# Mensuration Class-8 ML Aggarwal ICSE Maths Solutions

**Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions Chapter-18. We provide step by step Solutions of Exercise / lesson-18 **Mensuration** Class-8th ML Aggarwal ICSE Mathematics.

Our Solutions contain all type Questions with **Exe-18.1 , Exe-18.2 , Exe-18.3, Exe-18.4** , Objective Type Questions (including Mental Maths Multiple Choice Questions, Value Based Questions , HOTS ), and Check Your Progress to develop skill and confidence. Visit official Website CISCE for detail information about ICSE Board **Class-8** Mathematics.

**— : Select Topics : —**

**Objective Type Questions, **

**Multiple Choice Questions ,(MCQ)**

**Ex 18.1,** **Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions

Question 1.

The length and breadth of a rectangular field are in the ratio 9 : 5. If the area of the field is 14580 square metres, find the cost of surrounding the field with a fence at the rate of ₹3·25 per metre.

#### Answer

Let the length = 9x and the breadth = 5x

Area = l × b ⇒ 14580 = 9x × 5x

⇒ 45x^{2} = 14580

∴ x^{2} = = 324 ⇒ x =

⇒ x =

or x = 18

Length = 9 × 18 = 162 m

Breadth = 5 × 18 = 90 m

Perimeter = 2(l + b)

= 2 (162 + 90) = 2(252)

= 504 m.

∴ Cost for 504 m fencing the surrounding

at the rate of ₹3·25 per metre = ₹(504 × 3·25) = ₹1638

Question 2.

A rectangle is 16 m by 9 m. Find a side of the square whose area equals the area of the rectangle. By how much does the perimeter of the rectangle exceed the perimeter of the square?

#### Answer

Area of rectangle = (16 × 9) m^{2} = 144 m^{2}

Area of square = Area of rectangle (given)

∴ (side)^{2} = 144

Side = = 12 m

Perimeter of square = 4 × 12 = 48 m

Perimeter of rectangle = 2(l + b) = 2 (16 + 9) = 50 m

Difference in their perimeters = 50 – 48 = 2 m

Question 3.

Two adjacent sides of a parallelogram are 24 cm and 18 cm. If the distance between longer sides is 12 cm, find the distance between shorter sides.

#### Answer

Taking 24 cm as a base of parallelogram, its height is 12 cm.

∴ Area of parallelogram = b × h = 24 × 12 = 288 cm^{2}

Let d cm be the distance between the shortest sides.

∴ Area of parallelogram = (18 × d) cm^{2}

⇒ 18 × d = 288

⇒ d = = 16 cm

Question 4.

Rajesh has a square plot with the measurement as shown in the given figure. He 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 ₹50 per m^{2}.

#### Answer

Side of square plot = 24 m

Length of house (l) = 18 m

and breadth (b) = 12m

Now area of square plot = (24)^{2} m^{2} = 24 × 24 = 576 m^{2}

and area of hosue = 18 × 12 = 216 m^{2}

Remaining area of the garden = 576 – 216 = 360 m^{2}

Cost of developing the garden = ₹50 per m^{2}

Total cost = ₹50 × 360 = ₹18000

Question 5.

A flooring tile has a shape of a parallelogram whose base is 18 cm and the corresponding height is 6 cm. How many such tiles are required to cover a floor of area 540 m^{2}? (If required you can split the tiles in whatever way you want to fill up the comers).

#### Answer

Base of the parallelogram-shaped flooring tile = 18 cm

and height = 6 cm

∴ Area of one tile = Base × Height = 18 × 6 = 108 cm^{2}

Area of floor = 540 m^{2}

∴ Number of tiles =

= = 50000

Question 6.

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?

#### Answer

(a) Diameter of semicircle = 2.8 cm

∴ Perimeter = πr + 2r

= × 2.8 + 2 × 2.8

= 8.8 + 5.6 cm = 14.4 cm

(b) Total perimeters

= 1.5 + 1.5 + 2.8 + Semi circular

= 5.8 + 8.8 = 14.6 cm

(c) Total perimeter = 2 + 2 + Semi circumference

= 4 + 8.8 = 12.8 cm

It is clear that distance of (b) i.e. 14.6 is longer.

Question 7.

In the adjoining figure, the area enclosed between the concentric circles is 770 cm2. If the radius of the outer circle is 21 cm, calculate the radius of the inner circle.

#### Answer

Radius of outer circle (R) = 21 cm.

radius of inner circle (r) = r cm.

Area of shaded portion = 770 cm^{2}

⇒ π (R^{2} – r^{2}) = 770

⇒ (21^{2} – r^{2}) = 770

⇒ 441 – r^{2} = 770 × = 35 × 7 = 245

⇒ r^{2} = 441 – 245

⇒ r^{2} = 196

⇒ r^{2} = 196

⇒ r =

⇒ r = 14 cm

Question 8.

A copper wire when bent in the form of a square encloses an area of 121 cm2. If the same wire is bent into the form of a circle, find the area of the circle.

#### Answer

Area of the square = 121 cm^{2}

∴ Side = = 11 cm

Perimeter = 4 a = 4 × 11= 44 cm

Now, circumference of the circle = 44 cm

∴ Radius = = 7cm

and area of the circle = πr^{2} = (7)^{2}

= × 7 × 7 = 154 cm^{2}

Question 9.

From the given figure, find

(i) the area of ∆ ABC

(ii) length of BC

(iii) the length of altitude from A to BC

#### Answer

(i) Base = 3 cm, height = 4 cm.

Area = × base × height

= × 3 × 4 = 6 cm^{2}

(ii) By pythagoras theorem,

BC^{2} = AB^{2} + AC^{2}

∴ BC^{2} = (3)^{2} + (4)^{2}

= 9 + 16 = 25

⇒ BC = cm = 5 cm

(iii) Now, Base = BC = 5 cm., h = AD = ?

Area = × b × h

⇒ 6 = × 5 × h

[∵ Area = 6 cm^{2} as in part (i)]

⇒ h = = 2·4 cm.

Question 10.

A rectangular garden 80 m by 40 m is divided into four equal parts by two cross-paths 2.5 m wide. Find

(i) the area of the cross-paths.

(ii) the area of the unshaded portion.

#### Answer

Length of rectangular garden = 80 m

and breadth = 40 m

Width of crossing path 2.5 m

Area of length wise path

= 80 × 2.5 = 200 m^{2}

Area of breadth wise path

= 40 × 2.5 = 100 m^{2}

(i) Total area of both paths

= 200 + 100 – 2.5 × 2.5 m^{2}

= 300 – 6.25 = 293.75 m^{2}

(ii) Area of unshaded portion

= Area of garden – Area of paths

= 80 × 40 – 293.75 m^{2}

= 3200 – 293.75 m^{2}

= 2906.25 m^{2}

Question 11.

In the given figure, ABCD is a rectangle. Find the area of the shaded region.

#### Answer

In the given figure.

Length of rectangle = 18 cm

and breadth = 12 cm

∴ Area = l × b = 18 × 12 cm^{2} = 216 cm^{2}

Area of triangle I = × 12 × 10 = 60 cm^{2}

Area of triangle III = × 18 × 7 = 63 cm^{2}

∴ Area of shaded portion

= Area of rectangle – Area of 3 triangles

= 216 – (60 + 63 + 20)

= 216 – 143 cm2

= 73 cm^{2}

Question 12.

In the adjoining figure, ABCD is a square grassy lawn of area 729 m^{2}. A path of uniform width runs all around it. If the area of the path is 295 m^{2}, find

(i) the length of the boundary of the square field enclosing the lawn and the path.

