABC, O is a point in the interior of a triangle.

And OD ⊥ BC, OE ⊥ AC and OF ⊥ AB.

Join OA, OB and OC

(i) By Pythagoras theorem in ΔAOF, we have

OA 2  = OF 2  + AF 2

Similarly, in ΔBOD

OB 2  = OD 2  + BD 2

Similarly, in ΔCOE

OC 2  = OE 2  + EC 2

Adding these equations,

OA 2  + OB 2  + OC 2  = OF 2  + AF 2  + OD 2  + BD 2  + OE 2  + EC 2

OA 2  + OB 2  + OC 2  – OD 2  – OE 2  – OF 2  = AF 2  + BD 2  + CE 2 .

(ii) AF 2  + BD 2  + EC 2  = (OA 2  – OE 2 ) + (OC 2  – OD 2 ) + (OB 2  – OF 2 )

∴ AF 2  + BD 2  + CE 2  = AE 2  + CD 2  + BF 2 .

9. A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.

Given, a ladder 10 m long reaches a window 8 m above the ground.

Let BA be the wall and AC be the ladder,

Therefore, by Pythagoras theorem,

AC 2  =   AB 2  + BC 2

10 2  = 8 2  + BC 2

BC 2  = 100 – 64

Therefore, the distance of the foot of the ladder from the base of the wall is 6 m.

10. A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?

Given, a guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end.

Let AB be the pole and AC be the wire.

By Pythagoras theorem,

24 2  = 18 2  + BC 2

BC 2  = 576 – 324

BC 2  = 252

BC   = 6√7m

Therefore, the distance from the base is 6√7m.

Speed of first aeroplane = 1000 km/hr

Speed of second aeroplane = 1200 km/hr

In right angle ΔAOB, by Pythagoras Theorem,

AB 2  =   AO 2  + OB 2

⇒ AB 2  =   (1500) 2  + (1800) 2

⇒ AB = √(2250000 + 3240000)

⇒ AB = 300√61 km

Hence, the distance between two aeroplanes will be 300√61 km.

12. Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.

Given, Two poles of heights 6 m and 11 m stand on a plane ground.

And distance between the feet of the poles is 12 m.

Let AB and CD be the poles of height 6m and 11m.

Therefore, CP = 11 – 6 = 5m

From the figure, it can be observed that AP = 12m

By Pythagoras theorem for ΔAPC, we get,

AP 2  =   PC 2  + AC 2

(12m) 2  + (5m) 2  = (AC) 2

AC 2  = (144+25) m 2  = 169 m 2

Therefore, the distance between their tops is 13 m.

13. D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE 2  + BD 2  = AB 2  + DE 2 .

Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC 2  +   CE 2  = AE 2  …………………………………………. (i)

In ΔBCD, by Pythagoras theorem, we get

BC 2  +   CD 2  = BD 2  ……………………………….. (ii)

From equations  (i)  and  (ii) , we get,

AC 2  +   CE 2  + BC 2  +   CD 2  = AE 2  + BD 2  ………….. (iii)

In ΔCDE, by Pythagoras theorem, we get

DE 2  =   CD 2  + CE 2

In ΔABC, by Pythagoras theorem, we get

AB 2  =   AC 2  + CB 2

Putting the above two values in equation  (iii) , we get

DE 2  + AB 2  = AE 2  + BD 2 .

14. The perpendicular from A on side BC of a Δ ABC intersects BC at D such that DB = 3CD (see Figure). Prove that 2AB 2  = 2AC 2  + BC 2 .

Given, the perpendicular from A on side BC of a Δ ABC intersects BC at D such that;

AD ⊥BC and BD = 3CD

In right angle triangle, ADB and ADC, by Pythagoras theorem,

AB 2  =   AD 2  + BD 2  ………………………. (i)

AC 2  =   AD 2  + DC 2  …………………………….. (ii)

Subtracting equation  (ii)  from equation  (i) , we get

AB 2  – AC 2  = BD 2  – DC 2

= 9CD 2  – CD 2  [Since, BD = 3CD]

= 8(BC/4) 2  [Since, BC = DB + CD = 3CD + CD = 4CD]

Therefore, AB 2  – AC 2  = BC 2 /2

⇒ 2(AB 2  – AC 2 ) = BC 2

⇒ 2AB 2  – 2AC 2  = BC 2

∴ 2AB 2  = 2AC 2  + BC 2 .

