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Case Study Questions for Class 8 Maths Chapter 11 Mensuration

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Here we are providing Case Study questions for Class 8 Maths Chapter 11 Mensuration.

Mensuration Case Study Questions

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Unit 9: Mensuration

Mensuration 9.1.

Area of trapezoids (Opens a modal)
Area of composite shapes (Opens a modal)
Mensuration 9.1 Get 7 of 10 questions to level up!

Mensuration 9.2

Surface area of a box (cuboid) (Opens a modal)
Cylinder volume & surface area (Opens a modal)
Mensuration 9.2 Get 7 of 10 questions to level up!

Mensuration 9.3

Volume of a rectangular prism (Opens a modal)
Mensuration 9.3 Get 7 of 10 questions to level up!

Important Questions Class 8 Maths Chapter 11 Mensuration

Some of the most important mensuration class 8 questions, i.e. chapter 11 are given here. These chapter 11 class 8 maths questions cover several short answer type questions, long answer type questions and HOTS questions that are crucial for CBSE class 8 exams. Here, some of the important mensuration questions from NCERT class 8 are also included.

Also Check:

Important 2 Marks Questions for CBSE 8th Maths
Important 3 Marks Questions for CBSE 8th Maths
Important 4 Marks Questions for CBSE 8th Maths

Mensuration Important Questions For Class 8 (Chapter 11)

These class 8 mensuration questions are categorized into short answer type questions and long answer type questions. These questions cover various concepts which will help class 8 students to develop problem-solving skills for the exam.

Short Answer Type Questions:

1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.

Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides

= ½ × 15 × (12 + 20)

= 1/2 × 15 × 32

= 240 cm 2

2. Calculate the height of a cuboid which has a base area of 180 cm 2 and volume is 900 cm 3 .

Volume of cuboid = base area × height

900 = 180 × height

So, height = 900/180 = 5 cm

3. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.

Perimeter of square formula = 4 × side of the square

Hence, P (square) = 4 × 60 = 240 cm

Perimeter of rectangle formula = 2 × (Length + Breadth)

Hence, P (rectangle) = 2 (80 + Breadth)

= 160 + 2 × Breadth

According to the given question, 160 + 2 × Breadth = 240 cm 2 × Breadth = 240 – 160 Breadth = 80/2 The breadth of the rectangle = 40 cm

Now, the area of rectangle = Length × Breadth = 80 × 40 = 3200 cm 2

4. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and length is 1 m.

Given, length of lawnmower = 1m = 100cm

Its circumference = π × D = 22/7 × 84 = 264 cm

Length of field will be = 264 × 750 = 198000 cm

Here, the width of field = length of the lawnmower i.e. 100 cm

So, area of field = 198000 × 100 = 19,800,000 cm²

Or, 1980 m²

5. The area of a rhombus is 16 cm 2 and the length of one of its diagonal is 4 cm. Calculate the length of other the diagonal.

Area of rhombus = ½ × d 1 × d 2

⇒ 16 = ½ × 4 × d 2

So, d 2 = 32/4 = 8 cm

Long Answer Type Questions:

6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.

The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm

Area of circle = πr²

Area of the entire circle = π(4)² = 16π cm 2

Area of the circle with radius 3 cm which is cut out = π(3)² = 9π cm 2

Thus, the remaining area = 16π – 9π = 7π cm 2

7. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.

Length of the box, l = 2 m,

Breadth of box, b = 1 m

Height of box, h = 1.5 m

We know that the surface area of a cuboid = 2(lb + lh + bh)

But here the bottom part is not to be painted.

Surface area of box to be painted = lb + 2(bh + hl)

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0)

Hence, the required surface area of the cuboidal box = 11 m 2

8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.

From the question statement draw the diagram.

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Important questions class 8 maths chapter 11 mensuration Q8 sol

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Important questions class 8 maths chapter 11 mensuration Q8 sol.ii

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL 2 = CM 2 + LM 2 (Pythagoras Theorem)

26 2 = CM 2 + 10 2

CM 2 = 26 2 – 10 2

Using algebraic identities, we get; 26 2 – 10 2 = (26 – 10) (26 + 10)

CM 2 = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium, ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm 2

Class 8 Maths Chapter 11 Extra Question

A flooring tile is in the shape of a parallelogram with 24 cm base and the corresponding 10 cm height. Calculate the number of tiles required to cover a floor of area 1080 m 2 (If required you can split the tiles in whatever way you want to fill up the corners).
Two cubes are joined end to end. Now, calculate the volume of the resulting cuboid, if each side of the cubes is 6 cm.
How many bricks each 25 cm by 15 cm by 8 cm, are required for a wall 32 m long, 3 m high and 40 cm thick?
Find the area of a rhombus whose one side measures 5 cm and one diagonal as 8 cm.

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Chapter 9 Class 8 Mensuration

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Get solutions of all NCERT Questions and Examples of Chapter 9 Class 8 Mensuration free at teachoo. Answers to all questions have been solved in an easy way with detailed explanation of each and every solution.

In this chapter, we will

First revise our formulas for Perimeter and Area of Rectangle, Square, Triangle, Parallelogram, Circle
and do some questions using those formulas
Then, we will derive the formula for Area of Trapezium
and formula for Area of Quadrilateral
We also learn the formula for Area of Rhombus
and find Area of any general polygon by dividing it into different shapes and adding the area
Then, we see some common 3 Dimensional shapes
And see what area of 3 Dimensional Shapes is
and Curved and Total Surface Area of Cylinder
We also learn what Volume of 3 Dimensional shape is
And study formula of Volume of Cube, Cuboid and Cylinder
We also learn how to convert cm 3 into mL, L and m 3

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NCERT Solutions for Class 8 Maths Chapter 11 - Mensuration
NCERT Solutions

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration - Free PDF Download

Class 8 students have a lot on their plate in terms of academics. A subject like Maths requires them to spend enough time in solving assignments and preparing for exams. Vedantu’s team of scholars understand the needs and understanding level of Class 8 students and have designed NCERT Solution for Class 8 Maths Chapter 11 keeping all of this in mind. NCERT Solution Class 8 Maths Chapter 11 Mensuration Solution is entirely based on the latest CBSE curriculum. It will be able to clarify all the doubts that you might have in the Mensuration chapter. The problems in the Class 8 Maths Chapter 11 are well explained with diagrams and step-by-step explanations to make it simple for you to follow the solution.

You can also download Class 8 Science to help you to revise complete syllabus ans score more marks in your examinations.

Access NCERT Solutions for Class 8 Maths Chapter 11 – Mensuration

Exercise 11.1

Q1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?

Ans: Let Side of a Square be ‘ \[{\text{a}}\] ’.

Length and Breadth of Rectangle are ‘ \[{\text{l}}\] ’ and ‘ \[{\text{b}}\] ’ respectively.

Perimeter of Square = $4$ (side of square) = $4$ ( \[{\text{a}}\] )

= $4$ ( $60$${\text{m}}$ ) = $240{\text{m}}$

Perimeter of Rectangle = $2$ (Length+Breadth) = $2$ ( \[{\text{l + b}}\] )

= $2$ ( $80{\text{m}}$ + \[{\text{b}}\] )

Now, It is given that the perimeter of the square and the perimeter of the rectangle are equal.

So, $240{\text{m}}$ = $2$ ( $80{\text{m}}$ + \[{\text{b}}\] )

$240{\text{m}}$ = $160{\text{m}}$ + $2$\[{\text{b}}\]

$80{\text{m}}$ = 2 \[{\text{b}}\]

$40{\text{m}}$ = \[{\text{b}}\]

Area of square = $\mathop {{\text{(side)}}}\nolimits^{\text{2}} $ = $\mathop {{\text{(60m)}}}\nolimits^{\text{2}} $ = $\mathop {{\text{3600m}}}\nolimits^{\text{2}} $

Area of Rectangle = Length x Breadth = $(80 x 40)$$\mathop {\text{m}}\nolimits^{\text{2}} $ = $3200$${{\text{m}}^{\text{2}}}$

Thus, the area of the square field is larger than the area of the rectangular field.

Q2. Mrs. Kaushik has a square plot with the measurement as shown in the following figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of ${\text{Rs}}{\text{.}}\;{\text{55}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$?

Ans: Area of the Square Plot = ${{\text{(side)}}^{\text{2}}}$ = \[{{\text{(25m)}}^{\text{2}}}\] = ${\text{625}}{{\text{m}}^{\text{2}}}$

Area of the House = Length x Breadth = ${\text{15m x 20m}}$ = ${\text{300}}{{\text{m}}^{\text{2}}}$

Now, Area of the remaining portion = Area of the square plot – Area of the house

= ${\text{625}}{{\text{m}}^{\text{2}}}$ - ${\text{300}}{{\text{m}}^{\text{2}}}$

= ${\text{325}}{{\text{m}}^{\text{2}}}$

The cost of developing the garden around the house is ${\text{Rs}}{\text{.}}\;{\text{55}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$ .

Therefore, total cost of developing the area ${\text{325}}{{\text{m}}^{\text{2}}}$ is ${\text{Rs}}{\text{.}}\;{\text{(55 x 325)}}$ = ${\text{Rs}}{\text{.}}\;{\text{17875}}$ .

Q3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of the garden (Length of rectangle is ${\text{20 - (3}}{\text{.5 + 3}}{\text{.5)m}}$)?

Combination of a rectangle and two semicircles

Ans: As we have given that length of rectangle = ${\text{[20 - (3}}{\text{.5 + 3}}{\text{.5)]m}}$

= ${\text{[20 - 7]m}}$ = ${\text{13m}}$

Breadth = \[{\text{7m}}\] .

Now, we have to find the circumference of both semi circles.

As, Diameter = ${\text{7m}}$ , so, Radius(r) = $\dfrac{7}{{\text{2}}}{\text{m}}$ = ${\text{3}}{\text{.5m}}$

Circumference of one semicircle = ${\text{$\pi$ r}}$ = $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(3}}{\text{.5)m}}$ = ${\text{11m}}$

Circumference of both circles = ${\text{2 x 11m}}$ = ${\text{22m}}$

Combination of a rectangle ABCD and two semicircles

Now, Perimeter of the garden = AB + CD + Length of both semi-circular regions AD & BC

= ${\text{(13 + 13 + 22)m}}$

= ${\text{48m}}$

Area of the garden = Area of the rectangle + $2 x $ Area of two semi-circular regions

= (Length x Breadth) + ${\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{$\pi$ }}{{\text{r}}^{\text{2}}}$

= ${\text{[(13 x 7) + 2 x (}}\dfrac{{\text{1}}}{{\text{2}}}{\text{ x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5 x 3}}{\text{.5)]}}{{\text{m}}^{\text{2}}}$

= ${\text{(91 + 38}}{\text{.5)}}{{\text{m}}^{\text{2}}}$

= \[{\text{129}}{\text{.5}}{{\text{m}}^{\text{2}}}\]

Q4. A flooring tile has the shape of a parallelogram whose base is ${\text{24cm}}$and the corresponding height is ${\text{10cm}}$.How many such tiles are required to cover a floor of area ${\text{1080}}{{\text{m}}^{\text{2}}}$? (If required you can split the tiles in whatever way you want to fill up the corners).