(ii) the width of the path.

#### Answer

Area of square ABCD = 729 m^{2}

Side = = 27 m

Let the width of path = x m

Then side of outer field = 27 + x + x = (27 + 2x) m

Area of square PQRS = (27 + 2x)^{2} m^{2}

Area of PQRS – Area of ABCD = Area of path

∴ (27 + 2x)^{2} m^{2} – 729 m^{2} = 295 m^{2}

⇒ 729 + 4x^{2} + 108x – 729 = 295

⇒ 4x^{2} + 108x – 295 = 0

⇒

∴ Width of the path is 2.5 m

Now, side of square field PQRS

= 27 + 2x = (27 + 2 × 2·5) m = 32 m

Length of boundary = 4 × side = 32 × 4 = 128

**Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions **Ex 18.2**

Question 1.

Each sides of a rhombus is 13 cm and one diagonal is 10 cm. Find

(i) the length of its other diagonal

(ii) the area of the rhombus

#### Answer

(i) Side of rhombus = 13 cm.

Length of diagonal AC = 10 cm.

∴ OC = 5 cm.

Since the diagonals of rhombus are ± to each other

∴ ∆BOC is rt. angled.

Hence, BC^{2} = OC^{2} + OB^{2}

13^{2} = 5^{2} + OB^{2}

⇒ OB^{2} = 169 – 25 = 144

⇒ OB = = 12 cm

∴ Diagonal BD = 2 × OB = 2 × 12 = 24 cm

(ii) Area of rhombus = × d_{1} × d_{2}

= × 10 × 24 = 120 cm^{2}

Question 2.

The cross-section ABCD of a swimming pool is a trapezium. Its width AB = 14 m, depth at the shallow end is 1-5 m and at the deep end is 8 m. Find the area of the cross-section.

#### Answer

Here, two parallel sides of trapezium are AD and BC

and distance between them is 14 m.

∴ Area of trapezium = (1·5 + 8) × 14

= × 9·5 × 14 = 66 × 5 m^{2}

Question 3.

The area of a trapezium is 360 m^{2}, the distance between two parallel sides is 20 m and one of the parallel side is 25 m. Find the other parallel side.

#### Answer

Area of a trapezium = 360 m^{2}

Distance between two parallel lines = 20 m

One parallel side = 25 m

Let Second parallel side = 11 m

∴ Area = (25 + x) × 20

⇒ 360 = (25 + x) × 20

∴ x = 36 – 25 = 11 m

∴ Second parallel side = 11 m

Question 4.

Find the area of a rhombus whose side is 6.5 cm and altitude is 5 cm. If one of its diagonal is 13 cm long, find the length of other diagonal.

#### Answer

Side of rhombus = 6.5 cm

and altitude = 5 cm

Area of a rhombus = Side × Altitude = 6.5 × 5 = 32.5 cm^{2}

One diagonal = 13 cm

Length of other diagonal = = = 5 cm

Question 5.

From the given diagram, calculate

(i) the area of trapezium ACDE

(ii) the area of parallelogram ABDE

(iii) the area of triangle BCD.

#### Answer

(i) Area of trapezium ACDE

= (AC + DE) × h

= (13 + 7) × 6.5 = 65 m^{2}

(ii) Area of parallelogram ABDE = b × h = × 6 × 6·5 = 15·5 m^{2}

[∵ BC = AC – AB = 13 – 7 = 6 m]

Question 6.

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.

#### Answer

Base of triangle = 24.8 cm

and altitude = 16.5 cm

Area = base × altitude

= × 24.8 × 16.5 cm^{2} = 204.6 cm^{2}

Now, Area of ∆ = Area of rhombus

∴ Area of rhombus = 204.6 cm^{2}

Length of one diagonal = 22 cm

Area of rhombus = (First diagonal × Second diagonal)

∴ Second diagonal =

= = 18.6 cm

Question 7.

The perimeter of a trapezium is 52 cm. If its non-parallel sides are 10 cm each and its altitude is 8 cm, find the area of the trapezium.

#### Answer

Perimeter of a trapezium = 52 cm

Length of each non-parallel side = 10 cm

Altitude DL = 8 cm

In right ∆DAL (By Pythagoras Theorem)

DA^{2} = DL^{2} + AL^{2}

⇒ (10)^{2} = (8)^{2} + AL^{2}

⇒ 100 = 64 + AL^{2}

⇒ AL^{2} = 100 – 64 = 36 = (6)^{2}

∴ AL = 6 cm

Similarly BM = 6 cm

and DC = LM

Also, perimeter = AB + BC + CD + DA

and CD = DA

∴ CD + DA = 2DA

But AB + CD = Perimeter – 2 AD

= 52 – 2 × 10 = 52 – 20 = 32 cm

Now area of trapezium = (sum of parallel sides) × altitude

= × 32 × 8 = 128 cm^{2}

Question 8.

The area of a trapezium is 540 cm^{2}. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the lengths of parallel sides.

#### Answer

Let, the two parallel sides of trapezium are 7x and 5x.

Height = 18 cm

⇒ Area of trapezium = [(Sum of ||gm sides) × height]

540 = (7x + 5x) × 18

∴ 540 = × 12x × 18

or 108x = 540

⇒ x = 5 cm

Hence, two parallel sides are

7x = 7 × 5 = 35 cm

and 5x = 5 × 5 = 25 cm

Question 9.

Calculate the area enclosed by the given shapes. All measurements are in cm.

#### Answer

(i) Area of trapezium ABCD

= (Sum of opposite ||gm sides) × height

= [(AB + CD) × (AF + FD)]

= [(AB + CD) × (AF + FD)

= [(5 + 3) × (5 + 4)]

= (5 + 3) × 9 = 36 cm^{2}

Area of rectangle GAFE = Length × Breadth

= 2 × 5 = 10 cm^{2}

Total area of the figure

= Area of trapezium ABCD + Area of rectangle GAFE

= (36 + 10) cm^{2}

= 46 cm^{2}

(iii) Area of rectangle ABCD

Area of given figure = Area of rect. ABCD

+ Area of ||gm BIHJ + Area of rectangle EFGH.

= Length × Breadth

= AD × DC

= 9 × 2 = 18 cm^{2}

Area of rectangle EFGH = Length × Breadth

= (EJ + JH) × EF

= (7 + 2) × 2

= 9 × 2 = 18 cm^{2}

Area of parallelogram BIHJ = 2 × 5 = 10 cm^{2}

[∵ Distance between BI and HJ = 9 – 2 – 2 = 5 cm]

Total area of the figure = (18 + 18 + 10) cm^{2} = 46 cm^{2}

Question 10.

From the adjoining sketch, calculate

(i) the length AD

(ii) the area of trapezium ABCD

(iii) the area of triangle BCD

#### Answer

(i) In right angled ∆ ABD (By Pythagoras Theorem)

BD^{2} = AD^{2} + AB^{2}

⇒ AD^{2} = BD^{2} – AB^{2} = (41)^{2} – (40)^{2 }= 1681 – 1600 = 81

∴ AD = = 9 cm.

(ii) Area of trapezium ABCD

= (Sum of opposite ||gm lines) × height

= (AB + CD) × AD

= (40+ 15) × 9 = 247·5 cm^{2}

(iii) Area of triangle BCD = Area of trapezium ABCD – Area of ∆ ABD

= (247·5 – × 40 × 9 ) cm^{2}

= (247·5 – 180) cm^{2} = 67·5 cm^{2}

Question 11.