15.  In an equilateral triangle ABC, D is a point on side BC such that BD = 1/3BC. Prove that 9AD 2  = 7AB 2 .

Given, ABC is an equilateral triangle.

And D is a point on side BC such that BD = 1/3BC

Let the side of the equilateral triangle be  a , and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

And, AE = a√3/2

Given, BD = 1/3BC

DE = BE – BD = a/2 – a/3 = a/6

In ΔADE, by Pythagoras theorem,

AD 2  = AE 2  + DE 2 

⇒ 9 AD 2  = 7 AB 2

16. In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.

Given, an equilateral triangle say ABC,

Let the sides of the equilateral triangle be of length a, and AE be the altitude of ΔABC.

∴ BE = EC = BC/2 = a/2

In ΔABE, by Pythagoras Theorem, we get

AB 2  = AE 2  + BE 2

4AE 2  = 3a 2

⇒ 4 × (Square of altitude) = 3 × (Square of one side)

17. Tick the correct answer and justify: In ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm. The angle B is: (A) 120°

(B) 60° (C) 90° 

Given, in ΔABC, AB = 6√3 cm, AC = 12 cm and BC = 6 cm.

We can observe that,

AB 2  = 108

AC 2  = 144

And, BC 2  = 36

AB 2  + BC 2  = AC 2

The given triangle, ΔABC, is satisfying Pythagoras theorem.

Therefore, the triangle is a right triangle, right-angled at B.

Exercise 6.6 Page: 152

1. In Figure, PS is the bisector of ∠ QPR of ∆ PQR. Prove that QS/PQ = SR/PR

Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.

Given, PS is the angle bisector of ∠QPR. Therefore,

∠QPS = ∠SPR………………………………..(i)

As per the constructed figure,

∠SPR=∠PRT(Since, PS||TR)……………(ii)

∠QPS = ∠QRT(Since, PS||TR) …………..(iii)

From the above equations, we get,

In △QTR, by basic proportionality theorem,

QS/SR = QP/PT

Since, PT=TR

QS/SR = PQ/PR

• Let us join Point D and B.

BD ⊥AC, DM ⊥ BC and DN ⊥ AB

Now from the figure we have,

DN || CB, DM || AB and ∠B = 90 °

Therefore, DMBN is a rectangle.

So, DN = MB and DM = NB

The given condition which we have to prove, is when D is the foot of the perpendicular drawn from B to AC.

∴ ∠CDB = 90° ⇒ ∠2 + ∠3 = 90° ……………………. (i)

In ∆CDM, ∠1 + ∠2 + ∠DMC = 180°

⇒ ∠1 + ∠2 = 90° …………………………………….. (ii)

In ∆DMB, ∠3 + ∠DMB + ∠4 = 180°

⇒ ∠3 + ∠4 = 90° …………………………………….. (iii)

From equation (i) and (ii), we get

From equation (i) and (iii), we get

In ∆DCM and ∆BDM,

∠1 = ∠3 (Already Proved)

∠2 = ∠4 (Already Proved)

∴ ∆DCM ∼ ∆BDM (AA similarity criterion)

BM/DM = DM/MC

DN/DM = DM/MC (BM = DN)

⇒ DM 2 = DN × MC

(ii) In right triangle DBN,

∠5 + ∠7 = 90° ……………….. (iv)

In right triangle DAN,

∠6 + ∠8 = 90° ………………… (v)

D is the point in triangle, which is foot of the perpendicular drawn from B to AC.

∴ ∠ADB = 90° ⇒ ∠5 + ∠6 = 90° ………….. (vi)

From equation (iv) and (vi), we get,

From equation (v) and (vi), we get,

In ∆DNA and ∆BND,

∠6 = ∠7 (Already proved)

∠8 = ∠5 (Already proved)

∴ ∆DNA ∼ ∆BND (AA similarity criterion)

AN/DN = DN/NB

⇒ DN 2 = AN × NB

⇒ DN 2 = AN × DM (Since, NB = DM)

3. In Figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that

AC 2 = AB 2 + BC 2 + 2 BC.BD.

By applying Pythagoras Theorem in ∆ADB, we get,

AB 2 = AD 2 + DB 2 ……………………… (i)

Again, by applying Pythagoras Theorem in ∆ACD, we get,

AC 2 = AD 2 + DC 2

AC 2 = AD 2 + (DB + BC) 2

AC 2 = AD 2 + DB 2 + BC 2 + 2DB × BC

AC 2 = AB 2 + BC 2 + 2DB × BC

4. In Figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that

AC 2 = AB 2 + BC 2 – 2 BC.BD.

AB 2 = AD 2 + DB 2

We can write it as;

⇒ AD 2 = AB 2 − DB 2 ……………….. (i)

By applying Pythagoras Theorem in ∆ADC, we get,

AD 2 + DC 2 = AC 2

From equation (i),

AB 2 − BD 2 + DC 2 = AC 2

AB 2 − BD 2 + (BC − BD) 2 = AC 2

AC 2 = AB 2 − BD 2 + BC 2 + BD 2 −2BC × BD

AC 2 = AB 2 + BC 2 − 2BC × BD

5. In Figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that :

(i) AC 2 = AD 2 + BC.DM + 2 (BC/2) 2

(ii) AB 2 = AD 2 – BC.DM + 2 (BC/2) 2

(iii) AC 2 + AB 2 = 2 AD 2 + ½ BC 2

(i) By applying Pythagoras Theorem in ∆AMD, we get,

AM 2 + MD 2 = AD 2 ………………. (i)

Again, by applying Pythagoras Theorem in ∆AMC, we get,

AM 2 + MC 2 = AC 2

AM 2 + (MD + DC) 2 = AC 2

(AM 2 + MD 2 ) + DC 2 + 2MD.DC = AC 2

From equation(i), we get,

AD 2 + DC 2 + 2MD.DC = AC 2

Since, DC=BC/2, thus, we get,

AD 2 + (BC/2) 2 + 2MD.(BC/2) 2 = AC 2

AD 2 + (BC/2) 2 + 2MD × BC = AC 2

(ii) By applying Pythagoras Theorem in ∆ABM, we get;

AB 2 = AM 2 + MB 2

= (AD 2 − DM 2 ) + MB 2

= (AD 2 − DM 2 ) + (BD − MD) 2

= AD 2 − DM 2 + BD 2 + MD 2 − 2BD × MD

= AD 2 + BD 2 − 2BD × MD

= AD 2 + (BC/2) 2 – 2(BC/2) MD

= AD 2 + (BC/2) 2 – BC MD

(iii) By applying Pythagoras Theorem in ∆ABM, we get,

AM 2 + MB 2 = AB 2 ………………….… (i)

By applying Pythagoras Theorem in ∆AMC, we get,

AM 2 + MC 2 = AC 2 …………………..… (ii)

Adding both the equations (i) and (ii), we get,

2AM 2 + MB 2 + MC 2 = AB 2 + AC 2

2AM 2 + (BD − DM) 2 + (MD + DC) 2 = AB 2 + AC 2

2AM 2 +BD 2 + DM 2 − 2BD.DM + MD 2 + DC 2 + 2MD.DC = AB 2 + AC 2

2AM 2 + 2MD 2 + BD 2 + DC 2 + 2MD (− BD + DC) = AB 2 + AC 2

2(AM 2 + MD 2 ) + (BC/2) 2 + (BC/2) 2 + 2MD (-BC/2 + BC/2) 2 = AB 2 + AC 2

2AD 2 + BC 2 /2 = AB 2 + AC 2

6. Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Let us consider, ABCD be a parallelogram. Now, draw perpendicular DE on extended side of AB, and draw a perpendicular AF meeting DC at point F.

By applying Pythagoras Theorem in ∆DEA, we get,

DE 2 + EA 2 = DA 2 ……………….… (i)

By applying Pythagoras Theorem in ∆DEB, we get,

DE 2 + EB 2 = DB 2

DE 2 + (EA + AB) 2 = DB 2

(DE 2 + EA 2 ) + AB 2 + 2EA × AB = DB 2

DA 2 + AB 2 + 2EA × AB = DB 2 ……………. (ii)

By applying Pythagoras Theorem in ∆ADF, we get,

AD 2 = AF 2 + FD 2

Again, applying Pythagoras theorem in ∆AFC, we get,

AC 2 = AF 2 + FC 2 = AF 2 + (DC − FD) 2

= AF 2 + DC 2 + FD 2 − 2DC × FD

= (AF 2 + FD 2 ) + DC 2 − 2DC × FD AC 2

AC 2 = AD 2 + DC 2 − 2DC × FD ………………… (iii)

Since ABCD is a parallelogram,

AB = CD ………………….…(iv)

And BC = AD ………………. (v)

In ∆DEA and ∆ADF,

∠DEA = ∠AFD (Each 90°)

∠EAD = ∠ADF (EA || DF)

AD = AD (Common Angles)

∴ ∆EAD ≅ ∆FDA (AAS congruence criterion)

⇒ EA = DF ……………… (vi)

Adding equations (i) and (iii), we get,

DA 2 + AB 2 + 2EA × AB + AD 2 + DC 2 − 2DC × FD = DB 2 + AC 2

DA 2 + AB 2 + AD 2 + DC 2 + 2EA × AB − 2DC × FD = DB 2 + AC 2

From equation (iv) and (vi),

BC 2 + AB 2 + AD 2 + DC 2 + 2EA × AB − 2AB × EA = DB 2 + AC 2

AB 2 + BC 2 + CD 2 + DA 2 = AC 2 + BD 2

7. In Figure, two chords AB and CD intersect each other at the point P. Prove that :

(i) ∆APC ~ ∆ DPB

(ii) AP . PB = CP . DP

Firstly, let us join CB, in the given figure.

(i) In ∆APC and ∆DPB,

∠APC = ∠DPB (Vertically opposite angles)

∠CAP = ∠BDP (Angles in the same segment for chord CB)

∆APC ∼ ∆DPB (AA similarity criterion)

(ii) In the above, we have proved that ∆APC ∼ ∆DPB

We know that the corresponding sides of similar triangles are proportional.

∴ AP/DP = PC/PB = CA/BD

⇒AP/DP = PC/PB

∴AP. PB = PC. DP

8. In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that:

(i) ∆ PAC ~ ∆ PDB

(ii) PA . PB = PC . PD.

(i) In ∆PAC and ∆PDB,

∠P = ∠P (Common Angles)

As we know, exterior angle of a cyclic quadrilateral is ∠PCA and ∠PBD is opposite interior angle, which are both equal.

∠PAC = ∠PDB

Thus, ∆PAC ∼ ∆PDB(AA similarity criterion)

(ii) We have already proved above,

∆APC ∼ ∆DPB

AP/DP = PC/PB = CA/BD

AP/DP = PC/PB

∴ AP. PB = PC. DP

9. In Figure, D is a point on side BC of ∆ ABC such that BD/CD = AB/AC. Prove that AD is the bisector of ∠ BAC.

In the given figure, let us extend BA to P such that;

Now join PC.

Given, BD/CD = AB/AC

⇒ BD/CD = AP/AC

By using the converse of basic proportionality theorem, we get,

∠BAD = ∠APC (Corresponding angles) ……………….. (i)

And, ∠DAC = ∠ACP (Alternate interior angles) …….… (ii)

By the new figure, we have;

⇒ ∠APC = ∠ACP ……………………. (iii)

On comparing equations (i), (ii), and (iii), we get,

∠BAD = ∠APC

Therefore, AD is the bisector of the angle BAC.

10. Nazima is fly fishing in a stream. The tip of her fishing rod is 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away and 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Figure)? If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

Let us consider, AB is the height of the tip of the fishing rod from the water surface and BC is the

horizontal distance of the fly from the tip of the fishing rod. Therefore, AC is now the length of the string.

To find AC, we have to use Pythagoras theorem in ∆ABC, is such way;

AC 2 = AB 2 + BC 2

AB 2 = (1.8 m) 2 + (2.4 m) 2

AB 2 = (3.24 + 5.76) m 2

AB 2 = 9.00 m 2

⟹ AB = √9 m = 3m

Thus, the length of the string out is 3 m.

As its given, she pulls the string at the rate of 5 cm per second.

Therefore, string pulled in 12 seconds = 12 × 5 = 60 cm = 0.6 m

Let us say now, the fly is at point D after 12 seconds.

Length of string out after 12 seconds is AD.

AD = AC − String pulled by Nazima in 12 seconds

= (3.00 − 0.6) m

In ∆ADB, by Pythagoras Theorem,

AB 2 + BD 2 = AD 2

(1.8 m) 2 + BD 2 = (2.4 m) 2

BD 2 = (5.76 − 3.24) m 2 = 2.52 m 2

BD = 1.587 m

Horizontal distance of fly = BD + 1.2 m

= (1.587 + 1.2) m = 2.787 m

NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

NCERT Solutions Class 10 Maths Chapter 6 , Triangles, is part of the Unit Geometry, which constitutes 15 marks of the total marks of 80. On the basis of the updated CBSE Class 10 Syllabus for 2023-24, this chapter belongs to the Unit-Geometry and has the second-highest weightage. Hence, having a clear understanding of the concepts, theorems and problem-solving methods in this chapter is mandatory to score well in the board examination of Class 10 Maths.

Main topics covered in this chapter include:

6.1 introduction.

From your earlier classes, you are familiar with triangles and many of their properties. In Class 9, you have studied congruence of triangles in detail. In this chapter, we shall study about those figures which have the same shape, but not necessarily the same size. Two figures having the same shape (and not necessarily the same size) are called similar figures. In particular, we shall discuss the similarity of triangles and apply this knowledge in giving a simple proof of Pythagoras Theorem learnt earlier.

6.2 Similar Figures

In Class 9, you have seen that all circles with the same radii are congruent, all squares with the same side lengths are congruent and all equilateral triangles with the same side lengths are congruent. The topic explains similarity of figures by performing the relevant activity. Similar figures are two figures having the same shape, but not necessarily the same size.

6.3 Similarity of Triangles

The topic recalls triangles and its similarities. Two triangles are similar if (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). It explains Basic Proportionality Theorem and different theorems are discussed performing various activities.

6.4 Criteria for Similarity of Triangles

In the previous section, we stated that two triangles are similar (i) their corresponding angles are equal and (ii) their corresponding sides are in the same ratio (or proportion). The topic discusses the criteria for similarity of triangles referring to the topics we have studied in earlier classes. It also contains different theorems explained with proper examples.

6.5 Areas of Similar Triangles

You have learnt that in two similar triangles, the ratio of their corresponding sides is the same. The topic Areas of Similar Triangles consists of theorem and relatable examples to prove the theorem.

6.6 Pythagoras Theorem

You are already familiar with the Pythagoras Theorem from your earlier classes. You have also seen proof of this theorem in Class 9. Now, we shall prove this theorem using the concept of similarity of triangles. Hence, the theorem is verified through some activities, and you can make use of it while solving certain problems.

6.7 Summary

The summary contains the points you have studied in the chapter. Going through the points mentioned in the summary will help you to recollect all the important concepts and theorems of the chapter.

List of Exercises in NCERT Class 10 Maths Chapter 6:

Exercise 6.1 Solutions 3 Questions (3 Short Answer Questions)

Exercise 6.2 Solutions 10 Questions (9 Short Answer Questions, 1 Long Answer Question)

Exercise 6.3 Solutions 16 Questions (1 main question with 6 sub-questions, 12 Short Answer Questions, 3 Long Answer Questions)

Exercise 6.4 Solutions 9 Questions (2 Short Answer with Reasoning Questions, 5 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.5 Solutions 17 Questions (15 Short Answer Questions, 2 Long Answer Questions)

Exercise 6.6 Solutions 10 Questions (5 Short Answer Questions, 5 Long Answer Questions)

Triangle is one of the most interesting and exciting chapters of the unit Geometry as it takes us through the different aspects and concepts related to the geometrical figure triangle. A triangle is a plane figure that has three sides and three angles. This chapter covers various topics and sub-topics related to triangles, including a detailed explanation of similar figures, different theorems related to the similarities of triangles with proof, and the areas of similar triangles. The chapter concludes by explaining the Pythagoras theorem and the ways to use it in solving problems. Read and learn Chapter 6 of the Class 10 Maths NCERT textbook to learn more about Triangles and the concepts covered in it. Ensure to learn the NCERT Solutions for Class 10 effectively to score high in the board examinations.

Key Features of NCERT Solutions for Class 10 Maths Chapter 6 – Triangles

• Helps to ensure that the students use the concepts in solving the problems.
• Encourages the children to come up with diverse solutions to problems.
• Hints are given for those questions which are difficult to solve.
• Helps the students in checking if the solutions they gave for the questions are correct or not.

Disclaimer – 

Dropped Topics – 

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Case Study Class 10 Maths Questions

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Download the app to get CBSE Sample Papers 2023-24, NCERT Solutions (Revised), Most Important Questions, Previous Year Question Bank, Mock Tests, and Detailed Notes.

Now, CBSE will ask only subjective questions in class 10 Maths case studies. But if you search over the internet or even check many books, you will get only MCQs in the class 10 Maths case study in the session 2022-23. It is not the correct pattern. Just beware of such misleading websites and books.

We advise you to visit CBSE official website ( cbseacademic.nic.in ) and go through class 10 model question papers . You will find that CBSE is asking only subjective questions under case study in class 10 Maths. We at myCBSEguide helping CBSE students for the past 15 years and are committed to providing the most authentic study material to our students.

Here, myCBSEguide is the only application that has the most relevant and updated study material for CBSE students as per the official curriculum document 2022 – 2023. You can download updated sample papers for class 10 maths .

First of all, we would like to clarify that class 10 maths case study questions are subjective and CBSE will not ask multiple-choice questions in case studies. So, you must download the myCBSEguide app to get updated model question papers having new pattern subjective case study questions for class 10 the mathematics year 2022-23.

Class 10 Maths has the following chapters.

• Real Numbers Case Study Question
• Polynomials Case Study Question
• Pair of Linear Equations in Two Variables Case Study Question
• Quadratic Equations Case Study Question
• Arithmetic Progressions Case Study Question
• Triangles Case Study Question
• Coordinate Geometry Case Study Question
• Introduction to Trigonometry Case Study Question
• Some Applications of Trigonometry Case Study Question
• Circles Case Study Question
• Area Related to Circles Case Study Question
• Surface Areas and Volumes Case Study Question
• Statistics Case Study Question
• Probability Case Study Question

Format of Maths Case-Based Questions

CBSE Class 10 Maths Case Study Questions will have one passage and four questions. As you know, CBSE has introduced Case Study Questions in class 10 and class 12 this year, the annual examination will have case-based questions in almost all major subjects. This article will help you to find sample questions based on case studies and model question papers for CBSE class 10 Board Exams.

Maths Case Study Question Paper 2023

Here is the marks distribution of the CBSE class 10 maths board exam question paper. CBSE may ask case study questions from any of the following chapters. However, Mensuration, statistics, probability and Algebra are some important chapters in this regard.

Case Study Question in Mathematics

Here are some examples of case study-based questions for class 10 Mathematics. To get more questions and model question papers for the 2021 examination, download myCBSEguide Mobile App .

Case Study Question – 1

In the month of April to June 2022, the exports of passenger cars from India increased by 26% in the corresponding quarter of 2021–22, as per a report. A car manufacturing company planned to produce 1800 cars in 4th year and 2600 cars in 8th year. Assuming that the production increases uniformly by a fixed number every year.

• Find the production in the 1 st year.
• Find the production in the 12 th year.
• Find the total production in first 10 years. OR In which year the total production will reach to 15000 cars?

Case Study Question – 2

In a GPS, The lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.

• Find the distance between Lucknow (L) to Bhuj(B).
• If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3 : 2 then find the coordinate of Kota (K).
• Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P) OR Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri (P).

Case Study Question – 3

• Find the distance PA.
• Find the distance PB
• Find the width AB of the river. OR Find the height BQ if the angle of the elevation from P to Q be 30 o .

Case Study Question – 4

• What is the length of the line segment joining points B and F?
• The centre ‘Z’ of the figure will be the point of intersection of the diagonals of quadrilateral WXOP. Then what are the coordinates of Z?
• What are the coordinates of the point on y axis equidistant from A and G? OR What is the area of area of Trapezium AFGH?

Case Study Question – 5

The school auditorium was to be constructed to accommodate at least 1500 people. The chairs are to be placed in concentric circular arrangement in such a way that each succeeding circular row has 10 seats more than the previous one.

• If the first circular row has 30 seats, how many seats will be there in the 10th row?
• For 1500 seats in the auditorium, how many rows need to be there? OR If 1500 seats are to be arranged in the auditorium, how many seats are still left to be put after 10 th row?
• If there were 17 rows in the auditorium, how many seats will be there in the middle row?

Case Study Question – 6

• Draw a neat labelled figure to show the above situation diagrammatically.

• What is the speed of the plane in km/hr.

More Case Study Questions

We have class 10 maths case study questions in every chapter. You can download them as PDFs from the myCBSEguide App or from our free student dashboard .

As you know CBSE has reduced the syllabus this year, you should be careful while downloading these case study questions from the internet. You may get outdated or irrelevant questions there. It will not only be a waste of time but also lead to confusion.

Here, myCBSEguide is the most authentic learning app for CBSE students that is providing you up to date study material. You can download the myCBSEguide app and get access to 100+ case study questions for class 10 Maths.

How to Solve Case-Based Questions?

Questions based on a given case study are normally taken from real-life situations. These are certainly related to the concepts provided in the textbook but the plot of the question is always based on a day-to-day life problem. There will be all subjective-type questions in the case study. You should answer the case-based questions to the point.

What are Class 10 competency-based questions?

Competency-based questions are questions that are based on real-life situations. Case study questions are a type of competency-based questions. There may be multiple ways to assess the competencies. The case study is assumed to be one of the best methods to evaluate competencies. In class 10 maths, you will find 1-2 case study questions. We advise you to read the passage carefully before answering the questions.

Case Study Questions in Maths Question Paper

CBSE has released new model question papers for annual examinations. myCBSEguide App has also created many model papers based on the new format (reduced syllabus) for the current session and uploaded them to myCBSEguide App. We advise all the students to download the myCBSEguide app and practice case study questions for class 10 maths as much as possible.

Case Studies on CBSE’s Official Website

CBSE has uploaded many case study questions on class 10 maths. You can download them from CBSE Official Website for free. Here you will find around 40-50 case study questions in PDF format for CBSE 10th class.

10 Maths Case Studies in myCBSEguide App

You can also download chapter-wise case study questions for class 10 maths from the myCBSEguide app. These class 10 case-based questions are prepared by our team of expert teachers. We have kept the new reduced syllabus in mind while creating these case-based questions. So, you will get the updated questions only.

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Case Study Class 10 Maths

If you are looking for the CBSE Case Study class 10 Maths in PDF, then you are in the right place. CBSE 10th Class Case Study for the Maths Subject is available here on this website. These Case studies can help the students to solve the different types of questions that are based on the case study or passage.

CBSE Board will be asking case study questions based on Maths subjects in the upcoming board exams. Thus, it becomes an essential resource to study. 

The Case Study Class 10 Maths Questions cover a wide range of chapters from the subject. Students willing to score good marks in their board exams can use it to practice questions during the exam preparation. The questions are highly interactive and it allows students to use their thoughts and skills to solve the given Case study questions.

Download Class 10 Maths Case Study Questions and Answers PDF (Passage Based)

Download links of class 10 Maths Case Study questions and answers pdf is given on this website. Students can download them for free of cost because it is going to help them to practice a variety of questions from the exam perspective.

Case Study questions class 10 Maths include all chapters wise questions. A few passages are given in the case study PDF of Maths. Students can download them to read and solve the relevant questions that are given in the passage.

Students are advised to access Case Study questions class 10 Maths CBSE chapter wise PDF and learn how to easily solve questions. For gaining the basic knowledge students can refer to the NCERT Class 10th Textbooks. After gaining the basic information students can easily solve the Case Study class 10 Maths questions.

Case Study Questions Class 10 Maths Chapter 1 Real Numbers

Case Study Questions Class 10 Maths Chapter 2 Polynomials

Case Study Questions Class 10 Maths Chapter 3 Pair of Equations in Two Variables

Case Study Questions Class 10 Maths Chapter 4 Quadratic Equations

Case Study Questions Class 10 Maths Chapter 5 Arithmetic Progressions

Case Study Questions Class 10 Maths Chapter 6 Triangles

Case Study Questions Class 10 Maths Chapter 7 Coordinate Geometry

Case Study Questions Class 10 Maths Chapter 8. Introduction to Trigonometry

Case Study Questions Class 10 Maths Chapter 9 Some Applications of Trigonometry

Case Study Questions Class 10 Maths Chapter 10 Circles

Case Study Questions Class 10 Maths Chapter 12 Areas Related to Circles

Case Study Questions Class 10 Maths Chapter 13 Surface Areas & Volumes

Case Study Questions Class 10 Maths Chapter 14 Statistics

Case Study Questions Class 10 Maths Chapter 15 Probability

How to Solve Case Study Based Questions Class 10 Maths?

In order to solve the Case Study Based Questions Class 10 Maths students are needed to observe or analyse the given information or data. Students willing to solve Case Study Based Questions are required to read the passage carefully and then solve them. 

While solving the class 10 Maths Case Study questions, the ideal way is to highlight the key information or given data. Because, later it will ease them to write the final answers. 

Case Study class 10 Maths consists of 4 to 5 questions that should be answered in MCQ manner. While answering the MCQs of Case Study, students are required to read the paragraph as they can get some clue in between related to the topics discussed.

Also, before solving the Case study type questions it is ideal to use the CBSE Syllabus to brush up the previous learnings.

Features Of Class 10 Maths Case Study Questions And Answers Pdf

Students referring to the Class 10 Maths Case Study Questions And Answers Pdf from Selfstudys will find these features:-

• Accurate answers of all the Case-based questions given in the PDF.
• Case Study class 10 Maths solutions are prepared by subject experts referring to the CBSE Syllabus of class 10.
• Free to download in Portable Document Format (PDF) so that students can study without having access to the internet.

Benefits of Using CBSE Class 10 Maths Case Study Questions and Answers

Since, CBSE Class 10 Maths Case Study Questions and Answers are prepared by our maths experts referring to the CBSE Class 10 Syllabus, it provided benefits in various way:-

• Case study class 10 maths helps in exam preparation since, CBSE Class 10 Question Papers contain case-based questions.
• It allows students to utilise their learning to solve real life problems.
• Solving case study questions class 10 maths helps students in developing their observation skills.
• Those students who solve Case Study Class 10 Maths on a regular basis become extremely good at answering normal formula based maths questions.
• By using class 10 Maths Case Study questions and answers pdf, students focus more on Selfstudys instead of wasting their valuable time.
• With the help of given solutions students learn to solve all Case Study questions class 10 Maths CBSE chapter wise pdf regardless of its difficulty level.

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Important Questions for Class 10 Maths Chapter 6 Triangles

Chapter 6 triangles important questions for cbse class 10 maths board exams.

Important Questions for Chapter 6 Triangles Class 10 Maths

Triangles class 10 maths important questions very short answer (1 mark).

∆ABC ~ ∆PQR (Given)

Triangles Class 10 Maths Important Questions Short Answer-I (2 Marks)

More Study Materials for Class 10

• Important Questions Class 10
• NCERT Solutions Class 10 Maths
• Important Questions Class 10 Maths
• Study Materials for Class 10 Maths
• Maths Formula Class 10
• Solved MCQ Questions Class 10

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NCERT Solutions for Class 10 Maths Chapter 6 Triangles

NCERT Solutions Class 10 Maths Chapter 6 Triangles are provided here to help the students of CBSE class 10. Our expert teachers prepared all these solutions as per the latest CBSE syllabus and guidelines. In this chapter, we have discussed the similarity of triangles, criterion for similarity of triangles, areas of similar triangles and Pythagoras theorem. CBSE Class 10 Maths solutions provide a detailed and step-wise explanation of each answer to the questions given in the exercises of NCERT books.

CBSE Class 10 Maths Chapter 6 Triangles Solutions

Below we have given the answers to all the questions present in Triangles in our NCERT Solutions for Class 10 Maths chapter 6. In this lesson, students are introduced to a lot of important concepts that will be useful for those who wish to pursue mathematics as a subject in their future classes. Based on these solutions, students can prepare for their upcoming Board Exams. These solutions are helpful as the syllabus covered here follows NCERT guidelines.

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.1

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.2

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.3

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.4

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.5

NCERT Solutions for Class 10 Maths Chapter 6 Exercise 6.6

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