Ans: Given that the Base of a parallelogram is ${\text{24cm}}$ and Height of a parallelogram is ${\text{10cm}}$ .

Therefore, Area of Parallelogram = Base x Height

= ${\text{(24 x 10)c}}{{\text{m}}^2}$

= ${\text{240c}}{{\text{m}}^2}$ .

Therefore, Area of parallelogram = Area of one tile.

Now, we have to find the number of tiles.

Given that Area of floor = ${\text{1080}}{{\text{m}}^{\text{2}}}$

So, Number of tiles = $\dfrac{{{\text{Area}}\;{\text{of}}\;{\text{floor}}}}{{{\text{Area}}\;{\text{of}}\;{\text{one}}\;{\text{tile}}}}$

= $\dfrac{{{\text{1080}}{{\text{m}}^{\text{2}}}}}{{{\text{240c}}{{\text{m}}^{\text{2}}}}}$ = $\dfrac{{{\text{1080 x 10000}}}}{{{\text{240}}}}{\text{c}}{{\text{m}}^{\text{2}}}$ \[(\because \;1{\text{m}} = 100{\text{cm}})\]

= $45000$ tiles.

Hence, $45000$ tiles are required to cover a floor of area ${\text{1080}}{{\text{m}}^{\text{2}}}$

Q5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food − piece would the ant have to take a longer round? Remember,

circumference of a circle can be obtained by using the expression c =${\text{2$\pi$ r}}$, where r is

the radius of the circle.

Ans: (a) Diameter( \[{\text{d}}\] ) = ${\text{2}}{\text{.8cm}}$ , so radius( \[{\text{r}}\] ) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$ = ${\text{1}}{\text{.4cm}}$

Therefore, Perimeter = ${\text{d + $\pi$ r}}$

= ${\text{2}}{\text{.8cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$

= ${\text{[2}}{\text{.8 + (0}}{\text{.2 x 22)]cm}}$

= ${\text{(2}}{\text{.8 + 4}}{\text{.4)cm}}$

= ${\text{7}}{\text{.2cm}}$

(b) Diameter( \[{\text{d}}\] ) = ${\text{2}}{\text{.8cm}}$ , so radius( \[{\text{r}}\] ) = $\dfrac{{{\text{2}}{\text{.8}}}}{{\text{2}}}{\text{cm}}$ = ${\text{1}}{\text{.4cm}}$

Therefore, Perimeter = ${\text{1}}{\text{.5cm + 2}}{\text{.8cm + 1}}{\text{.5cm + $\pi$ (1}}{\text{.4cm)}}$

= ${\text{5}}{\text{.8cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$

= ${\text{5}}{\text{.8cm + 4}}{\text{.4cm}}$

= ${\text{10}}{\text{.2cm}}$ .

Perimeter = ${\text{2cm + $\pi$ r + 2cm}}$

= ${\text{4cm + }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{(1}}{\text{.4cm)}}$

= ${\text{4cm + 4}}{\text{.4cm}}$

= ${\text{8}}{\text{.4cm}}$ .

Thus, the ant will have to take a longer round for the food piece (b) because its perimeter is ${\text{10}}{\text{.2cm}}$ which is the greatest among all.

Exercise 11.2

Q1. The shape of the top surface of a table is a trapezium. Find its area if its parallel sides are ${\text{1m}}$and ${\text{1}}{\text{.2m}}$and perpendicular distance between them is${\text{0}}{\text{.8m}}$.

Ans: Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ ( Sum of parallel sides) x (Distances between parallel sides) }}$

= $\dfrac{{\text{1}}}{{\text{2}}}{\text{[(1m + 1}}{\text{.2m) x 0}}{\text{.8m]}}$

= $\dfrac{{\text{1}}}{{\text{2}}}{\text{(1}}{\text{.76}}{{\text{m}}^{\text{2}}}{\text{)}}$ = ${\text{0}}{\text{.88}}{{\text{m}}^{\text{2}}}$

Q2. The area of a trapezium is ${\text{34c}}{{\text{m}}^{\text{2}}}$and the length of one of the parallel sides is ${\text{10cm}}$and its height is ${\text{4cm}}$. Find the length of the other parallel side.

Ans: It is given that Area of Trapezium = ${\text{34c}}{{\text{m}}^{\text{2}}}$ .

Length of one parallel side = ${\text{10cm}}$

Height = ${\text{4cm}}$

Now, let length of other parallel side = ‘ \[{\text{a}}\] ’ cm

Therefore, Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ ( Sum of parallel sides) x (Distances between parallel sides) }}$

${\text{34c}}{{\text{m}}^{\text{2}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}{\text{[(10 + a) x 4]cm}}$

${\text{34c}}{{\text{m}}^{\text{2}}}$ = ${\text{(20 + 2a)cm}}$

$\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm}}$ = \[{\text{a}}\] cm

Thus, Length of the other parallel side ( \[{\text{a}}\] ) = ${\text{7cm}}$ .

Q3. Length of the fence of a trapezium shaped field ABCD is ${\text{120m}}$. If BC = ${\text{48m}}$, CD =${\text{17m}}$ and AD =${\text{40m}}$, find the area of this field. Side AB is perpendicular to the parallel sides AD and BC.

Ans: Length of the fence of Trapezium ABCD = AB+BC+CD+DA

${\text{120m}}$ =${\text{(AB + 48 + 17 + 40)m}}$

${\text{120m}}$ = ${\text{AB + 105m}}$

Therefore, AB = ${\text{15m}}$

Area of the field ABCD = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(AD + BC) x AB}}$

=$\dfrac{{\text{1}}}{{\text{2}}}{\text{(40 + 48) x 15}}$

=$\dfrac{{\text{1}}}{{\text{2}}}{\text{(88) x 15}}$

= ${\text{660}}{{\text{m}}^{\text{2}}}$

Q4. The diagonal of a quadrilateral shaped field is 24m and the perpendiculars dropped on it from the remaining opposite vertices are 8m and 13m. Find the area of the field.

Ans: It is given that Length of diagonal = ${\text{24m}}$

Length of the perpendiculars ${{\text{h}}_{\text{1}}}$ and ${{\text{h}}_{\text{2}}}$ from the opposite vertices to the diagonal are

${{\text{h}}_{\text{1}}}$ = ${\text{8m}}$ and ${{\text{h}}_{\text{2}}}$ = ${\text{13m}}$

Area of Quadrilateral = $\dfrac{{\text{1}}}{{\text{2}}}{\text{d}}\left( {{{\text{h}}_{\text{1}}}{\text{ + }}{{\text{h}}_{\text{2}}}} \right)$

= $\dfrac{{\text{1}}}{{\text{2}}}{\text{(24m) x (13m + 8cm)}}$

= $\dfrac{{\text{1}}}{{\text{2}}}{\text{(24m)(21m)}}$

= ${\text{252}}{{\text{m}}^{\text{2}}}$

Thus, Area of field = ${\text{252}}{{\text{m}}^{\text{2}}}$

Q5. The diagonals of a rhombus are ${\text{7}}{\text{.5cm}}$and ${\text{12cm}}$. Find its area.

Ans: Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$

= $\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 7}}{\text{.5cm x 12cm}}$

= ${\text{45c}}{{\text{m}}^{\text{2}}}$

Q6. Find the area of a rhombus whose side is ${\text{6cm}}$and whose altitude is ${\text{4cm}}$. If one of its diagonals is ${\text{8cm}}$long, find the length of the other diagonal.

Ans: Let the length of other diagonal of Rhombus be ‘ \[{\text{X}}\] ’

A Rhombus is a special case of Parallelogram

The area of Parallelogram is given by its base and height

Thus, Area of Rhombus = Base x Height

= ${\text{6cm x 4cm}}$ = ${\text{24c}}{{\text{m}}^{\text{2}}}$

So,

Area of Rhombus = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Product of its diagonals) }}$

${\text{24c}}{{\text{m}}^{\text{2}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(8cm x X)}}$

${\text{X = }}\;{\text{6cm}}$

Therefore, Length of other diagonal of Rhombus = ${\text{6cm}}$

Q7. The floor of a building consists of $3000$tiles which are rhombus shaped and each of its diagonals are ${\text{45cm}}$and ${\text{30cm}}$in length. Find the total cost of polishing the floor, if the cost per ${{\text{m}}^{\text{2}}}$is Rs$4$.

Ans: Given that each diagonals of Rhombus are ${\text{45cm}}$ and ${\text{30cm}}$

= $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 45 x 30}}} \right){\text{c}}{{\text{m}}^{\text{2}}}$

= ${\text{675c}}{{\text{m}}^{\text{2}}}$

So, Area of $3000$ tiles = $(675 x 3000){\text{c}}{{\text{m}}^2} = 2025000\;{\text{c}}{{\text{m}}^2} = 202.5\;{{\text{m}}^2}$

Now, it is given that cost of polishing is ${\text{Rs}}{\text{.}}\;{\text{4}}\;{\text{per}}\;{{\text{m}}^{\text{2}}}$

So, Cost of Polishing ${\text{202}}{\text{.5}}{{\text{m}}^{\text{2}}}$ area = ${\text{Rs(4 x 202}}{\text{.5) = Rs 810}}$

Hence, Cost of polishing the floor is ${\text{Rs}}{\text{.}}\;{\text{810}}$

Q8. Mohan wants to buy a trapezium shaped field. Its side along the river is parallel to and twice the side along the road. It the area of this field is ${\text{10500}}{{\text{m}}^{\text{2}}}$and the perpendicular distance between the two parallel sides is ${\text{100m}}$, find the length of the side along the river.

Ans: Let the length of the side along the road = ‘ \[{\text{l}}\] ’

And Let the length of the side along the river = ‘ ${\text{2l}}$ ’

It is given that distance between two parallel sides = ${\text{100m}}$

and Area of Trapezium = ${\text{10500}}{{\text{m}}^{\text{2}}}$

Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$

${\text{10500}}{{\text{m}}^{\text{2}}}$ = $\dfrac{{\text{1}}}{{\text{2}}}{\text{(l + 2l) x (100m)}}$

${\text{3l = }}\left( {\dfrac{{{\text{2 x 10500}}}}{{{\text{100}}}}} \right){\text{m = 210m}}$

${\text{l = }}\;\dfrac{{{\text{210}}}}{{\text{3}}}{\text{m}}\;{\text{ = }}\;{\text{70m}}$

Therefore, Length of the side along the river ‘ ${\text{2l}}$ ’ = ${\text{140m}}$

Q9. Top surface of a raised platform is in the shape of a regular octagon as shown in the figure. Find the area of the octagonal surface.

It is given in the figure that side of octagon = ${\text{5cm}}$

Area of Trapezium ABCH = Area of Trapezium DEFG

Area of Trapezium = $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$

= $\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{(4)(11 + 5)}}} \right]{{\text{m}}^{\text{2}}}$

= $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 4 x 16}}} \right){{\text{m}}^{\text{2}}}{\text{ = 32}}{{\text{m}}^{\text{2}}}$

In rectangle HCDG, Length( \[{\text{l}}\] ) = ${\text{11m}}$ and Breadth( \[{\text{b}}\] ) = ${\text{5m}}$

So, Area of rectangle = ${\text{(11 x 5)}}{{\text{m}}^{\text{2}}}$ = ${\text{55}}{{\text{m}}^{\text{2}}}$

Therefore, Area of octagon = Area of Trapezium ABCH + Area of Trapezium DEFG +

Area of Rectangle

= ${\text{(32 + 32 + 55)}}{{\text{m}}^{\text{2}}}$

= ${\text{119}}{{\text{m}}^{\text{2}}}$

Q10. There is a pentagonal shaped park as shown in the figure. For finding its area Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some other way of finding its area?

Ans: From Jyoti’s Way of finding area ,

Area of Pentagon = $2$ (Area of Trapezium ABCF)

= $2$ [ $\dfrac{{\text{1}}}{{\text{2}}}{\text{ (Sum of parallel sides) (Distance between the parallel sides) }}$ ]

= $\left[ {{\text{2 x }}\dfrac{{\text{1}}}{{\text{2}}}{\text{(15 + 30)}}\left( {\dfrac{{{\text{15}}}}{{\text{2}}}} \right)} \right]{{\text{m}}^{\text{2}}}$

= ${\text{337}}{\text{.5}}{{\text{m}}^{\text{2}}}$

From Kavita’s Way of finding area ,

Area of Pentagon = Area of Triangle ABE + Area of Square BCDE

= [ $\dfrac{1}{2}$ (basexheight)] + (sidexside)

= $\left[ {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 15 x (30 - 15) + (15}}{{\text{)}}^{\text{2}}}} \right]{{\text{m}}^{\text{2}}}$

= $\left( {\dfrac{{\text{1}}}{{\text{2}}}{\text{ x 15 x 15 + 225}}} \right){{\text{m}}^{\text{2}}}$

= ${\text{(112}}{\text{.5 + 225)}}{{\text{m}}^{\text{2}}}$

= ${\text{337}}{\text{.5}}{{\text{m}}^{\text{2}}}$

Q11. Diagram of the adjacent picture frame has outer dimensions = ${\text{24cm}}$× ${\text{28cm}}$and inner dimensions ${\text{16cm}}$× ${\text{20cm}}$ Find the area of each section of the frame, if the width of each section is same.

Given that, the width of each section is the same.

IB = BJ = CK = CL = DM = DN = AO = AP

IL = IB + BC + CL

\[28\ =\ \text{IB}\ +\ 20\ +\ \text{CL}\]

$\begin{align} & 28-20\ \text{=}\ \text{IB+}\ \text{CL} \\ & 8\ \text{=}\ \text{IB+}\ \text{CL} \\ \end{align}$

IB = CL = \[{\text{4cm}}\]

Hence, IB = BJ = CK = CL = DM = DN = AO = AP = $\text{4}\,\text{cm}$

Area of section BEFC = Area of section DGHA = Area of Trapezium

$\left[ \frac{\text{1}}{\text{2}}\text{(bas}{{\text{e}}_{1}}\text{ + bas}{{\text{e}}_{2}}\text{)(h)} \right]=\left[ \frac{1}{2}(20\text{ }+\text{ }28)(4) \right]\text{c}{{\text{m}}^{\text{2}}}\text{=}96\ \text{c}{{\text{m}}^{\text{2}}}$

Therefore, Area of section ABEH = Area of section CDGF.

Hence area of each section of frame is $96\ \text{c}{{\text{m}}^{\text{2}}}$

Exercise 11.3

Q1. There are two cuboidal boxes as shown in the adjoining figure. Which box requires the lesser amount of material to make?

Ans: From the given figure; Length( \[{\text{l}}\] ), Breadth( \[{\text{b}}\] ) and Height( \[{\text{h}}\] ) of the Cuboid is

\[{\text{60cm}}\] , ${\text{40cm}}$ , ${\text{50cm}}$ respectively.

And, Side of Cube is ${\text{50cm}}$ .

Now, Total Surface area of Cuboid(a) = $2$ ( \[{\text{l}}\] \[{\text{h}}\] + \[{\text{b}}\] \[{\text{h}}\] + \[{\text{l}}\] \[{\text{b}}\] )

= \[{\text{[2\{ (60)(40) + (40)(50) + (50)(60)\} ]c}}{{\text{m}}^{\text{2}}}\]

= ${\text{[2(2400 + 2000 + 3000)]c}}{{\text{m}}^{\text{2}}}$

= ${\text{(2 x 7400)c}}{{\text{m}}^{\text{2}}}$

= ${\text{14800c}}{{\text{m}}^{\text{2}}}$

Total Surface area of Cube(b) = \[{\text{6}}{{\text{l}}^{\text{2}}}\]

= $6{(50\;{\text{cm}})^2} = 15000\;{\text{c}}{{\text{m}}^2}$

Therefore, Cuboidal box(a) requires a lesser amount of material for making.

Q2. A suitcase with measure ${\text{80cm x 48cm x 24cm}}$is to be covered with a tarpaulin cloth. How many metres of tarpaulin of width ${\text{96cm}}$is required to cover $100$ such suitcases?

Ans: Given that length( \[{\text{l}}\] ),breadth( \[{\text{b}}\] ),height( \[{\text{h}}\] ) of suitcase is \[{\text{(80,48,24)cm}}\] respectively.

Total surface area of suitcase = $2$ ( \[{\text{l}}\] \[{\text{h}}\] + \[{\text{b}}\] \[{\text{h}}\] + \[{\text{l}}\] \[{\text{b}}\] )

= ${\text{2[(80)(48) + (48)(24) + (24)(80)]}}$

= \[{\text{2[3840 + 1152 + 1920]}}\]

= ${\text{13824}}{{\text{m}}^{\text{2}}}$

Total surface area of $100$ suitcases = ${\text{100 x 13824c}}{{\text{m}}^{\text{2}}}$

= ${\text{1382400c}}{{\text{m}}^{\text{2}}}$

= required Tarpaulin

We have given that breadth of Tarpaulin is ${\text{96cm}}$ and we have to find Length

of tarpaulin.

Required Tarpaulin = (Length x Breadth) of Tarpaulin

${\text{1382400c}}{{\text{m}}^{\text{2}}}$ = Length x ${\text{96cm}}$

Length = $\left( {\dfrac{{{\text{1382400}}}}{{{\text{96}}}}} \right){\text{cm = 14400cm}}$

Therefore, Length = \[{\text{144m}}\;\;\;\;\;(\because \;{\text{1m = 100cm)}}\]

Thus, ${\text{144m}}$ of tarpaulin is required to cover $100$ suitcases.

Q3. Find the side of a cube whose surface area is ${\text{600c}}{{\text{m}}^{\text{2}}}$ .

Ans: It is given that the surface area of the cube is ${\text{600c}}{{\text{m}}^{\text{2}}}$ .

We have to find the side of the cube ( \[{\text{a}}\] ).

Surface area of cube = ${\text{6}}{{\text{a}}^{\text{2}}}$

${\text{600c}}{{\text{m}}^{\text{2}}}$ = ${\text{6}}{{\text{a}}^{\text{2}}}$

$\therefore \;{{\text{a}}^{\text{2}}}{\text{ = 100c}}{{\text{m}}^{\text{2}}}$

\[{\text{a}}\] = ${\text{10cm}}$

Thus, the side of the cube is ${\text{10cm}}$ .

Q4. Rukhsar painted the outside of the cabinet of measure${\text{1m x 2m x 1}}{\text{.5m}}$. How much surface area did she cover if she painted all except the bottom of the cabinet?

Ans: It is given that length ( \[{\text{l}}\] ), breadth( \[{\text{b}}\] ), height( \[{\text{h}}\] ) of the cabinet is

${\text{2m,1m,1}}{\text{.5m}}$ respectively.

Area of the surface = ${\text{2h(1 + b) + lb}}$

= ${\text{[2 x 1}}{\text{.5 x (2 + 1) + (2)(1)]}}{{\text{m}}^{\text{2}}}$

= ${\text{[3(3) + 2]}}{{\text{m}}^{\text{2}}}$

= ${\text{11}}{{\text{m}}^{\text{2}}}$

Q5. Daniel is painting the walls and ceiling of a cuboidal hall with length, breadth and height of ${\text{15m,10m and 7m}}$respectively. From each can of paint ${\text{100}}{{\text{m}}^{\text{2}}}$ of area is painted. How many cans of paint will she need to paint the room?

Ans: Given that length ( \[{\text{l}}\] ), breadth( \[{\text{b}}\] ), height( \[{\text{h}}\] ) of Cuboid is

${\text{15m,10m and 7m}}$ respectively.

Area of the hall to be painted = Area of the wall + Area of ceiling

= $2$ \[{\text{h}}\] ( \[{\text{l}}\] + \[{\text{b}}\] ) + \[{\text{l}}\]\[{\text{b}}\]

= ${\text{[2 x 7(15 + 10) + (15 x 10)]}}{{\text{m}}^{\text{2}}}$

= ${\text{[(2 x 175) + 150]}}{{\text{m}}^{\text{2}}}$

= ${\text{500}}{{\text{m}}^{\text{2}}}$

It is given that ${\text{100}}{{\text{m}}^{\text{2}}}$ area is to be painted from each can.

Therefore, Number of cans required to paint an area of ${\text{500}}{{\text{m}}^{\text{2}}}$

= $\dfrac{{500}}{{100}}{\kern 1pt} \; = \;5$

Hence, $5$ cans of paint are required to paint the room.

Q6. Describe how the two figures at the right are alike and how they are different. Which box has a larger lateral surface area?

Ans: The above given two figures alike for same height( \[{\text{h}}\] )= ${\text{7cm}}$

And the difference between in these two figures is that one is cylinder and the other

One is a cube.

Now, we have to find the lateral surface area for both of the given figures.

Given that Side of cube( \[{\text{l}}\] )= ${\text{7cm}}$

Height and Diameter of cylinder = ${\text{7cm}}$ each

Radius of cylinder = $\dfrac{{{\text{Diameter}}}}{{\text{2}}}\;{\text{ = }}\;\dfrac{{\text{7}}}{{\text{2}}}{\text{cm = }}\;{\text{3}}{\text{.5cm}}$

First, Lateral surface area of Cube = ${\text{4}}{{\text{l}}^{\text{2}}}$ = ${\text{4(}}{{\text{7}}^{\text{2}}}{\text{)c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{4 x 49c}}{{\text{m}}^{\text{2}}}\;{\text{ = }}\;{\text{196c}}{{\text{m}}^{\text{2}}}$

Second, Lateral surface area of cylinder = ${\text{2$\pi$ rh}}$

= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 3}}{\text{.5cm x 7cm}}\;{\text{ = }}\;{\text{154c}}{{\text{m}}^{\text{2}}}$

Therefore, the Cube has a larger lateral surface area.

Q7. A closed cylindrical tank of radius ${\text{7m}}$ and height ${\text{3m}}$ is made from a sheet of metal. How much sheet of metal is required?

Ans: Given that the radius and height of Cylinder is ${\text{7m}}$ and ${\text{3m}}$ respectively.

Therefore, Total surface area of Cylinder = ${\text{2$\pi$ }}$\[{\text{r}}\] ( \[{\text{r}}\] + \[{\text{h}}\] )

= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7m(7m + 3m)}}$

= ${\text{440}}{{\text{m}}^2}$

Thus, ${\text{440}}{{\text{m}}^2}$ of metal sheet is required.

Q8. The lateral surface area of a hollow cylinder is ${\text{4224c}}{{\text{m}}^{\text{2}}}$. It is cut along its height and formed a rectangular sheet of width ${\text{33cm}}$. Find the perimeter of the rectangular?

Ans: It is given that Hollow cylinder is cut along its height and formed a

Rectangular sheet.

Area of cylinder = ${\text{4224c}}{{\text{m}}^{\text{2}}}$

And, Breadth of rectangular sheet = ${\text{33cm}}$

So, Area of Cylinder = Area of Rectangular Sheet

${\text{4224c}}{{\text{m}}^{\text{2}}}$ = Length x Breadth

${\text{4224c}}{{\text{m}}^{\text{2}}}$ = Length x ${\text{33cm}}$

Length = \[\dfrac{{{\text{4224}}}}{{{\text{33}}}}{\text{cm}}\;{\text{ = }}\;{\text{128cm}}\]

Now, Perimeter of Rectangle = $2$ (length + breadth)

= ${\text{2(128 + 33)cm}}\;{\text{ = }}\;{\text{2(161cm)}}\;{\text{ = }}\;{\text{322cm}}$

Q9. A road roller takes ${\text{750}}$complete revolutions to move once over to level a road. Find the area of the road if the diameter of a road roller is ${\text{84cm}}$ and length is ${\text{1m}}$.

Ans: In one revolution, the roller will cover an area equal to its lateral surface area.

Here, Radius = $\dfrac{{{\text{diameter}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{84}}}}{{\text{2}}}{\text{cm = 42cm = }}\dfrac{{{\text{42}}}}{{{\text{100}}}}{\text{m}}\;\;\;\;\;\;(\because {\text{1m = 100cm}})$

Height = ${\text{1m}}$

Thus, In One Revolution,

Area of the road covered = ${\text{2$\pi$ }}$ rh

= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{{\text{42}}}}{{{\text{100}}}}{\text{ x 1}}{{\text{m}}^{\text{2}}}$

= $\dfrac{{{\text{264}}}}{{{\text{100}}}}{{\text{m}}^{\text{2}}}$

In ${\text{750}}$ revolutions, area of road covered = ${\text{750 x }}\dfrac{{{\text{264}}}}{{{\text{100}}}}{{\text{m}}^{\text{2}}}$

= \[{\text{1980}}{{\text{m}}^{\text{2}}}\]

Q10. A company packages its milk powder in a cylindrical container whose base has a diameter of ${\text{14cm}}$and height ${\text{20cm}}$. Company places a label around the surface of the container (as shown in the figure). If the label is placed ${\text{2cm}}$from top and bottom, what is the area of the label?

Ans: It is given that Radius = $\dfrac{{{\text{diameter}}}}{{\text{2}}}{\text{ = }}\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm = 7cm}}$

Height of label = ${\text{(20 - 2 - 2)cm}}\;\;\;\;\;\;(\because {\text{2cm}}\;{\text{deducted}}\;{\text{from}}\,{\text{top}}\;{\text{bottom}}\;{\text{each)}}$

= ${\text{16cm}}$

As shown in the figure, the label is in the shape of a cylinder.

So, Area of label(cylinder)= ${\text{2$\pi$ }}$ rh

= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7cm x 16cm}}\;{\text{ = }}\;{\text{44 x 16c}}{{\text{m}}^{\text{2}}}$

= ${\text{704c}}{{\text{m}}^{\text{2}}}$

Exercise 11.4

Q1. Given a cylindrical tank, in which situation will you find surface area and in which situation volume.

(a) To find how much it can hold

(b) Number of cement bags required to plaster it

Ans: (a) In this situation, we will find the volume.

(b) Number of cement bags required to plaster cylindrical bags so for that

situation, we will find the surface area.

find the volume.

Q2. Diameter of cylinder A is ${\text{7cm}}$, and the height is ${\text{14cm}}$. Diameter of cylinder B is ${\text{14cm}}$ and height is ${\text{7cm}}$. Without doing any calculations can you suggest whose volume is greater? Verify it by finding the volume of both the cylinders. Check whether the cylinder with greater volume also has greater surface area?

Ans: The heights and diameters of these cylinders A and B are interchanged.

We know that,

Volume of cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

If measures of radius(r) and height(h) are same, then the cylinder with greater

radius will have greater area.

Here, Radius of cylinder A = $\dfrac{{\text{7}}}{{\text{2}}}{\text{cm}}$

Radius of cylinder B = $\dfrac{{{\text{14}}}}{{\text{2}}}{\text{cm = 7cm}}$

As the radius of cylinder B is greater, therefore, the volume of cylinder B will

be greater.

Let us verify it by calculating the volume of both the cylinders.

Volume of Cylinder A = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x 14c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{11 x 49c}}{{\text{m}}^{\text{3}}}\;{\text{ = 539c}}{{\text{m}}^{\text{3}}}$

Volume of Cylinder B = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7 x 7 x 7c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{22 x 49c}}{{\text{m}}^{\text{3}}}\;{\text{ = 1078c}}{{\text{m}}^{\text{3}}}$

Therefore, the volume of cylinder B is greater.

Now, Surface area of cylinder A = ${\text{2$\pi$ r(r + h)}}$

= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{\text{7}}}{{\text{2}}}{\text{ x (}}\dfrac{{\text{7}}}{{\text{2}}}{\text{ + 14)c}}{{\text{m}}^{\text{2}}}$

= \[{\text{22 x }}\dfrac{{{\text{35}}}}{{\text{2}}}{\text{c}}{{\text{m}}^{\text{2}}}{\text{ = }}\;{\text{385c}}{{\text{m}}^{\text{2}}}\]

Surface area of cylinder B = ${\text{2$\pi$ r(r + h)}}$

= ${\text{2 x }}\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 7 x (7 + 7)c}}{{\text{m}}^{\text{2}}}$

= \[{\text{44 x 14c}}{{\text{m}}^{\text{2}}}{\text{ = }}\;{\text{616c}}{{\text{m}}^{\text{2}}}\]

Q3. Find the height of a cuboid whose base area is ${\text{180c}}{{\text{m}}^{\text{2}}}$ and volume is ${\text{900c}}{{\text{m}}^{\text{3}}}$?

Ans. Here, we have given that Base Area of Cuboid = Length x Breadth

= ${\text{180c}}{{\text{m}}^{\text{2}}}$

Volume of Cuboid = Length x Breadth x Height

${\text{900c}}{{\text{m}}^{\text{3}}}$ = ${\text{180c}}{{\text{m}}^{\text{2}}}$ x Height

Height = $\dfrac{{{\text{900}}}}{{{\text{180}}}}{\text{cm = 5cm}}$

Q4. A cuboid is of dimensions ${\text{60cm x 54cm x 30cm}}$. How many small cubes with side${\text{6cm}}$can be placed in the given cuboid?

Ans. From given condition,

Volume of Cuboid = ${\text{60cm x 54cm x 30cm}}$

= ${\text{97200c}}{{\text{m}}^{\text{3}}}$

Given that side of cube = ${\text{6cm}}$

So, Volume of cube = ${\text{(6 x 6 x 6)c}}{{\text{m}}^{\text{3}}}\;{\text{ = }}\;{\text{216c}}{{\text{m}}^{\text{3}}}$

Required number of cubes = \[\dfrac{{{\text{volume}}\;{\text{of}}\;{\text{cuboid}}}}{{{\text{volume}}\;{\text{of}}\;{\text{cube}}}}{\text{ = }}\dfrac{{{\text{97200}}}}{{{\text{216}}}}{\text{ = 450}}\]

Therefore, ${\text{450}}$ cubes can be placed in the given Cuboid.

Q5. Find the height of the cylinder whose volume is ${\text{1}}{\text{.54}}{{\text{m}}^{\text{3}}}$and diameter of the base is ${\text{140cm}}$?

Ans. It is given that Radius of Cylinder = $\dfrac{{{\text{140}}}}{{\text{2}}}{\text{cm = 70cm}}$

Volume of Cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

${\text{1}}{\text{.54}}{{\text{m}}^{\text{3}}}$ = $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x }}\dfrac{{{\text{70}}}}{{{\text{100}}}}{\text{m x }}\dfrac{{{\text{70}}}}{{{\text{100}}}}{\text{m x h}}$

Height = $\dfrac{{{\text{1}}{\text{.54 x 100}}}}{{{\text{22 x 7}}}}{\text{m = 1m}}$

Hence, the height of cylinder = ${\text{1m}}$

Q6. A milk tank is in the form of cylinder whose radius is ${\text{1}}{\text{.5m}}$and length is ${\text{7m}}$. Find the quantity of milk in litres that can be stored in the tank?

Ans. It is given that Radius and height of the cylinder is ${\text{1}}{\text{.5m}}$ and ${\text{7m}}$ respectively .

Therefore, Volume of Cylinder = ${\text{$\pi$ }}{{\text{r}}^{\text{2}}}{\text{h}}$

= $\dfrac{{{\text{22}}}}{{\text{7}}}{\text{ x 1}}{\text{.5m x 1}}{\text{.5m x 7m}}$

= ${\text{22 x 2}}{\text{.25}}{{\text{m}}^{\text{3}}}$

= ${\text{49}}{\text{.5}}{{\text{m}}^{\text{3}}}$

As, ${\text{1}}{{\text{m}}^{\text{3}}}{\text{ = 1000}}\;{\text{litre}}$

So, Required Quantity = ${\text{49}}{\text{.5 x 1000}}\;{\text{litre}}\;{\text{ = }}\;{\text{49500}}\;{\text{litre}}$

Therefore, ${\text{49500}}\;{\text{litre}}$ can be stored in the tank.

Q7. If each edge of a cube is doubled,

(i) how many times will its surface area increase?

(ii) how many times will its volume increase?

Ans. (i) Let the edge of the cube be ‘ \[{\text{a}}\] ’.

Surface area of cube = ${\text{6}}{{\text{a}}^{\text{2}}}$

If each edge of the cube is doubled, then it becomes ${\text{2a}}$

Therefore, New surface area = ${\text{6(2a}}{{\text{)}}^{\text{2}}}{\text{ = 24}}{{\text{a}}^{\text{2}}}{\text{ = 4(6}}{{\text{a}}^{\text{2}}}{\text{)}}$

Clearly, the surface area will be increased by ${\text{4}}$ times.

(ii) Let Volume of the cube = ${{\text{a}}^{\text{3}}}$

When each edge of the cube is doubled, it becomes ${\text{2a}}$ .

New volume = \[{{\text{(2a)}}^{\text{3}}}{\text{ = 8}}{{\text{a}}^{\text{3}}}{\text{ = 8 x }}{{\text{a}}^{\text{3}}}\]

Clearly, the volume of the cube will be increased by ${\text{8}}$ times.

Q8. Water is pouring into a cubiodal reservoir at the rate of $60$litres per minute. If the volume of reservoir is ${\text{108}}{{\text{m}}^{\text{3}}}$, find the number of hours it will take to fill the reservoir.

Ans. Volume of cuboidal reservoir = ${\text{108}}{{\text{m}}^{\text{3}}}$ = $(108 x 1000)$ \[{\text{L}}\] = $108000$ \[{\text{L}}\]

It is given that water is being poured at the rate of $60$ L per minute.

That is, $(60 x 60)$ \[{\text{L}}\] = $3600$ \[{\text{L}}\] per hour

Required number of hours = $30$ hours

Thus, it will take $30$ hours to fill the reservoir.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration - PDF Download

It is not compulsory to be connected to the internet to access our solutions as Vedantu has made NCERT Solutions for Class 8 Maths Mensuration available in the PDF format on its official website. Now download the NCERT Solutions Class 8 Maths Chapter 11 PDF on your devices or get a printout and access it from anywhere, anytime. This mode of revision is quick and easy and can be done at your pace during a crucial exam period.

All Topics of NCERT Class 8 Maths Chapter 11 - Mensuration

The topics covered under chapter 11 Mensuration are given below.

Table of Important Formulas

Below given are the list of important formulas that you must remember to solve the exercise of chapter 11 NCERT Maths Class 8 th .

You should also remember some of the basic conversion parameters. For your ease, the most important parameters are given below.

1 m 3 = 1000000 cm 3 = 1000 L

1 cm 3 = 1 mL

1 L = 1000 cm 3

List of Exercises in class 8 Maths Chapter 11:

Chapter 11 – mensuration.

Introduction

Mensuration is the process applied to different 2-D and 3-D solids of various shapes and figures to measure their lengths, volumes, area, heights, perimeters and several other dimensions. In this section of Ch 11 Maths Class 8, you will recall the areas of plane figures like triangles, circles, rectangles, etc., that you learned in the previous chapter. In NCERT Solutions for Class 8 Maths Ch 11, you would learn how to calculate perimeter and areas of other closed figures like Quadrilaterals.

Let us Recall

In this section of Mensuration Class 8 NCERT, students would recall the formula for calculating the perimeter and area of a park, flower bed, and amount of cement required to cover a given area. All the problems are based on areas of the following shape:

Rectangle - The area of a rectangle is x*y where x is the length and y is the breadth of the rectangle.

Square - The area of a square is x 2 where x is the length of one side of a square.

Triangle - The area of a triangle is ½ *b*h where b is the length of the base and h is the height of the triable

Circle - The area of a circle is πr 2 where r is the radius of the circle.

Area of a Trapezium

A trapezium is a quadrilateral where two of the sides are parallel to each other. In this portion of Mensuration Class 8 NCERT Solutions, students will learn how to derive the area of a trapezium which is given by:

Area of a trapezium = h * (x + y)/2, where h is the height of the trapezium, x and y are the lengths of its two sides.

Area of a general Quadrilateral

A simple definition of a general quadrilateral is a closed 2-D shape having 4 straight sides. If we break the word Quadrilateral, Quad means 4, and lateral means sides. The area of a quadrilateral in Class 8th Maths Chapter 11 is calculated by splitting it into two triangles and then calculating and adding the areas of the two triangles.

(Image to be added soon)

So if PQRS is a quadrilateral, then its area = (area of triangle PQR) + (area of triangle PRS) = ½ * d * (h 1 + h 2 ), where d is the length of the diagonal from P to R and h 1 and h 2 are heights of perpendiculars dropped from Q and Pr and S or PR respectively.

11.8 Volume of Cube, Cuboid, and Cylinder

The amount of space that a 3-D object occupies gives the volume of that object. To take examples from real life, the volume of a cupboard in a room is less than the volume of the room where it is placed.

11.8.1 - Volume of a Cuboid - A 3-D structure with 6 rectangular faces is a cuboid. Its volume is given by “l * h* b”, here l = length, b = breadth, and h= height.

11.8.2 - Volume of a Cube - A cube is a special type of cuboid where its length, breadth, and height are the same. Hence the volume of a cube = length 3 .

11.8.3 - Volume of a Cylinder - A cylinder has two circular bases that are parallel to each other and separated by a distance. To measure the volume of a cylinder we use the formula πr 2 * h. Here r is the radius of the circular base and h is the distance between the bases.

Key Features of NCERT Solutions for Class 8 Maths Chapter 11

You will find the NCERT Solutions of Class 8 Maths Chapter 11 by Vedantu extremely beneficial for your exams. The key features are:

Comprehensive explanations for each exercise and questions, promoting a deeper understanding of the subject.

Clear and structured presentation for easy comprehension.

Accurate answers aligned with the curriculum, boosting students' confidence in their knowledge.

Visual aids like diagrams and illustrations to simplify complex concepts.

Additional tips and insights to enhance students' performance.

Chapter summaries for quick revision.

Online accessibility and downloadable resources for flexible study and revision.

NCERT Solutions play a crucial role in Class 8 exam prep. Start by thoroughly reading the textbook chapter. After that, solve the NCERT questions for Class 8 Chapter 11. You can find detailed solutions on Vedantu, aligning with CBSE guidelines. Download the free NCERT Solutions for Class 8 Chapter 11 to guide your exam preparation with expert-reviewed answers.

FAQs on NCERT Solutions for Class 8 Maths Chapter 11 - Mensuration

1. What are some of the applications of mensuration in our everyday lives?

In real life, mensuration is applied in many fields like:

Measuring floor and site areas required for purchasing or selling land.

Measuring agricultural fields.

Measuring surface areas of a house required for estimating painting cost.

To know the volume of the level of water in rivers and tanks.

To find out the amount of carpet required for covering a specific room.

The volume of soil that is needed to fill a ditch.

2. What are solid shapes and what is the method of measuring the surface area of a solid shape?

Any 3-D shapes occupying some space are called solid shapes like cube, sphere, cylinder, etc. To measure the surface area of a solid shape, we need to draw the net of that solid shape. From the net, we can see all the faces of the solid clearly. Then, we calculate the areas of each of the faces and add them up to get the total surface area of the solid shape. Surface area is measured in a square unit.

3. What are the concepts covered under Chapter 11 of Class 8 Maths?

The ideas or topics that are included in Chapter 11 “Mensuration” of Class 8 Maths are given below:

Let Us Recall

Area of Trapezium

Area of a General Quadrilateral

Area of Special Quadrilaterals

Area of a Polygon

Solid Shapes

Surface Area of Cylinder, Cube and Cuboid

Cuboid Cube Cylinders

Volume of Cuboid, Cube and Cylinder

Cuboid Cube Cylinder

Volume and Capacity

What Have We Discussed?

4. The base area of the cuboid is 25cm sq. Its volume is 275 cubic cm. What will be the height of the cuboid?

In the question, we are given the base area of the cuboid which is equal to 25cm sq.

The volume of the cuboid is 275 cubic cm.

We know that according to the formula,

The volume of a cuboid = Height × Base Area

Therefore, the height of the cuboid will be = Volume of cuboid/ Base Area

Height = 275/25 = 11cm

Thus, 11cm is the height of the cuboid.

5. The distance between two parallel sides is 15m and the length of one parallel side is 20m. 480m sq. Is the area of the trapezium-shaped field. What is the length of the other parallel side?

Let one parallel side be a = 20m and the other parallel side is ‘b’.

The height of the field is 15m.

Given, the area of the trapezium is 480m sq.

The formula of trapezium is,

Area of trapezium = ½ (a + b) * h

480 = ½ (20 + b) * 15

20 + b = 480 × 2/ 15

64 = 20 + b

Thus, the length of the other parallel side is 44m.

6. What are the perks of NCERT Solutions of Chapter 11 of Class 8 Maths?

The perks of the NCERT Solutions of Chapter 11 of Class 8 Maths are given below:

The NCERT Solutions of Chapter 11 of Class 8 Maths offers comprehensive learning.

It enables students to develop their reasoning and logical skills.

These solutions assist students in understanding the difficult concepts.

By practising these, students will have a strong grip over the chapter.

You will get a hint of how to answer the questions in the proper format.

It helps in scoring good marks in Chapter 11 of Class 8 Maths.

The NCERT Solutions for Chapter 11 of Class 8 Maths are available free of cost on the Vedantu website and on the Vedantu app.

7. How can I make the best study plan for Chapter 11 of Class 8 Maths?

Keep the following points in mind while making an effective study plan for Chapter 11 of Class 8 Maths:

Have a timetable or schedule to manage your time.

Centralize the NCERT book to read Chapter 11 of Class 8 Maths.

Practice the NCERT Solutions to comprehend the Chapter.

Give yourself a break.

Do meditation and exercise to keep your body and mind fit.

Attend all your school lectures.

NCERT Solutions for Class 8 Maths

Ncert solutions for class 8.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration are provided below. Our solutions covered each questions of the chapter and explains every concept with a clarified explanation. To score good marks in Class 8 Mathematics examination, it is advised to solve questions provided at the end of each chapter in the NCERT book.

NCERT Solutions for Class 8 Maths Chapter 11 Mensuration are prepared based on Class 8 NCERT syllabus, taking the types of questions asked in the NCERT textbook into consideration. Further, all the CBSE Class 8 Solutions Maths Chapter 11 are in accordance with the latest CBSE guidelines and marking schemes.

Class 8 Maths Chapter 11 Exercise 11.1 Solutions

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Class 8 Maths Chapter 11 Exercise 11.2 Solutions

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Class 8 Maths Chapter 11 Exercise 11.3 Solutions

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Class 8 Maths Chapter 11 Exercise 11.4 Solutions

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CBSE Class 8th Maths Value Based Questions Chapter 11 Mensuration PDF Download

CBSE Class 8th Maths Value Based Questions Chapter 11 Mensuration are the easiest questions which you see in your question paper and the scoring one all student who attempt it surely get they are just little bit difficult and examine your basic knowledge regarding the particular chapter. Maths Value Based Questions for Class 8th are available here at Free of cost. These questions are expected to be asked in the Class 8th board examination. These Maths Value Based Questions are from complete CBSE Syllabus.

CBSE Class 8th Maths Value Based Questions Chapter 11 Mensuration

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CBSE Maths Value Based Questions Class 8th Chapter 11 Mensuration PDF

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RD Sharma Solutions
Chapter 20 Area Of Trapezium And Polygon

RD Sharma Solutions for Class 8 Maths Chapter 20 Mensuration - I (Area of a Trapezium and a Polygon)

In this chapter, students shall study methods of finding the area of a trapezium and also problems on finding areas of some polygons by using the formulae for the area of a triangle and that of a trapezium. Our subject experts have solved the problems using shortcut methods to help students in solving the problems effortlessly. RD Sharma Class 8 is the best study material, provided in accordance with the latest CBSE syllabus. One can use this to prepare well for the final exam. Students can refer to and download the PDF easily from the links provided below.

Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) contains three exercises, and the RD Sharma Class 8 Solutions present in this page provide solutions to the questions given in each exercise. Now, let us have a look at the concepts discussed in this chapter.

Area of a trapezium
Area of a polygon
RD Sharma Solutions for Class 8 Maths Chapter 1 Rational Numbers
RD Sharma Solutions for Class 8 Maths Chapter 2 Powers
RD Sharma Solutions for Class 8 Maths Chapter 3 Squares and Square Roots
RD Sharma Solutions for Class 8 Maths Chapter 4 Cubes and Cube Roots
RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers
RD Sharma Solutions for Class 8 Maths Chapter 6 Algebraic Expressions and Identities
RD Sharma Solutions for Class 8 Maths Chapter 7 Factorization
RD Sharma Solutions for Class 8 Maths Chapter 8 Division of Algebraic Expressions
RD Sharma Solutions for Class 8 Maths Chapter 9 Linear Equations in One Variable
RD Sharma Solutions for Class 8 Maths Chapter 10 Direct and Inverse Variations
RD Sharma Solutions for Class 8 Maths Chapter 11 Time and Work
RD Sharma Solutions for Class 8 Maths Chapter 12 Percentage
RD Sharma Solutions for Class 8 Maths Chapter 13 Profit, Loss, Discount and Value Added Tax (VAT)
RD Sharma Solutions for Class 8 Maths Chapter 14 Compound Interest
RD Sharma Solutions for Class 8 Maths Chapter 15 Understanding Shapes – I (Polygons)
RD Sharma Solutions for Class 8 Maths Chapter 16 Understanding Shapes – II (Quadrilaterals)
RD Sharma Solutions for Class 8 Maths Chapter 17 Understanding Shapes – II (Special Types of Quadrilaterals)
RD Sharma Solutions for Class 8 Maths Chapter 18 Practical Geometry (Constructions)
RD Sharma Solutions for Class 8 Maths Chapter 19 Visualising Shapes

RD Sharma Solutions for Class 8 Maths Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

RD Sharma Solutions for Class 8 Maths Chapter 21 Mensuration – II (Volumes and Surface Areas of a Cuboid and a Cube)
RD Sharma Solutions for Class 8 Maths Chapter 22 Mensuration – III (Surface Area and Volume of a Right Circular Cylinder)
RD Sharma Solutions for Class 8 Maths Chapter 23 Data Handling – I (Classification and Tabulation of Data)
RD Sharma Solutions for Class 8 Maths Chapter 24 Data Handling – II (Graphical Representation of Data as Histograms)
RD Sharma Solutions for Class 8 Maths Chapter 25 Data Handling – III (Pictorial Representation of Data as Pie Charts)
RD Sharma Solutions for Class 8 Maths Chapter 26 Data Handling – IV (Probability)
RD Sharma Solutions for Class 8 Maths Chapter 27 Introduction to Graphs
Exercise 20.1 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)
Exercise 20.2 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)
Exercise 20.3 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

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Access answers to Maths RD Sharma Solutions for Class 8 Chapter 20 Mensuration – I (Area of a Trapezium and a Polygon)

EXERCISE 20.1 PAGE NO: 20.13

1. A flooring tile has the shape of a parallelogram whose base is 24 cm, and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m 2 ?

Given that,

Base of parallelogram = 24cm

Height of parallelogram = 10cm

Area of floor = 1080m 2

We know that,

Area of parallelogram = Base × Height

Area of 1 tile = 24 × 10 = 240cm 2

We know that, 1m = 100cm

So for 1080m 2 = 1080 × 100 × 100 cm 2

To calculate the Number of tiles required = Area of floor/Area of 1 tile

i.e., Number of tiles required = (1080 × 100 × 100) / (24 × 10) = 45000

∴ Number of tiles required = 45000

2. A plot is in the form of a rectangle ABCD having semi-circle on BC, as shown in Fig. 20.23. If AB = 60 m and BC = 28 m, Find the area of the plot.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 1

Area of the plot = Area of the rectangle + Area of semi-circle

Radius of semi-circle = BC/2 = 28/2 = 14m

Area of the Rectangular plot = Length × Breadth = 60 × 28 = 1680 m 2

Area of the Semi-circular portion = πr 2 /2

= 1/2 × 22/7 × 14 × 14

= 308 m 2

∴ The total area of the plot = 1680 + 308 = 1988 m 2

3. A playground has the shape of a rectangle, with two semi-circles on its smaller sides as diameters, added to its outside. If the sides of the rectangle are 36 m and 24.5 m, find the area of the playground. (Take π= 22/7.)

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 2

Area of the plot = Area of the Rectangle + 2 × area of one semi-circle

Radius of semi-circle = BC/2 = 24.5/2 = 12.25m

Area of the Rectangular plot = Length × Breadth = 36 × 24.5 = 882 m 2

Area of the Semi-circular portions = 2 × πr 2 /2

= 2 × 1/2 × 22/7 × 12.25 × 12.25 = 471.625 m 2

Area of the plot = 882 + 471.625 = 1353.625 m 2

4. A rectangular piece is 20 m long and 15 m wide. From its four corners, quadrants of radii 3.5 m have been cut. Find the area of the remaining part.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 3

Area of the plot = Area of the rectangle – 4 × area of one quadrant

Radius of semi-circle = 3.5 m

Area of four quadrants = area of one circle

Area of the plot = Length × Breadth – πr 2

Area of the plot = 20 × 15 – (22/7 × 3.5 × 3.5)

Area of the plot = 300 – 38.5 = 261.5 m 2

5. The inside perimeter of a running track (shown in Fig. 20.24) is 400 m. The length of each of the straight portion is 90 m and the ends are semi-circles. If track is everywhere 14 m wide, find the area of the track. Also, find the length of the outer running track.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 4

Perimeter of the inner track = 2 × Length of rectangle + perimeter of two semi-circular ends

Perimeter of the inner track = Length + Length + 2πr

400 = 90 + 90 + (2 × 22/7 × r)

(2 × 22/7 × r) = 400 – 180

(2 × 22/7 × r) = 220

44r = 220 × 7

r = 1540/44 = 35

So, the radius of inner circle = 35 m

Now, let’s calculate the radius of outer track

Radius of outer track = Radius of inner track + width of the track

Radius of outer track = 35 + 14 = 49m

Length of outer track = 2× Length of rectangle + perimeter of two outer semi-circular ends

Length of outer track = 2× 90 + 2πr

Length of outer track = 2× 90 + (2 × 22/7 × 49)

Length of outer track = 180 + 308 = 488

So, Length of outer track = 488m

Area of inner track = Area of inner rectangle + Area of two inner semi-circles

Area of inner track = Length × Breadth + πr 2

Area of inner track = 90 × 70 + (22/7 × 35 × 35)

Area of inner track = 6300 + 3850

So, Area of inner track = 10150 m 2

Area of outer track = Area of outer rectangle + Area of two outer semi-circles

Breadth of outer track = 35 + 35 +14 + 14 = 98 m

Area of outer track = length× breadth + πr 2

Area of outer track = 90 × 98 + (22/7 × 49 × 49)

Area of outer track = 8820 + 7546

So, Area of outer track = 16366 m 2

Now, let’s calculate area of path

Area of path = Area of outer track – Area of inner track

Area of path = 16366 – 10150 = 6216

So, Area of path = 6216 m 2

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 5

Area of the Figure = Area of square + Area of semi-circle – Area of right angled triangle

Area of the Figure = side × side + πr 2 /2 – (1/2 × base × height)

Area of the Figure = 10 × 10 + (1/2 × 22/7 × 5 × 5) – (1/2 × 8 × 6)

Area of the Figure = 100 + 39.28 – 24

Area of the Figure = 115.3

So, Area of the Figure = 115.3 cm 2

7. The diameter of a wheel of a bus is 90 cm which makes 315 revolutions per minute. Determine its speed in kilometres per hour. (Take π=22/7)

Given that, Diameter of a wheel = 90 cm

We know that, Perimeter of wheel = πd

Perimeter of wheel = 22/7 × 90 = 282.857

So, Perimeter of a wheel = 282.857 cm

Distance covered in 315 revolutions = 282.857× 315 = 89099.955 cm

One km = 100000 cm

Therefore, Distance covered = 89099.955/100000 = 0.89 km

Speed in km per hour = 0.89 × 60 = 53.4 km per hour

8. The area of a rhombus is 240 cm 2 and one of the diagonal is 16 cm. Find another diagonal.

Area of rhombus = 1/2 × d 1 × d 2

240 = 1/2 × 16 × d 2

240 = 8 × d 2

d 2 = 240/8 = 30

So, the other diagonal is 30 cm

9. The diagonals of a rhombus are 7.5 cm and 12 cm. Find its area.

Area of rhombus = 1/2 × 7.5 × 12

Area of rhombus = 6 × 7.5 = 45

So, Area of rhombus = 45 cm 2

10. The diagonal of a quadrilateral shaped field is 24 m and the perpendiculars dropped on it from the remaining opposite vertices are 8 m and 13 m. Find the area of the field.

Area of quadrilateral = 1/2 × d 1 × (p 1 + p 2 )

Area of quadrilateral = 1/2 × 24 × (8 + 13)

Area of quadrilateral = 12 × 21 = 252

So, Area of quadrilateral is 252 cm 2

11. Find the area of a rhombus whose side is 6 cm and whose altitude is 4 cm. If one of its diagonals is 8 cm long, find the length of the other diagonal.

Side of rhombus = 6 cm

Altitude of rhombus = 4 cm

Since rhombus is a parallelogram, therefore area of parallelogram = base × altitude

i.e., Area of parallelogram = 6 × 4 = 24 cm 2

Area of parallelogram = Area of rhombus

24 = 1/2 × 8 × d 2

24 = 4 × d 2

d 2 = 24/4 = 6

So, length of other diagonal of rhombus is 6 cm

12. The floor of a building consists of 3000 tiles which are rhombus shaped and each of its diagonals are 45 cm and 30 cm in length. Find the total cost of polishing the floor, if the cost per m 2 is Rs. 4.

Area of rhombus = 1/2 × 45 × 30

Area of rhombus = 1350/2 = 675

So, Area of rhombus = 675 cm 2

∴ Area of one tile = 675 cm 2

Now, Area of 3000 tiles = 675× 3000 = 2025000 cm 2

Area of tiles in m 2 = 2025000/10000 = 202.5 m 2

Total cost for polishing the floor = 202.5× 4 = Rs 810

13. A rectangular grassy plot is 112 m long and 78 m broad. It has gravel path 2.5 m wide all around it on the side. Find the area of the path and the cost of constructing it at Rs. 4.50 per square metre.

Outer area of rectangle = length × breadth

Outer area of rectangle = 112 × 78 = 8736 m 2

Width of path = 2.5 m

Length of inner rectangle = 112 – (2.5 + 2.5) = 107 m

Breadth of inner rectangle = 78 – (2.5 + 2.5) = 73 m

Inner area of rectangle = length × breadth

Inner area of rectangle = 107 × 73 = 7811 m 2

Now let’s calculate Area of path,

Area of path = Outer area of rectangle – Inner area of rectangle

Area of path = 8736 – 7811 = 925 m 2

Also given that,

Cost of construction for 1 m 2 = Rs 4.50

∴ Cost of construction for 925 m 2 = 925 × 4.50 = Rs 4162.5

14. Find the area of a rhombus, each side of which measures 20 cm and one of whose diagonals is 24 cm.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 6

Length of side of rhombus = 20 cm

Length of one diagonal = 24 cm

In ΔAOB,

Using Pythagoras theorem:

AB 2 = OA 2 + OB 2

20 2 = 12 2 + OB 2

OB 2 = 20 2 – 12 2

OB 2 = 400 – 144

OB 2 = 256

So, length of the other diameter = 16 × 2 = 32 cm

Area of rhombus = 1/2 × 24 × 32

Area of rhombus = 384 cm 2

15. The length of a side of a square field is 4 m. What will be the altitude of the rhombus, if the area of the rhombus is equal to the square field and one of its diagonal is 2 m?

Length of a side of a square = 4 m

Area of square = side 2

Area of square = 4 × 4 = 16 m 2

Area of square = Area of rhombus

So, Area of rhombus = 16 m 2

16 = 1/2 × 2 × d 2

16 = d 2

∴ the diagonal of rhombus = 16 m

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 7

AB 2 = 8 2 + 1 2

AB 2 = 65

AB = √65

Area of parallelogram = AB × DE

16 = √65 × DE

DE = 16/√65

i.e., Altitude of Rhombus = 16/√65 cm

16. Find the area of the field in the form of a rhombus, if the length of each side be 14 cm and the altitude be 16 cm.

Side of rhombus = 14 cm

Altitude of rhombus = 16 cm

Since rhombus is a parallelogram, therefore

Area of parallelogram = base × altitude

Area of parallelogram = 14 × 16 = 224 cm 2

17. The cost of fencing a square field at 60 paise per metre is Rs. 1200. Find the cost of reaping the field at the rate of 50 paise per 100 sq. metres.

Perimeter of square field = Cost of fencing / rate of fencing

Perimeter of square field = 1200/0.6 = 2000

So, Perimeter of square field = 2000 m

Perimeter of square = 4 × side

Side of square = Perimeter / 4 = 2000/4 = 500

So, Side of square = 500 m

We know that, Area of square = side 2

Area of square = 500 × 500 = 250000 m 2

Cost of reaping = (250000 × 0.5) / 100 = 1250

∴ Cost of reaping the field is Rs 1250

18. In exchange of a square plot one of whose sides is 84 m, a man wants to buy a rectangular plot 144 m long and of the same area as of the square plot. Find the width of the rectangular plot.

Area of square = 84 × 84 = 7056

Since, Area of square = Area of rectangle

7056 = 144 × width

Width = 7056/144 = 49

∴ Width of rectangle = 49 m

19. The area of a rhombus is 84 m 2 . If its perimeter is 40 m, then find its altitude.

Area of rhombus = 84 m 2

Perimeter = 40 m

Perimeter of rhombus = 4 × side

∴ Side of rhombus = Perimeter / 4 = 40/4 = 10

So, Side of rhombus = 10 m

Since rhombus is a parallelogram, therefore Area of parallelogram = base × altitude

84 = 10 × altitude

Altitude = 84/10 = 8.4

So, Altitude of rhombus = 8.4 m

20. A garden is in the form of a rhombus whose side is 30 metres and the corresponding altitude is 16 m. Find the cost of levelling the garden at the rate of Rs. 2 per m 2 .

Side of rhombus = 30 m

Altitude of rhombus = 16 m

Area of parallelogram = 30 × 16 = 480 m 2

Cost of levelling the garden = area × rate

Cost of levelling the garden = 480 × 2 = 960

So, Cost of levelling the garden is Rs 960

21. A field in the form of a rhombus has each side of length 64 m and altitude 16 m. What is the side of a square field which has the same area as that of a rhombus?

Side of rhombus = 64 m

Area of parallelogram = 64 × 16 = 1024 m 2

Since Area of rhombus = Area of square

Therefore, Area of square = side 2

Or side 2 = Area of square

Side of a square = √square

Side of square = √1024 = 32

∴ Side of square = 32 m

22. The area of a rhombus is equal to the area of a triangle whose base and the corresponding altitude are 24.8 cm and 16.5 cm respectively. If one of the diagonals of the rhombus is 22 cm, find the length of the other diagonal.

Length of base of triangle = 24.8 cm

Length of altitude of triangle= 16.5 cm

∴ Area of triangle = 1/2 × base × altitude

Area of triangle = 1/2 × 24.8 × 16.5 = 204.6

So, Area of triangle = 204.6 cm

Since, Area of triangle = Area of rhombus

∴ Area of rhombus = 1/2 × d 1 × d 2

204.6 = 1/2 × 22 × d 2

204.6 = 11 × d 2

d 2 = 204.6/11 = 18.6

∴ The length of other diagonal is 18.6 cm

EXERCISE 20.2 PAGE NO: 20.22

1. Find the area, in square metres, of the trapezium whose bases and altitudes are as under: (i) bases = 12 dm and 20 dm, altitude = 10 dm (ii) bases = 28 cm and 3 dm, altitude = 25 cm (iii) bases = 8 m and 60 dm, altitude = 40 dm (iv) bases = 150 cm and 30 dm, altitude = 9 dm

(i) Given that,

Length of bases of trapezium = 12 dm and 20 dm

Length of altitude = 10 dm

We know that, 10 dm = 1 m

∴ Length of bases in m = 1.2 m and 2 m

Similarly, length of altitude in m = 1 m

Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude

Area of trapezium = 1/2 (1.2 + 2.0) × 1

Area of trapezium = 1/2 × 3.2 = 1.6

So, Area of trapezium = 1.6m 2

(ii) Given that,

Length of bases of trapezium = 28 cm and 3 dm

Length of altitude = 25 cm

∴ Length of bases in m = 0.28 m and 0.3 m

Similarly, length of altitude in m = 0.25 m

Area of trapezium = 1/2 (0.28 + 0.3) × 0.25

Area of trapezium = 1/2 × 0.58× 0.25 = 0.0725

So, Area of trapezium = 0.0725m 2

(iii) Given that,

Length of bases of trapezium = 8 m and 60 dm

Length of altitude = 40 dm

∴ Length of bases in m = 8 m and 6 m

Similarly, length of altitude in m = 4 m

Area of trapezium = 1/2 (8 + 6) × 4

Area of trapezium = 1/2 × 56 = 28

So, Area of trapezium = 28m 2

(iv) Given that,

Length of bases of trapezium = 150 cm and 30 dm

Length of altitude = 9 dm

∴ Length of bases in m = 1.5 m and 3 m

Similarly, length of altitude in m = 0.9 m

Area of trapezium = 1/2 (1.5 + 3) × 0.9

Area of trapezium = 1/2 × 4.5 × 0.9 = 2.025

So, Area of trapezium = 2.025m 2

2. Find the area of trapezium with base 15 cm and height 8 cm, if the side parallel to the given base is 9 cm long.

Length of bases of trapezium = 15 cm and 9 cm

Length of altitude = 8 cm

Area of trapezium = 1/2 (15 + 9) × 8

Area of trapezium = 1/2 × 192 = 96

So, Area of trapezium = 96m 2

3. Find the area of a trapezium whose parallel sides are of length 16 dm and 22 dm and whose height is 12 dm.

Length of bases of trapezium = 16 dm and 22 dm

Length of altitude = 12 dm

∴ Length of bases in m = 1.6 m and 2.2 m

Similarly, length of altitude in m = 1.2 m

Area of trapezium = 1/2 (1.6 + 2.2) × 1.2

Area of trapezium = 1/2 × 3.8 × 1.2 = 2.28

So, Area of trapezium = 2.28m 2

4. Find the height of a trapezium, the sum of the lengths of whose bases (parallel sides) is 60 cm and whose area is 600 cm 2 .

Length of bases of trapezium = 60 cm

Area = 600 cm 2

600 = 1/2 (60) × altitude

600 = 30 × altitude

Which implies, altitude = 600/30 = 20

∴ Length of altitude is 20 cm

5. Find the altitude of a trapezium whose area is 65 cm 2 and whose base are 13 cm and 26 cm.

Length of bases of trapezium = 13 cm and 26 cm

Area = 65 cm 2

65 = 1/2 (13 + 26) × altitude

65 = 39/2 × altitude

Which implies, altitude = (65×2) /39 = 130/39 = 10/3

∴ Length of altitude = 10/3 cm

6. Find the sum of the lengths of the bases of a trapezium whose area is 4.2 m 2 and whose height is 280 cm.

Height of trapezium = 280 cm = 2.8m

Area = 4.2 m 2

Area of trapezium = 1/2 (Sum of lengths of parallel sides) × altitude To calculate the length of parallel sides we can rewrite the above equation as,

Sum of lengths of parallel sides = (2 × Area) / altitude

Sum of lengths of parallel sides = (2 × 4.2) / 2.8 = 8.4/2.8 = 3

∴ Sum of lengths of parallel sides = 3 m

7. Find the area of a trapezium whose parallel sides of lengths 10 cm and 15 cm are at a distance of 6 cm from each other. Calculate this area as, (i) the sum of the areas of two triangles and one rectangle. (ii) the difference of the area of a rectangle and the sum of the areas of two triangles.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 8

We know that, Area of a trapezium ABCD

= area (∆DFA) + area (rectangle DFEC) + area (∆CEB)

= (1/2 × AF × DF) + (FE × DF) + (1/2 × EB × CE)

= (1/2 × AF × h) + (FE × h) + (1/2 × EB × h)

= 1/2 × h × (AF + 2FE + EB)

= 1/2 × h × (AF + FE + EB + FE)

= 1/2 × h × (AB + FE)

= 1/2 × h × (AB + CD) [Opposite sides of rectangle are equal]

= 1/2 × 6 × (15 + 10)

= 1/2 × 6 × 25 = 75

∴ Area of trapezium = 75 cm 2

8. The area of a trapezium is 960 cm 2 . If the parallel sides are 34 cm and 46 cm, find the distance between them.

Area of trapezium = 1/2 (Sum of lengths of parallel sides) × distance between parallel sides

i.e., Area of trapezium = 1/2 (Sum of sides) × distance between parallel sides

To calculate the distance between parallel sides we can rewrite the above equation as,

Distance between parallel sides = (2 × Area) / Sum of sides

= (2 × 960) / (34 + 46)

= (2 × 960) / 80 = 1920/80 = 24

∴ Distance between parallel sides = 24 cm

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 9

From the figure we can write,

Area of figure = Area of two trapeziums + Area of rectangle

Length of rectangle = 50 cm

Breadth of rectangle = 10 cm

Length of parallel sides of trapezium = 30 cm and 10 cm

Distance between parallel sides of trapezium = (70–50)/2 = 20/2 = 10

So, Distance between parallel sides of trapezium = 10 cm

Area of figure = 2 × 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth

Area of figure = 2 × 1/2 (30+10) × 10 + 50 × 10

Area of figure = 40 × 10 + 50 × 10

Area of figure = 400 + 500 = 900

∴ Area of figure = 900 cm 2

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 10

Length of parallel sides of trapezium = 1.2m and 1m

Distance between parallel sides of trapezium = 0.8m

Area of trapezium = 1/2 (1.2 + 1) × 0.8

Area of trapezium = 1/2 × 2.2 × 0.8 = 0.88

So, Area of trapezium = 0.88m 2

11. The cross-section of a canal is a trapezium in shape. If the canal is 10 m wide at the top 6 m wide at the bottom and the area of cross-section is 72 m 2 determine its depth.

Length of parallel sides of trapezium = 10m and 6m

Area = 72 m 2

Let the distance between parallel sides of trapezium = x meter

72 = 1/2 (10 + 6) × x

72 = 8 × x

x = 72/8 = 9

∴ The depth is 9m.

12. The area of a trapezium is 91 cm 2 and its height is 7 cm. If one of the parallel sides is longer than the other by 8 cm, find the two parallel sides.

Let the length of one parallel side of trapezium = x meter

Length of other parallel side of trapezium = (x+8) meter

Area of trapezium = 91 cm 2

Height = 7 cm

91 = 1/2 (x+x+8) × 7

91 = 1/2(2x+8) × 7

91 = (x+4) × 7

(x+4) = 91/7

x = 13 – 4

∴ Length of one parallel side of trapezium = 9 cm

And, Length of other parallel side of trapezium = x+8 = 9+8 = 17 cm

13. The area of a trapezium is 384 cm 2 . Its parallel sides are in the ratio 3:5 and the perpendicular distance between them is 12 cm. Find the length of each one of the parallel sides.

Let the length of one parallel side of trapezium = 3x meter

Length of other parallel side of trapezium = 5x meter

Area of trapezium = 384 cm 2

Distance between parallel sides = 12 cm

384 = 1/2 (3x + 5x) × 12

384 = 1/2 (8x) × 12

4x = 384/12

∴ Length of one parallel side of trapezium = 3x = 3× 8 = 24 cm

And, Length of other parallel side of trapezium = 5x = 5× 8 = 40 cm

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 11

Let the length of side of trapezium shaped field along road = x meter

Length of other side of trapezium shaped field along road = 2x meter

Area of trapezium = 10500 cm 2

Distance between parallel sides = 100 m

10500 = 1/2 (x + 2x) × 100

10500 = 1/2 (3x) × 100

3x = 10500/50

x = 210/3 = 70

∴ Length of side of trapezium shaped field along road = 70 m

And, Length of other side of trapezium shaped field along road = 2x = 70× 2 = 140 m

15. The area of a trapezium is 1586 cm 2 and the distance between the parallel sides is 26 cm. If one of the parallel sides is 38 cm, find the other.

Let the length of other parallel side of trapezium = x cm

Length of one parallel side of trapezium = 38 cm

Area of trapezium = 1586 cm 2

Distance between parallel sides = 26 cm

1586 = 1/2 (x + 38) × 26

1586 = (x + 38) × 13

(x + 38) = 1586/13

x = 122 – 38

∴ Length of the other parallel side of trapezium = 84 cm

16. The parallel sides of a trapezium are 25 cm and 13 cm; its nonparallel sides are equal, each being 10 cm, find the area of the trapezium.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 12

In ΔCEF,

CE = 10 cm and EF = 6cm

CE 2 = CF 2 + EF 2

CF 2 = CE 2 – EF 2

CF 2 = 10 2 – 6 2

CF 2 = 100-36

CF 2 = 64

Area of trapezium = Area of parallelogram AECD + Area of area of triangle CEF

Area of trapezium = base × height + 1/2 (base × height)

Area of trapezium = 13 × 8 + 1/2 (12 × 8)

Area of trapezium = 104 + 48 = 152

∴ Area of trapezium = 152 cm 2

17. Find the area of a trapezium whose parallel sides are 25 cm, 13 cm and the other sides are 15 cm each.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 13

CF 2 = 15 2 – 6 2

CF 2 = 225-36

CF 2 = 189

CF = √189

= √ (9×21)

= 3√21 cm

Area of trapezium = height + 1/2 (sum of parallel sides)

Area of trapezium = 3√21 × 1/2 (25 + 13)

Area of trapezium = 3√21 × 19 = 57√21

∴ Area of trapezium = 57√21 cm 2

18. If the area of a trapezium is 28 cm 2 and one of its parallel sides is 6 cm, find the other parallel side if its altitude is 4 cm.

Length of one parallel side of trapezium = 6 cm

Area of trapezium = 28 cm 2

Length of altitude of trapezium = 4 cm

28 = 1/2 (6 + x) × 4

28 = (6 + x) × 2

(6 + x) = 28/2

(6 + x) = 14

x = 14 – 6

∴ Length of the other parallel side of trapezium = 8 cm

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 14

Area of parallelogram = 80 cm 2

Area of trapezium = 10 × 8 + 1/2 (12 × 8)

Area of trapezium = 80 + 48 = 128

∴ Area of trapezium = 128 cm 2

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 16

Area of given figure = Area of square ABCD + Area of rectangle DEFG + Area of rectangle GHIJ + Area of triangle FHI

i.e., Area of given figure = side × side + length × breadth + length × breadth + 1/2 × base × altitude

Area of given figure = 4×4 + 8×4 + 3×4 + 1/2×5×5

Area of given figure = 16 + 32 + 12 + 10 = 70

∴ Area of given figure = 70 cm 2

EXERCISE 20.3 PAGE NO: 20.28

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 18

GH = AG – AH = 8 – 6 = 2 cm

HF = AH – AF = 6 – 5 = 1 cm

GD = AD – AG = 10 – 8 = 2 cm

Area of given figure = Area of triangle AFB + Area of trapezium BCGF + Area of triangle CGD + Area of triangle AHE + Area of triangle EGD

Area of right angled triangle = 1/2 × base × altitude

Area of given pentagon = 1/2 × AF × BF + 1/2 (CG + BF) × FG + 1/2 × GD × CG + 1/2 × AH × EH + 1/2 × HD × EH

Area of given pentagon = 1/2 × 5 × 5 + 1/2 (7 + 5) × 3 + 1/2 × 2 × 7 + 1/2 × 6 × 3 + 1/2 × 4 × 3

Area of given pentagon = 12.5 + 18 + 7 + 9 + 6 = 52.5

∴ Area of given pentagon = 52.5 cm 2

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 19

Area of figure = Area of trapezium + Area of rectangle

Area of figure = 1/2 (Sum of lengths of parallel sides) × altitude + Length × Breadth

Area of figure = 1/2 (18 + 7) × 8 + 18 × 18

Area of figure = 1/2 (25) × 8 + 18 × 18

https://gradeup-question-images.grdp.co/liveData/PROJ6494/1500879336795394.png

∴ Area of figure is 424 cm 2

Figure (ii)

Area of given figure = 1/2 (15 + 6) × 8 + 15 × 20

Area of given figure = 84 + 300 = 384

∴ Area of figure is 384 cm 2

Figure (iii)

Using Pythagoras theorem in the right angled triangle,

5 2 = 4 2 + x 2

x 2 = 25 – 16

x 2 = 9

Area of given figure = 1/2 (14 + 6) × 3 + 4 × 6

Area of given figure = 30 + 24 = 54

∴ Area of figure is 54 cm 2

3. There is a pentagonal shaped park as shown in Fig. 20.50. Jyoti and Kavita divided it in two different ways. Find the area of this park using both ways. Can you suggest some another way of finding its area?

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 20

Area of Jyoti’s diagram = 2 × 1/2 (Sum of lengths of parallel sides) × altitude

Area of figure = 2 × 1/2 × (15 + 30) × 7.5

Area of figure = 45 × 7.5 = 337.5

Therefore, Area of figure = 337.5 cm 2

We also know that,

Area of Pentagon = Area of triangle + area of rectangle

Area of Pentagon = 1/2 × Base × Altitude + Length × Breadth

Area of Pentagon = 1/2 × 15 × 15 + 15 × 15

Area of Pentagon = 112.5 + 225 = 337.5

∴ Area of pentagon is 337.5 m 2

4. Find the area of the following polygon, if AL = 10 cm, AM = 20 cm, AN = 50 cm. AO = 60 cm and AD = 90 cm.

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 21

AL = 10 cm; AM = 20 cm; AN = 50 cm; AO = 60 cm; AD = 90 cm

LM = AM – AL = 20 – 10 = 10 cm

MN = AN – AM = 50 – 20 = 30 cm

OD = AD – AO = 90 – 60 = 30 cm

ON = AO – AN = 60 – 50 = 10 cm

DN = OD + ON = 30 + 10 = 40 cm

OM = MN + ON = 30 + 10 = 40 cm

LN = LM + MN = 10 + 30 = 40 cm

Area of figure = Area of triangle AMF + Area of trapezium FMNE + Area of triangle END + Area of triangle ALB + Area of trapezium LBCN + Area of triangle DNC

Area of given hexagon = 1/2 × AM × FM + 1/2 (MF + OE) × OM + 1/2 × OD × OE + 1/2 × AL × BL + 1/2 × (BL + CN) × LN + 1/2 × DN × CN

Area of given hexagon = 1/2 × 20 × 20 + 1/2 (20 + 60) × 40 + 1/2 × 30 × 60 + 1/2 × 10 × 30 + 1/2 × (30 + 40) × 40 + 1/2 × 40 × 40

Area of given hexagon = 200 + 1600 + 900 + 150 + 1400 + 800 = 5050

∴ Area of given hexagon is 5050 cm 2

RD Sharma Solutions for Class 8 Maths Chapter 20 – Mensuration – I (Area of a Trapezium and a Polygon) image - 22

NA = BQ = 10/2 = 5 cm

MR = OP = 13 cm

In the right triangle BPQ

PQ 2 = BQ 2 + BP 2

Substituting the values

(13) 2 = (5) 2 + BP 2

169 = 25 + BP 2

BP 2 = 169 – 25 = 144

PR = MO = 2 × 12 = 24 cm

Area of rectangle RPOM = RP × PO = 24 × 13 = 321 cm 2

Area of triangle PRQ = 1/2 × PR × BQ

= 1/2 × 24 × 5

Area of triangle MON = 60 cm 2

Area of hexagon = 312 + 60 + 60 = 432 cm 2

∴ Area of given hexagon is 432 cm 2

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