Diagram of the adjacent picture frame has outer dimensions = 28 cm × 32 cm and inner dimensions 20 cm × 24 cm. Find the area of each section of the frame, if the width of each section is same.

#### Answer

Outer length of the frame = 32 cm

and outer breadth = 28 cm

Inner length = 24 cm

and outer breadth = 20 cm

∴ Width of the frame = = 4 cm

i. e., height = 4 cm

Now area of each portion of length side

= (24 + 32) × 4

= × 56 × 4 = 112 cm^{2}

and area of each portion of breadth side

= (20 + 28) × 4

= × 48 × 4 = 96 cm^{2}

∴ Area each sections =112 cm^{2}, 96 cm^{2}, 112 cm^{2}, 96 cm^{2}

Question 12.

In the given quadrilateral ABCD, ∠BAD = 90° and ∠BDC = 90°. All measurements are in centimetres. Find the area of the quadrilateral ABCD.

#### Answer

In right angled triangle ABD, (By Pythagoras Theorem)

BD^{2} = AB^{2} + AD^{2 }= (6)^{2} + (8)^{2} = 36 + 64 = 100 cm^{2}

BD =

∴ BD = 10 cm.

Now, Area of A ABD = × b × h

= × 6 × 8 = 24cm^{2} …(i)

In ∆ BDC, BD = 10 cm.,

BC = 26 cm., DC = ?

By Pythagoras theorem,

BC^{2} = BD^{2} + DC^{2}

(26)^{2} = (10)^{2} + DC^{2}

676 – 100 = DC^{2}

⇒ DC = = 24 cm.

Now, Area of ∆ BDC = × b × h

= × 24 × 10 = 12 cm^{2} …(ii)

Add (i) and (ii), we get

Area of ∆ABD + Area of ∆BDC = (24 + 120) cm^{2}

Area of quadrilateral ABCD = 144 cm^{2}

Question 13.

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

#### Answer

Raised surface of platform is in the shape of regular octagon ABCDEFGH.

Each side = 8 cm, join HC

GD = HC = 15 cm, FL = AM = 6 cm

Now in each trapezium parallel sides are 15 cm and 6 cm

and height = 6 cm

∴ Area of each trapezium FEDG

= (GD + FE) × FL

= (15 + 8) × 6

= 23 × 3 cm^{2} = 69 cm^{2}

Area of trapezium FEDG = Area of trapezium ABCH = 69 cm^{2}

and area of rectangle HCDG

= HC × CD = 15 × 8 = 120 cm^{2}

Total area = Area of trapezium FEDG + Area of trapezium ABCH

+ Area of rectangle HCDG.

Total area = 69 + 69 + 120 = 258 cm^{2}

Question 14.

There is a pentagonal shaped park as shown in the following figure:

For finding its area Jaspreet and Rahul 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?

#### Answer

The pentagonal shaped park is shown in the given figure,

in which DL ⊥ CE and is produced to M.

∴. DM = 32 m

LM = CB = 18 m

∴ DL = 32 – 18 = 14 m

(i) According to Jaspreet’s the figure is divided into

two equal trapezium in area: DEAM and DCBM

Area of trapezium DEAM

= (AE + DM) × AM

= (32 + 18 ) × 9

= = 225m^{2}

According to Rahul’s the figure is divided into shapes

one square and on isoscles triangle.

Area of square ABCE = (Side)^{2 }= (18)^{2} = 324 m^{2}

and area of isosceles ∆EDC

= × EC × DC

= × 18 × 14 = 126 m^{2}

∴ Total area = 225 × 2 = 450 m^{2}

Third way to find out the area of given figure is as follow:

Here, DL ⊥ ED, DL = 14 m

Area of ∆DEC = × EC × LD

= × 18 × 14 = 126 m^{2}

Area of ∆AEB = × AB × AE

= × 18 × 18= 162 m^{2}

Area of ∆BEC = × BC × EC

= × 18 × 18 = 162 m^{2}

Now, area of pentagon ABCDE = Area ∆DEC

+ Area of ∆AEB + Area of ∆BEC

= (126 + 162 + 162) m^{2} = 450 m^{2}

Question 15.

In the diagram, ABCD is a rectangle of size 18 cm by 10 cm. In ∆ BEC, ∠E = 90° and EC = 8 cm. Find the area enclosed by the pentagon ABECD.

#### Answer

Area of rectangle ABCD = Length × Breadth

= 18 × 10 = 180 cm^{2}

In right angled ∆ BEC,

BC^{2} = CE^{2} + BE^{2} (By Pythagoras theorem)

(10)^{2} = 8^{2} + BE^{2}

∴ BE^{2} = 100 – 64 = 36

⇒ BE = ⇒ BE = 6 cm.

∴ Area of rt. ∆ BEC = × 6 × 8 = 24cm^{2}

Area of pentagon ABECD = Area of rectangle – area of ∆

= (180 – 24) cm^{2} = 156 cm^{2}

Question 16.

Polygon ABCDE is divided into parts as shown in the given figure. Find its area if AD = 8 cm, AH = 6 cm, AG = 4 cm, AF = 3 cm and perpendiculars BF = 2 cm, CH = 3 cm, EG = 2.5 cm.

#### Answer

In the given figure, ABCDE, AD = 8 cm, AH = 6 cm, AG = 4 cm,

AF = 3cm ⊥ BF = 2 cm CH = 3 cm and ⊥ EG = 2.5 cm

The given figure, consists of 3 triangles and one trapezium.

Now area of ∆AED = AD × GE

= × 8 × 2.5 = 10 cm^{2}

Area of ∆ABF = AF × BF

= × 3 × 2 = 3 cm^{2}

Area of ∆CDH = × HD × CH

= (AD – AH) × 3

= (8 – 6) × 3

= × 2 × 3 = 3 cm^{2}

Area of trapezium BFHC

= (BF + CH) × FH

= (2 + 3) × (AH – AF)

= × 5 × (6 – 3)

= × 5 × 3 = 7.5 cm^{2}

∴ Total area of the figure = Area of ∆AED + Area of ∆ABF

+ Area of ∆CDH + Area of trapezium BFHC

= 10 + 3 + 3 + 7.5 = 23.5 cm^{2}

Question 17.

Find the area of polygon PQRSTU shown in 1 the given figure, if PS = 11 cm, PY = 9 cm, PX = 8 cm, PW = 5 cm, PV = 3 cm, QV = 5 cm, UW = 4 cm, RX = 6 cm, TY = 2 cm.

#### Answer

In the figure, PQRSTU, in which PS = 11 cm, PY = 9 cm, PX = 8 cm,

PW = 5 cm, PV = 3 cm, QV = 5 cm, UW = 4 cm, RX = 6 cm, TY = 2 cm

The figure, consists of 4 triangle and 2 trapeziums

VX = PX – PV

= 8 – 3 = 5 cm

XS = PS – PX

= 11 – 8 = 3 cm

YS = PS – PY

= 11 – 9 = 2 cm

WY = PY – PW

= 9 – 5 = 4 cm

Now area ∆PQV = PV + QV

= × 3 × 5 = =7.5 cm^{2}

Area of ∆RXS = × S × R

= 3 × 6 = 9 cm^{2}

Area ∆PUW = × PW × UW

= × 5 × 4 = 10 cm^{2}

Area ∆YTS = × YS × TY

= × 2 × 2 = 2 cm^{2}

Area of trapezium ∆VX R

= (QV + RX) × VX

= (5 + 6) × 5 = × 11 × 5 cm^{2}

= = 27.5 cm^{2}

Area of trapezium WUTY

= (UW + TY) × WY

= (4 + 2) × 4 = × 6 × 4 = 12 cm^{2}

Now area of the figure = 7.5 + 9 + 10 + 2 + 27.5 + 12 cm^{2} = 68 cm^{2}

**ML Aggarwal Solutions Mensuration Class-8 **ICSE Mathematics **Ex 18.3**

Question 1.

The volume of a cube is 343 cm^{3}, find the length of an edge of cube.

#### Answer

Volume of a cube = 343 cm^{3}

Let a be the edge of cube, then

V = a^{3} = 343 = (7)^{3} .

∴ a = 7 cm

Question 2.

Fill in the following blanks:

Solution:

Question 3.

Find the height of a cuboid whose volume is 312 cm^{3} and base area is 26 cm^{2}.

#### Answer

Volume of a cuboid = 312 cm^{3}

Base area = l × b = 26 cm^{2}

∴ Height= = 12cm

Question 4.

A godown is in the form of a cuboid of measures 55 m × 45 m × 30 m. How many cuboidal boxes can be stored in it if the volume of one box is 1.25 m^{3}?

#### Answer

Length of a godown (l) = 55 m

Breadth (b) = 45 m

Height (h) = 30 m

Volume = lbh = 55 × 45 × 30 m^{3}

Volume of one box = 1.25 m^{3}

Number of box = = 59400 boxes

Question 5.

A rectangular pit 1.4 m long, 90 cm broad and 70 cm deep was dug and 1000 bricks of base 21 cm by 10.5 cm were made from the earth dug out. Find the height of each brick.

#### Answer

Here l = 1·4 m = 140 cm, b = 90 cm, h = 70 cm

Volume of rectangular pit = l × b × h

= (140 × 90 × 70) cm^{3} = 882000 cm^{3}

Volume of brick = 21 × 10.5 × h

Question 6.

If each edge of a cube is tripled, then find how many times will its volume become?

#### Answer

Let edge of a cube = x

Then volume = x^{3}

If the edge is trippled, then

Edge = 3x

Now, volume = (3x)^{3} = 27x^{3}

∴ Its volume is 27 times the volume of the given cube.

Question 7.

A milk tank is in the form of cylinder whose radius is 1.4 m and height is 8 m. Find the quantity of milk in litres that can be stored in the tank.

#### Answer

Radius of the milk cylindrical tank = 1.4 m and height (h) = 8 m

∴ Volume of milk in the tank = πr^{2}h

= × 1.4 × 1.4 × 8 m^{3}

= 49.28 m^{3}

= 49.28 × 1000 litres

= 49280 litres

Question 8.

A closed box is made of 2 cm thick wood with external dimension 84 cm × 75 cm × 64 cm. Find the volume of the wood required to make the box.

#### Answer

Thickness of the wood used in a closed box = 2 cm

External length of box (L) = 84 cm

Breadth (b) = 75 cm

and height (h) = 64 cm

∴ Internal length (l) = 84 – (2 × 2) = 84 – 4 = 80 cm

Breadth (b) = 75 – (2 × 2) = 75 – 4 = 71 cm

and height (h) = 64 – (2 × 2) = 64 – 4 = 60 cm

∴ Volume of wood used

= 84 × 75 × 64 – 80 × 71 × 60 cm^{3}

= 403200 – 340800 cm^{3}

= 62400 cm^{3}

Question 9.

Two cylindrical jars contain the same amount of milk. If their diameters are in the ratio 3 : 4, find the ratio of their heights.

#### Answer

Ratio in diameters of two cylindrical jars = 3 : 4

But their volume is same.

Let h_{1} and h_{2} be the heights of the two jars respectively.

Let radius of the first jar (r_{1}) =

and radius of the second jar (r_{2}) =

According to the condition,

∴ Ratio in their heights =16 : 9

Question 10.

The radius of the base of a right circular cylinder is halved and the height is doubled. What is the ratio of the volume of the new cylinder to that of the original cylinder?

#### Answer

Let radius of a cylinder = r

and height = h

Volume = πr^{2}h

Its radius is halved and height is doubled, then

∴ Ratio in the volumes of the new cylinder to old one

= = 1 : 2

Question 11.

A rectangular piece of tin of size 30 cm × 18 cm is rolled in two ways, once along its length (30 cm) and once along its breadth. Find the ratio of volumes of two cylinders so formed.

#### Answer

Size of rectangular tin plate = 30 cm × 18 cm

(i) When rolled along its length (30 cm),

then Circumference of the circle so formed = 30 cm

Circumference = 2πr

Question 12.

Water flows through a cylindrical pipe of internal diameter 7 cm at 5 m per sec. Calculate

(i) the volume in litres of water discharged by the pipe in one minute.

(ii) the time in minutes, the pipe would take to fill an empty rectangular tank of size 4 m × 3 m × 2.31 m.

#### Answer

Speed of water flow through cylindrical pipe = 5 m/sec.

Internal diameter of the pipe = 7 cm

∴ Radius (r) = cm

∴ Length of water flow in 1 minutes (h)

= 5 × 60 = 300 m

∴ Volume of water = πr^{2}h

= 1155 litres

Now volume of water = 1155000 cm^{3}

Volume of rectangular tank of size

= 4m × 3m × 2.31m

= 27.72 m^{3}

Speed of water 4 m/sec.

Radius of pipe = cm

Volume of water in 1 sec

Question 13.

Two cylindrical vessels are filled with milk. The radius of one vessel is 15 cm and height is 40 cm, and the radius of other vessel is 20 cm and height is 45 cm. Find the radius of another cylindrical vessel of height 30 cm which may just contain the milk which is in the two given vessels.

#### Answer

Radius of one cylinder (r_{1}) = 15 cm

and height (h_{1}) = 40 cm

and radius of second cylinder (r_{2}) = 20 cm

and height (h_{2}) = 45 cm

∴ Radius of the third cylinder = 30 cm

Question 14.

A wooden pole is 7 m high and 20 cm in diameter. Find its weight if the wood weighs 225 kg per m^{3} .

#### Answer

Height of pole (h) = 7 m

Diameter = 20 cm

Weight of wood = 225 kg per m^{3}

Total weight = 225 × = 49.5 kg

Question 15.

A cylinder of maximum volume is cut from a wooden cuboid of length 30 cm and cross-section a square of side 14 cm. Find the volume of the cylinder and the volume of the wood wasted.

#### Answer

A cylinder of the maximum volume is cut from a wooden cuboid

of length 30 cm and cross-section a square side 14 cm.

∴ Diameter of the cylinder = 14 cm

∴ Radius (r) = = 7 cm

and height (h) = 30 cm

Volume of cuboid = 30 × 14 × 14 = 5880 cm^{3}

Volume of cylinder = πr^{2}h

= × 7 × 7 × 30 = 4620 cm^{3}

and wastage of wood = 5880 – 4620 = 1260 cm^{3}

**Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions **Exe-18.4**

Question 1.

The surface area of a cube is 384 cm^{2}. Find

(i) the length of an edge

(ii) volume of the cube.

#### Answer

Surface area of a cube = 384 cm^{2}

(ii) Volume = (Edge)^{3} = (8)^{3 }= 8 × 8 × 8 cm^{3} = 512 cm^{3}

Question 2.

Find the total surface area of a solid cylinder of radius 5 cm and height 10 cm. Leave your answer in terms of n.

#### Answer

Radius of a solid cylinder (r) = 5 cm

Height (h) = 10 cm

Total surface area = 2πrh + 2πr^{2}

= 2rπ(h + r)

= 2π × 5(10 + 5)

= π × 10 × 15 = 150π cm^{2}

Question 3.

An aquarium is in the form of a cuboid whose external measures are 70 cm × 28 cm × 35 cm. The base, side faces and back face are to be covered with coloured paper. Find the area of the paper needed.

#### Answer

A cuboid shaped aquarium,

Length (l) = 70 cm

Breadth (b) = 28 cm

and height (h) = 35 cm

Area of base = 70 × 28 cm^{3} = 1960 cm^{3}

Area of side face = (28 × 35) × 2 cm^{2 }= 1960 cm^{2}

Area of back face = 70 × 35 cm^{2} = 2450 cm

∴. Total area = 1960 + 1960 + 2450 = 6370 cm^{2}

∴ Area of paper required = 6370 cm^{2}

Question 4.

The internal dimensions of rectangular hall are 15 m × 12 m × 4 m. There are 4 windows each of dimension 2 m × 1.5 m and 2 doors each of dimension 1.5 m × 2.5 m. Find the cost of white washing all four walls of the hall, if the cost of white washing is ₹5 per m^{2}. What will be the cost of white washing if the ceiling of the hall is also white washed?

#### Answer

Internal dimension of rectangular hall = 15m × 12 m × 4 m

Area of 4-walls = 2(l + b) × h

= 2(15 + 12) × 4

= 2 × 27 × 4 m^{2}

= 216 m^{2}

Area of 4 windows of size = 2 × 1.5 = 2 × 1.5 × 4 = 12 m^{2}

Area of 2 door of size = 1.5 × 2.5 = 2 × 1.5 × 2.5 = 7.5 m^{2}

∴ Area of remaining hall = 216 – (12 + 7.5) = 216 – 19.5 m^{2} = 196.5 m^{2}

Area of ceiling = l × b = 15 × 12= 180 m^{2}

Cost of white washing the walls at the rate of ₹5 per m^{2
}= 196.5 × 5 = ₹982.50

Area of ceiling = l × b = 15 × 12= 180 m^{2}

Cost of white washing = 180 × 5 = ₹900

∴ Total cost = ₹982.50 + 900.00 = ₹1882.50

Question 5.

A swimming pool is 50 m in length, 30 m in breadth and 2·5 m in depth. Find the cost of cementing its floor and walls at the rate of ₹27 per square metre.

#### Answer

Length of swimming pool = 50 m

Breadth of swimming pool = 30 m

Depth (Height) of swimming pool = 2·5 m

Area of floor = 50 × 30 = 1500 m^{2}

Area of four walls = 2 (50 + 30) × 2·5 = 160 × 2·5 = 400 m^{2}

Area to be cemented = 1500 m^{2} + 400 m^{2} = 1900 m^{2}

Cost of cementing 1m2 = ₹27

Cost of cementing 1900m^{2 }= ₹27 × 1900 = ₹51300

Question 6.

The floor of a rectangular hall has a perimeter 236 m. Its height is 4·5 m. Find the cost of painting its four walls (doors and windows be ignored) at the rate of Rs. 8.40 per square metre.

#### Answer

Perimeter of Hall = 236 m.

Height = 4·5 m

Perimeter = 2 (l + b) = 236 m

Area of four walls = 2 (l + b) × h = 236 × 4·5 = 1062 m^{2}

Cost of painting 1 m^{2} = ₹8·40

Cost of painting 1062 m^{2} = ₹8·40 × 1062 = ₹8920·80

Question 7.

A cuboidal fish tank has a length of 30 cm, a breadth of 20 cm and a height of 20 cm. The tank is placed on a horizontal table and it is three-quarters full of water. Find the area of the tank which is in contact with water.

#### Answer

Length of tank = 30 cm

Breadth of tank = 20 cm

Height of tank = 20 cm

As the tank is three-quarters full of water

∴ Height of water in the tank = = 15 cm

Area of the tank in contact with the water = Area of floor of Tank

+ Area of 4 walls upto 15 cm

= 30 × 20 + 2 (30 + 20) × 15

= 600 + 2 × 50 × 15

= 600 + 1500 = 2100 cm^{2}

Question 8.

The volume of a cuboid is 448 cm^{3}. Its height is 7 cm and the base is a square. Find

(i) a side of the square base

(ii) surface area of the cuboid.

#### Answer

Volume of a cuboid = 448 cm^{3}

Height = 7 cm

∴ Area of base = = 64 cm^{2}

∵ Base is a square.

(i) ∴ Side of square base = = 8 cm

(ii) Surface area of the cuboid = 2 [lb + bh + hl]

= 2[8 × 8 + 8 × 7 + 7 × 8] cm^{2}

= 2[64 + 56 + 56]

= 2 × 176 = 352 cm^{2}

Question 9.

The length, breadth and height of a rectangular solid are in the ratio 5 : 4 : 2. If its total surface area is 1216 cm^{2}, find the volume of the solid.

#### Answer

Ratio in length, breadth and height of a rectangular solid = 5 : 4 : 2

Total surface area =1216 cm^{2}

Let Length = 5x,

Breadth = 4x

and height = 2x

Total surface area = 2[5x × 4x + 4x × 2x + 2x × 5x]

= 2[20x^{2} + 8x^{2} + 10x^{2} ]

= 2 × 38x^{2} = 76x^{2}

∴ 94x^{2} = 1216

⇒ x^{2} = = 16 = (4)^{2}

∴ x = 4

∴ Length = 5 × 4 = 20 cm

Breadth = 4 × 4 = 16 cm

Height = 2 × 4 = 8 cm

and volume = lbh = 20 × 16 × 8 = 2560 cm^{3}

Question 10.

A rectangular room is 6 m long, 5 m wide and 3·5 m high. It has 2 doors of size 1·1 m by 2 m and 3 windows of size 1·5 m by 1·4 m. Find the cost of whitewashing the walls and the ceiling of the room at the rate of ₹5·30 per square metre.

#### Answer

Length of room = 6 m

Breadth of room = 5 m

Height of room = 3·5 m

Area of four walls = 2 (l + b) × h

= 2 (6 + 5) × 3·5 = 77 m^{2}

Area of 2 doors and 3 windows

= (2 × 1·1 × 2 + 3 × 1·5 × 1·4)

= (44 + 6·3) m^{2} = 10·7 m^{2}

Area of ceiling = l × b = 6 × 5 = 30 m^{2}

Total area for white washing

= (77 – 10·7 + 30) m^{2} = 96·3 m^{2}

Cost of white washing = ₹(96·3 × 5·30) = ₹510·39

Question 11.

A cuboidal block of metal has dimensions 36 cm by 32 cm by 0·25 m. It is melted and recast into cubes with an edge of 4 cm.

(i) How many such cubes can be made?

(ii) What is the cost of silver coating the surfaces of the cubes at the rate of ₹0·75 per square centimetre?

#### Answer

(i) Length of cuboid = 36 cm

Breadth of cuboid = 32 cm

Height of cuboid = 0·25 × 100 = 25 cm

Volume of cubiod = lbh = (36 × 32 × 25) cm^{3} = 28800 cm^{3}.

Volume of cube = (side)^{2 }= (4)^{2} = 64 cm^{2}

Number of cubes recasting from cubiod = = 450

(ii) Surface area of 1 cube = 6 × a^{2 }= 6 × 16 = 96 cm^{2}

Surface area of 450 cubes = 96 × 450 = 43200 cm^{2}

Cost of silver coating on cubes = ₹0·75 × 43200 = ₹32400

Question 12.

Three cubes of silver with edges 3 cm, 4 cm and 5 cm are melted and recast into a single cube, find the cost of coating the surface of the new cube with gold at the rate of ₹3·50 per square centimetre?

#### Answer

Let a cm be the edge of new cube

∴ According to question,

a^{3} = 3^{3} + 4^{3} + 5^{3} = 27 + 64 + 125 = 216 cm^{3}

a =

⇒ a = 6 cm.

Surface area of new cube = 6 × (side)^{2 }= 6 × (6)^{2} = 216 cm^{2}

Cost of coating the surface of new cube = ₹3·50 × 216 = ₹156

Question 13.

The curved surface area of a hollow cylinder is 4375 cm^{2}, it is cut along its height and formed a rectangular sheet of width 35 cm. Find the perimeter of the rectangular sheet.

#### Answer

Curved surface area of a hollow cylinder = 4375 cm^{2}

By cutting it from the height,

it becomes a rectangular sheet whose width = 35 cm

∴Height of cylinder = 35 cm

∴ Length of sheet =

= = 125 cm

Now perimeter of the sheet = 2(l + b)

= 2 × (125 + 35)

= 2 × 160 = 320 cm

A road roller has a diameter 0.7 m and its width is 1.2 m. Find the least number of revolutions that the roller must take in order to level a playground of size 120 m × 44 m.

#### Answer

Diameter of a road roller = 0.7 m = 70 cm

∴ Radius (r) = cm = 35 cm = m

and width (h) = 1.2 m

Curved surface area = 2πrh

=

=

Area of polyground = 120 m × 44 m

= 120 × 44 m^{2}

= 5280 m^{2}

Number of revolution made by the road roller

= × 100

= 2000 revolutions

Question 15.

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

Solution:

Diameter of cylindrical container = 14 cm

∴ Radius (r) = = 7 cm

and height (h) = 20 cm

Width of lable = 2 0 – (2 + 2) cm = 20 – 4 = 16 cm

∴ Area of lable = 2πrh = 2 × × 7 × 16 = 704 cm^{2}

Question 16.

The sum of the radius and height of a cylinder is 37 cm and the total surface area of the cylinder is 1628 cm^{2}. Find the height and the volume of the cylinder.

#### Answer

Sum of height and radius of a cylinder = 37 cm

Total surface area = 1628 cm^{2}

Let radius be r, then height = (37 – r) cm

Total surface area = 2π(h + r)

⇒ r = 7 cm

Height = 37 – 7 = 30 cm

Now, volume = πr^{2}h = × 7 × 7 × 30 cm^{3 }= 4620 cm^{3}

Question 17.

The ratio between the curved surface and total surface of a cylinder is 1 : 2. Find the volume of the cylinder, given that its total surface area is 616 cm^{3}.

#### Answer

Ratio between curved surface and total surface area of a cylinder = 1 : 2

Total surface area = 616 cm^{2}

Question 18.

The given figure shown a metal pipe 77 cm long. The inner diameter of cross section is 4 cm and the outer one is 4.4 cm.

Find its

(i) inner curved surface area

(ii) outer curved surface area

(iii) total surface area.

#### Answer

Length of metal pipe (h) = 77 cm

Inner diameter = 4 cm

and outer diameter = 4.4 cm

Total surface area = 968 + 1064.8 + 5.28 = 2038.08 cm^{3}

### Objective Type Questions, **Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions Chapter-18

**Mental Maths**

Question 1.

Fill in the blanks:

(i) Area of a parallelogram = base × …….

(ii) Area of a trapezium = × ……….. × distance between parallel sides.

(iii) Area of a rhombus = × product of ……..

(iv) Area is measured in ……….. units.

(v) Volume of a solid is the measurement of ………… occupied by it.

(vi) Volume is measured in ………… units.

(vii) The volume of a unit cube is ……….

(viii) 1 litre = …………… cm^{3}

(ix) 1 m^{3} = ………… litres

(x) Volume of a cuboid = ……….. × height.

(xi) Cylinders in which line segment joining the centres of the circular faces is perpendicular to the base are called ……….

(xii) Volume of a cylinder = area of base × ………..

(xiii) Area of four walls = perimeter of floor × …….

(xiv) Lateral surface area of a cube = 4 × (…………)^{2}

(xv) Total surface area of a cylinder of radius r and height h is ………..

#### Answer

(i) Area of a parallelogram = base × height.

(ii) Area of a trapezium = × (sum of parallel sides)

× distance between parallel sides.

(iii) Area of a rhombus = × product of its diagonals.

(iv) Area is measured in square units.

(v) Volume of a solid is the measurement of the space occupied by it.

(vi) Volume is measured in cubic units.

(vii) The volume of a unit cube is 1 cubic unit.

(viii) l litre = 1000 cm^{3}

(ix) 1 m^{3} = 1000 litres.

(x) Volume of a cuboid = length × breadth × height.

OR

Volume of a cuboid = area of the base × height.

(xi) Cylinders in which line segment joining the centres of the circular faces

is perpendicular to the base are called right circular cylinders.

(xii) Volume of a cylinder = area of base × height.

(xiii) Area of four walls = perimeter of floor × height of the room.

(xiv) Lateral surface area of a cube = 4 × (edge)^{2}

(xv) Total surface area of a cylinder of radius r and height h is 2πr(h + r).

#### Question 2.

State which of the following statements are true (T) or false (F):

(i) Perimeter of a rectangle is the sum of lengths of its four sides.

(ii) Area of a quadrilateral can be found by splitting it into two triangles.

(iiii) Perimeter of a circle of radius r = πr^{2}.

(iv) Volume of a cube = 6 × (side)^{2}

(v) 1 m^{3} = 100000 cm^{3}

(vi) Total surface area of a cuboid

= 2 (lb + bh + hl)

(vii) There is no difference between volume and capacity.

(viii)Total surface area of a cylinder = lateral surface area + area of two circular ends.

(ix) Surface area of a cube = 4 × (side)^{2}

(x) Lateral surface area of a cuboid = perimeter of base × height.

#### Answer

(i) Perimeter of a rectangle is the sum of lengths of its four sides. True

(ii) Area of a quadrilateral can be found by splitting it into two triangles. True

(iii) Perimeter of a circle of radius r = πr^{2}. False

Correct :

It is area of a circle perimeter is 2πr.

(iv) Volume of a cube = 6 × (side)^{2} False Correct :

It is surface area not volume, volume is (side)^{3}.

(v) 1 m^{3} = 100000 cm^{3} False

Correct:

1 m^{3} = 1000000 cm^{3}

(vi) Total surface area of a cuboid

= 2 (lb + bh +hl) True

(vii) There is no difference between volume and capacity. False

Correct :

Volume refers to the amount of space occupied by an object

whereas capacity refers to the quantity that a container holds.

(viii)Total surface area of a cylinder = lateral surface area

+ area of two circular ends. True

(ix) Surface area of a cube = 4 × (side)^{2} False Correct :

It is 6 × (side)^{2}

(x) Lateral surface area of a cuboid = perimeter of base × height. True

**Multiple Choice Questions**

**MCQs, Mensuration Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-18**

**Choose the correct answer from the given four options (3 to 17):**

Question 3.

Area of a triangle is 30 cm^{2}. If its base is 10 cm, then its height is

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

#### Answer

Area of a triangle = 30 cm^{2}

Base = 10 cm

Area = × Base × Height

Height = =6 cm (b)

Question 4.

If the perimeter of a square is 80 cm, then its area is

(a) 800 cm^{2}

(b) 600 cm^{2}

(c) 400 cm^{2}

(d) 200 cm^{2}

#### Answer

Perimeter of a square = 80 cm

Perimeter of square = 4(Side)

∴ Side =

∴ Side = = 20 cm

Area = (side)^{2} = (20)^{2} = 400 cm^{2} (c)

Question 5.

Area of a parallelogram is 48 cm^{2}. If its height is 6 cm then its base is

(a) 8 cm

(b) 4 cm

(c) 16 cm

(d) none of these

#### Answer

Area of parallelogram = 48 cm^{2}

Height = 6 cm

Area of ||gm = Base × Height

Base = = 8 cm (a)

Question 6.

If d is the diameter of a circle, then its area is

(a) πd^{2}

(b)

(c)

(d) 2πd^{2}

#### Answer

d is the diameter a circle

∴ Radius = r =

Area = πr^{2} = π = π (c)

Question 7.

If the area of a trapezium is 64 cm^{2} and the distance between parallel sides is 8 cm, then sum of its parallel sides is

(a) 8 cm

(b) 4 cm

(c) 32 cm

(d) 16 cm

#### Answer

Area of trapezium = 64 cm^{2}

Distance between parallelogram (h) = 8 cm

Area of trapezium = × (Sum of ||gm sides) × h

Sum of parallel lines =

= = 16 cm (d)

Question 8.

Area of a rhombus whose diagonals are 8 cm and 6 cm is

(a) 48 cm^{2}

(b) 24 cm^{2}

(c) 12 cm^{2}

(d) 96 cm^{2}

#### Answer

Area of rhombus =

= = 24 cm^{2} (b)

Question 9.

If the lengths of diagonals of a rhombus is doubled, then area of rhombus will be

(a) doubled

(b) tripled

(c) four times

(d) remains same

#### Answer

Area of rhombus_{1} = × d_{1} × d_{2}

Now the diagonals are doubled

Area of rhombus = × 2d_{1} × 2d_{2} = 2d_{1}d_{2}

with doubled diagonals.

Lengths of diagonals are doubled, then the area will be four times. (c)

Question 10.

If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is

(a) 100 cm^{2}

(b) 200 cm^{2}

(c) 50 cm^{2}

(d) none of these

#### Answer

Length of a diagonal of a quadrilateral = 10 cm

and lengths of perpendicular on it from the

opposite vertices = 4 cm and 6 cm

∴ Area = (4 + 6) × 10 cm^{2}

= × 10 × 10 = 50 cm^{2} (c)

Question 11.

Area of a rhombus is 90 cm^{2}. If the length of one diagonal is 10 cm then the length of other diagonal is

(a) 18 cm

(b) 9 cm

(c) 36 cm

(d) 4.5 cm

#### Answer

Area of a rhombus = 90 cm^{2}

Length of one diagonal = 10 cm

Area of rhombus = × d_{1} × d_{2}

∴ Length of second diagonal =

= = 18cm (a)

Question 12.

If the volume of a cube is 729 cm^{3}, then its surface area is

(a) 486 cm^{2}

(b) 324 cm^{2}

(c) 162 cm^{2}

(d) none of these

#### Answer

Volume of a cube = 729 cm^{3}

V = (Side)^{3}

∴ Side (edge) = = 9 cm

Then surface area = 6 × (side)^{2}

= 6 × (9)^{2} = 6 × 81 cm^{2}

= 486 cm^{2} (a)

Question 13.

If the lateral surface area of a cube is 100 cm^{2}, then its volume is

(a) 25 cm^{3}

(b) 125 cm^{3}

(c) 625 cm^{3}

(d) none of these

#### Answer

Lateral surface area of a cube = 4(Edge)^{2} = 100 cm^{2}

∴ 4 × (edge)^{2} = 100

⇒ (edge)^{2} = = 25 = (5)^{2}

∴ Edge of cube = 5 cm

Volume = (edge)^{3 }= (5)^{3} = 125 cm^{3} (b)

Question 14.

If the length of side of a cube is doubled, then the ratio of volumes of new cube and original cube is

(a) 1 : 2

(b) 2 : 1

(c) 4 : 1

(d) 8 : 1

#### Answer

Let original side of a cube = x

Then volume = x^{3}

If edge is doubled i.e. edge = 2x

Then volume = (2x)^{3} = 8x^{3}

∴ Ratio between new cube and original cube

= 8x^{3} : x = 8 : 1 (d)

Question 15.

If the dimensions of a rectangular room are 10m × 12m × 9m, then the cost of painting its four walls at the rate of ₹8 per m^{2} is

(a) ₹3186

(b) ₹3618

(c) ₹3168

(d) none of these

#### Answer

Dimensions of a room = 10m× 12 × 9m

Area of 4 walls = 2(l + b)h

= 2(10 + 12) × 9 = 2 × 22 × 9 m^{2}

= 396 cm^{2}

Cost of painting = ₹8 per m^{2}

∴ Total cost = 396 × 8 = ₹3168 (c)

Question 16.

Volume of a cylinder is 1848 cm^{2}. If the diameter of its base is 14 cm, then the height of the cylinder is

(a) 12 cm

(b) 6 cm

(c) 3 cm

(d) none of these

#### Answer

Volume of a cylinder = 1848 cm^{2}

Diameter of base = 14 cm

∴ Radius (r) = = 7 cm

V = πr^{2}h

∴ Height =

= = 12 cm

Question 17.

If the radius of a cylinder is doubled and height is halved, then new volume is

(a) same

(b) 2 times

(c) 4 times

(d) 8 times

#### Answer

Let radius = r

and height = h

Then volume = πr^{2}h

If radius is doubled i. e. 2 r and height is halved

i.e. , then

Volume = π(2r)^{2} ×

= π × 4r^{2} × = 2πr^{2}h

∴ Its volume is doubled (2 times) (b)

**Value Based Questions**

**Mensuration Class-8 ML Aggarwal ICSE Mathematics Solutions Chapter-18**

Question 1.

Pulkit painted four walls and roof of a rectangular room of size 10m × 12m × 12m. He got ₹10 per m^{2} for his work. How much money he earned? He always give one fourth of his income to an orphanage. Find how much money he gave to orphanage? What values are being promoted?

#### Answer

Dimensions of a room = 10m × 12m × 10m

∴ Area of 4-walls = 2(l + b) × h

= 2(10 + 12) × 10 m^{2}

= 2 × 22 × 10 = 440 m^{3}

and area of cielings = l x b = 10 × 12 = 120 m^{2}

Total area = 440 + 120 = 560 m^{2}

Rate of painting = ₹10 per m^{2}

Total changes for painting = ₹560 × 10 = ₹5600

Money gave to an orphanage = of ₹5600 = ₹1400

Remaining money = ₹5600 – 1400 = ₹4200

Amount given to an orphanage is a good and noble deed.

He help the poor and needy.

Question 2.

In a slogan writing competition in a school, Rama wrote the slogan ‘Truth pays, never betrays’ on a trapezium shaped cardboard. If the lengths of parallel sides of trapezium are 60 cm and 80 cm and the distance between them is 50 cm, find the area of trapezium. What are the advantages of speaking truth?

#### Answer

Parallel sides of a trapezium = 60 cm and 80 cm

and distance between then = 50 cm

∴ Area of trapezium = (sum of parallel sides) × height

= (60 + 80) × 50

= × 140 × 50 cm^{3}

= 3500 cm^{3}

Rama wrote on it a slogan: Truth pays, never betrays’

Always speek the truth. It pays in the long run on speaking truth,

people will believe you, as truth is like a God.

**Higher Order Thinking Skills **

**HOTS, Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions Chapter-18

Question 1.

The length of a room is 50% more than its breadth. The cost of carpeting the room at the rate of ₹38.50 m^{2} is ₹924 and the cost of papering the walls at ₹3.30 m^{2} is ₹214.50. If the room has one door of dimensions 1 m × 2 m and two windows each of dimensions 1 m × 1.5 m, find the dimensions of the room.

#### Answer

Length of a room is 50% more than its breadth

Let breadth (b) = xm

Then length (l) = x +

Cost of carpeting the room at the rate of ₹38.50 = ₹924

Area of floor = ₹

= = 24 m^{2}

∴ l × b = 24 m^{2} …(i)

Cost of papering the walls at the rate of ₹33.30 per m^{2} = ₹3214.50

∴ Area of paper

Area of one door of dimension 1 m × 2 m = 2 m^{2}

and area of two windows of size

= 1 × 1.5 m = 1 × 1.5 × 2 = 3 m^{2}

∴ Area of 4-walls = 65 + 2 + 3 = 70 m^{2}

Now, l × b = 24 ⇒ x × x = 24

⇒ x^{2} = = 16 = (4)^{2}

∴ x = 4

∴ Length = 4 × = 6 m

and breadth = x = 4 m

∴ Length = 6 m, breadth = 4 m and height = 3.5 m

Question 2.

The barrel of a fountain pen, cylindrical in shape, is 7 cm long and 5 mm in diameter. A full barrel of ink in the pen will be used up when writing 310 words on an average. How many words would use up a bottle of ink containing one-fifth of a litre? Answer correct to the nearest 100 words.

#### Answer

Height of cylindrical shaped barrel (h) = 7 cm

Diameter = 5 mm

∴ Radius (r) = mm

One-fifth of litre = 200 ml

∴ In 200 ml, words will be written

Question 3.

A cylindrical jar is 20 cm high with internal diameter 7 cm. An iron cube of edge 5 cm is immersed in the jar completely in the water which was originally 12 cm high. Find the rise in the level of water.

#### Answer

Height of cylindrical jar = 20 cm

and diameter = 7 cm

Question 4.

Squares each of side 6 cm are cut off from the four comers of a sheet of tin measuring 42 cm by 30 cm. The remaining portion of the tin sheet is made into an open box by folding up the flaps. Find the capacity of the box.

#### Answer

From a sheet, squares of 6 cm sides are cut and cutout

Length of sheet = 32 cm

and breadth = 30 cm

Remaining sheet is folded into a box whose length

= 42 – 6 × 2 = 42 – 12 = 30 cm

Breadth = 30 – 6 × 2 = 30 – 12 = 18 cm

and height = 6 cm

Capacity of the box = 30 × 18 × 6 = 3240 cm^{3}

**Check Your Progress**

**Mensuration Class-8 ML Aggarwal** ICSE Mathematics Solutions Chapter-18

Question 1.

A square field of side 65 m and rectangular field of length 75 m have the same perimeter. Which field has a larger area and by how much?

#### Answer

Side of a square field = 65 m

∴ Perimeter = 4 × Side = 4 × 65 = 260 m

Now perimeter of a rectangular field = 260 m

and length = 75 m

Perimeter of rectangle = 2(l + B)

∴ Breadth = – 75 = 130 – 75 = 55 m

Area of square field = (side)^{2} = (65)^{2} m^{2 }= 4225 m^{2}

and area of rectangular field = l × b = 75 × 55 m = 4125 m^{2}

It is clear that area of square field is greater difference = 4225 – 4125 = 100 m^{2}

Question 2.

The shape of a top surface of table is a trapezium. Find the area if its parallel sides are 1.5 m and 2.5 m and perpendicular distance between them is 0.8 m.

#### Answer

Shape of the top of a table is trapezium and parallel sides are

1.5 m and 2.5 m and the perpendicular distance between then = 0.8m

Area = (Sum of parallel sides) × height

= (1.5 + 2.5) × 0.8

= × 4 × 0.8 m^{2} = 1.6 m^{2}

Question 3.

The length and breadth of a hall of a school are 26 m and 22 m respectively. If one student requires 1.1 sq. m area, then find the maximum number of students to be seated in this hall.

#### Answer

Length of a school hall (l) = 26 m

and breadth (b) = 22 m

∴ Area = l × b = 26 × 22 m^{2} = 572 m^{2}

One student requires 1.1 sq. m area

∴ Number of students =

= = 520 students

Question 4.

It costs ₹936 to fence a square field at ₹7·80 per metre. Find the cost of levelling the field at ₹2.50 per square metre.

#### Answer

Cost of fencing the square field at ₹7·80 per metre = ₹936.

∴ Total fence required = = 120

⇒ Perimeter of the field = 120 m

⇒ 4 × Side = 120 m (∵ Field is square)

Side =

∴ Side = 30 m

Hence, Area of square field

= (30)^{2} = 900 m^{2}

= 900 × 2.50 = ₹2250

Question 5.

Find the area of the shaded portion in the following figures all measurements are given in cm.

#### Answer

(i) Outer length = 30 cm

Breadth = 10 cm

Side of each rectangle of the corner (l)

= = 6 cm

and b = 10 – 6 = = 2 cm

∴ Area of 4 comer = 6 × 2 × 4 = 48 cm^{2}

and area of inner rectangle = 18 × 6 = 108 cm^{2}

∴ Area of shaded portion = 108 + 48 = 156cm^{2}

(ii) Area of rectangle I = 4 × 2 = 8 cm^{2}

Area of rectangle II = 4 × 1 = 4 cm^{2}

Area of rectangle III = 6 × l = 6 cm^{2}

and area of square IV = 1 × 1 = 1 Cm^{2}

∴ Total area of shaded portion = 8 + 4 + 6 + 1 = 19 cm^{2}

Question 6.

Area of a trapezium is 160 sq. cm. Lengths of parallel sides are in the ratio 1:3. If smaller of the parallel sides is 10 cm in length, then find the perpendicular distance between them.

#### Answer

Area of trapezium = 160 cm^{2}

Rate in length of parallel sides = 1 : 3

Smaller paralell side = 10 cm

Then length of greater side = = 30 cm

Now, distance between them = h

Question 7.

The area of a trapezium is 729 cm^{2} and the distance between two parallel sides is 18 cm. If one of its parallel sides is 3 cm shorter than the other parallel side, find the lengths of its parallel sides.

#### Answer

Area of a trapezium = 729 cm^{2}

Distance between two parallel sides (Altitude) = 18 cm

∴ Sum of parallel sides =

= = 81 cm

One parallel side is shorter than the second by 3 cm

Let longer side = x

Then shorter side = x – 3

∴ x + x – 3 = 81

⇒ 2x = 81 + 3 = 84

x = = 42

∴ Longer side = 42 cm

and shorter side = 42 – 3 = 39 cm

Question 8.

Find the area of the polygon given in the figure:

#### Answer

In the given figure,

AC = 60 m, AH = 46 m, AF = 16 m, EF = 24 m,

DH = 14 m, BG = 16 m

∴ FH = AH – AF = 46 – 16 = 30

HC = AC – AH = 60 – 46 = 14

In the figure, there are 3 triangles and one trapezium.

Area of trapezium EFHD

= (EF + DH) × FH

= (24 + 14) × 30

= × 38 × 30 = 570 m^{2}

∴ Total area of the figure,

= Area of ∆ABC + area ∆AEF + area ∆DHC + area trapezium EFHD

= 480 + 192 + 98 + 570 = 1340 m^{2}

Question 9.

The diagonals of a rhombus are 16 m and 12 m, find:

(i) its area

(ii) length of a side

(iii) perimeter.

#### Answer

Diagonals of a rhombus are d_{1} = 16 cm

and d_{2} = 12 cm

(i) