## Mensuration Class 8 Extra Questions Maths Chapter 11

**Extra Questions for Class 8 Maths Chapter 11 Mensuration**

### Mensuration Class 8 Extra Questions Very Short Answer Type

Question 1.

Find the perimeter of the following figures:

Solution:

(i) Perimeter of the rectangle = 2(l + b) = 2(8 + 6) = 2 × 14 = 28 cm

(ii) Perimeter of the square = 4 × side = 4 × 6 = 24 cm

(iii) Perimeter of the circle = 2πr = 2 × \(\frac { 22 }{ 7 }\) × 7 = 44 cm.

Question 2.

The length and breadth of a rectangle are 10 cm and 8 cm respectively. Find its perimeter if the length and breadth are (i) doubled (ii) halved.

Solution:

Length of the rectangle = 10 cm

Breadth of the rectangle = 8 cm

(i) When they are doubled,

l = 10 × 2 = 20 cm

and b = 8 × 2 = 16 cm

Perimeter = 2(l + b) = 2(20 + 16) = 2 × 36 = 72 cm

(ii) When they are halved,

l = \(\frac { 10 }{ 2 }\) = 5 cm

b = \(\frac { 8 }{ 2 }\) = 4 cm

Perimeter = 2(l + b) = 2(5 + 4) = 2 × 9 = 18 cm

Question 3.

A copper wire of length 44 cm is to be bent into a square and a circle. Which will have a larger area?

Solution:

(i) When the wire is bent into a square.

Side = \(\frac { 44 }{ 4 }\) = 11 cm

Area of the square = (side)^{2} = (11)^{2} = 121 cm^{2}

(ii) When the wire is bent into a circle.

Circumference = 2πr

44 = 2πr

So, the circle will have a larger area.

Question 4.

The length and breadth of a rectangle are in the ratio 4 : 3. If its perimeter is 154 cm, find its length and breadth.

Solution:

Let the length of the rectangle be 4x cm and that of breadth = 3x cm

Perimeter = 2(l + b) = 2(4x + 3x) = 2 × 7x = 14x cm

14x = 154

x = 11

Length = 4 × 11 = 44 cm

and breadth = 3 × 11 = 33 cm

Question 5.

The area of a rectangle is 544 cm^{2}. If its length is 32 cm, find its breadth.

Solution:

Area = 544 cm^{2}

Length = 32 cm

Breadth of the rectangle = \(\frac { Area }{ Length }\)

= \(\frac { 544 }{ 32 }\)

= 17 cm

Hence, the required breadth = 17 cm

Question 6.

If the side of a square is doubled then how much time its area becomes?

Solution:

Let the side of the square be x cm.

Area = (side)^{2} = x^{2} sq. cm

If its side becomes 2x cm then area = (2x)^{2} = 4x^{2} sq. cm

Ratio is x^{2} : 4x^{2} = 1 : 4

Hence, the area would become four times.

Question 7.

The areas of a rectangle and a square are equal. If the length of the rectangle is 16 cm and breadth is 9 cm, find the side of the square.

Solution:

Area of the square = Area of the rectangle = 16 × 9 = 144 cm^{2}

Side of the square = √Area of the square = √144 = 12 cm

Hence, the side of square = 12 cm.

Question 8.

If the lengths of the diagonals of a rhombus are 16 cm and 12 cm, find its area.

Solution:

Given:

First diagonal d_{1} = 16 cm

Second diagonal d_{2}= 12 cm

Hence, the required area = 96 cm^{2}.

Question 9.

The area of a rhombus is 16 cm^{2}. If the length of one diagonal is 4 cm, find the length of the other diagonal.

Solution:

Given: Area of the rhombus = 16 cm^{2}

Length of one diagonal = 4 cm

Hence, the required length = 8 cm.

Question 10.

If the diagonals of a rhombus are 12 cm and 5 cm, find the perimeter of the rhombus.

Solution:

Given: d_{1} = 12 cm, d_{2} = 5 cm

The perimeter = 4 × side = 4 × 6.5 = 26 cm

Hence, the perimeter = 26 cm.

### Mensuration Class 8 Extra Questions Short Answer Type

Question 11.

The volume of a box is 13400 cm^{3}. The area of its base is 670 cm^{2}. Find the height of the box.

Solution:

Volume of the box = 13400 cm^{3}

Area of the box = 670 cm^{2}

Hence, the required height = 20 cm.

Question 12.

Complete the following table; measurement in centimetres.

Solution:

Question 13.

Two cubes are joined end to end. Find the volume of the resulting cuboid, if each side of the cubes is 6 cm.

Solution:

Length of the resulting cuboid = 6 + 6 = 12 cm

Breadth = 6 cm

Height = 6 cm

Volume of the cuboid = l × b × h = 12 × 6 × 6 = 432 cm^{3}

Question 14.

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?

Solution:

Converting into same units, we have,

Length of the wall = 32 m = 32 × 100 = 3200 cm

Breadth of the wall = 3 m = 3 × 100 = 300 cm

and the height = 40 cm

v, length of the brick = 25 cm

breadth = 15 cm

and height = 8 cm

Number of bricks required

Hence, the required number of bricks = 12800.

Question 15.

MNOPQR is a hexagon of side 6 cm each. Find the area of the given hexagon in two different methods.

Solution:

Method I: Divide the given hexagon into two similar trapezia by joining QN.

Area of the hexagon MNOPQR = 2 × area of trapezium MNQR

= 2 × \(\frac { 1 }{ 2 }\) (6 + 11) × 4

= 17 × 4

= 68 cm^{2}

Method II: The hexagon MNOPQR is divided into three parts, 2 similar triangles and 1 rectangle by joining MO, RP.

Question 16.

The area of a trapezium is 400 cm^{2}, the distance between the parallel sides is 16 cm. If one of the parallel sides is 20 cm, find the length of the other side.

Solution:

Given: Area of trapezium = 400 cm^{2}

Height = 16 cm

Hence, the required length = 30 cm.

Question 17.

Find the area of the hexagon ABCDEF given below. Given that: AD = 8 cm, AJ = 6 cm, AI – 5 cm, AH = 3 cm, AG = 2.5 cm and FG, BH, EI and CJ are perpendiculars on diagonal AD from the vertices F, B, E and C respectively.

Solution:

Given:

AD = 8 cm

FG = 3 cm

AJ = 6 cm

EI = 4 cm

AI = 5 cm

BH = 3 cm

AH = 3 cm

CJ = 2 cm

AG = 2.5 cm

Area of hexagon ABCDEF = Area of ΔAGF + Area of trapezium FGIE + Area of ΔEID + Area of ΔCJD + Area of trapezium HBCJ + Area of ΔAHB

= 3.75 cm^{2} + 8.75 cm^{2} + 6 cm^{2} + 2 cm^{2} + 7.5 cm^{2} + 4.5 cm^{2}

= 32.50 cm^{2}.

Question 18.

Three metal cubes of sides 6 cm, 8 cm and 10 cm are melted and recast into a big cube. Find its total surface area.

Solution:

Volume of the cube with side 6 cm = (side)^{3} = (6)^{3} = 216 cm^{3}

Volume of the cube with side 8 cm = (side)^{3} = (8)^{3} = 512 cm^{3}

Volume of the cube with side 10 cm = (side)^{3} = (10)^{3} = 1000 cm^{3}

Volume of the big cube = 216 cm^{3} + 512 cm^{3} + 1000 cm^{3} = 1728 cm^{3}

Side of the resulting cube = \(\sqrt [ 3 ]{ 1728 }\) = 12 cm

Total surface area = 6 (side)^{2} = 6(12)^{2} = 6 × 144 cm^{2} = 864 cm^{2}.

Question 19.

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2.

Solution:

Given: Diameter of the roller = 84 cm

Radius = \(\frac { 84 }{ 2 }\) = 42 cm

Height = 120 cm

Curved surface area of the roller = 2πrh

Area covered by the roller in one complete revolution = 3.168 m^{2}

Area covered in 500 complete revolutions = 500 × 3.168 = 1584 m^{2}

Hence, the required area = 1584 m^{2}.

Question 20.

A rectangular metal sheet of length 44 cm and breadth 11 cm is folded along its length to form a cylinder. Find its volume.

Solution:

Circumference of the base = 2πr

= 423.5 cm^{3}

Hence, the required volume = 423.5 cm^{3}.

Question 21.

160 m^{3} of water is to be used to irrigate a rectangular field whose area is 800 m^{2}. What will be the height of the water level in the field? (NCERT Exemplar)

Solution:

Volume of water = 160 m^{3}

Area of rectangular field = 800 m^{2}

Let h be the height of water level in the field.

Now, the volume of water = volume of cuboid formed on the field by water.

160 = Area of base × height = 800 × h

⇒ h = 0.2

So, required height = 0.2 m

Question 22.

Find the area of a rhombus whose one side measures 5 cm and one diagonal as 8 cm. (NCERT Exemplar)

Solution:

Let ABCD be the rhombus as shown below.

DO = OB = 4 cm, since diagonals of a rhombus are perpendicular bisectors of each other.

Therefore, using Pythagoras theorem in ΔAOB, AO^{2} + OB^{2} = AB^{2}

Question 23.

The parallel sides of a trapezium are 40 cm and 20 cm. If its non-parallel sides are both equal, each being 26 cm, find the area of the trapezium.

Solution:

Let ABCD be the trapezium such that

AB = 40 cm and CD 20 cm and AD = BC = 26 cm.

Now, draw CL || AD

Then ALCD is a parallelogram.

So AL = CD = 20 cm

and CL = AD = 26 cm.

In ΔCLB, we have CL = CB = 26 cm

Therefore, ΔCLB is an isosceles triangle.

Draw altitude CM of ΔCLB.

Since ΔCLB is an isosceles triangle. So, CM is also the median.

Then LM = MB = \(\frac { 1 }{ 2 }\) BL = \(\frac { 1 }{ 2 }\) × 20 cm = 10 cm

[as BL = AB – AL = (40 – 20) cm = 20 cm].

Applying Pythagoras theorem in ΔCLM, we have

CL^{2} = CM^{2} + LM^{2}

26^{2} = CM^{2 }+ 10^{2}

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

CM = √576 = 24 cm

Hence, the area of the trapezium = \(\frac { 1 }{ 2 }\) (sum of parallel sides) × height

= \(\frac { 1 }{ 2 }\) (20 + 40) × 24

= 30 × 24

= 720 cm^{2}.

Question 24.

Find the area of polygon ABCDEF, if AD = 18 cm, AQ = 14 cm, AP = 12 cm, AN = 8 cm, AM = 4 cm, and FM, EP, QC and BN are perpendiculars to diagonal AD. (NCERT Exemplar)

Solution:

In the figure

MP = AP – AM = (12 – 4) cm = 8 cm

PD = AD – AP = (18 – 12) cm = 6 cm

NQ = AQ – AN = (14 – 8) cm = 6 cm

QD = AD – AQ = (18 – 14) cm = 4 cm

Area of the polygon ABCDEF = area of ∆AFM + area of trapezium FMPE + area of ∆EPD + area of ∆ANB + area of trapezium NBCQ + area of ∆QCD.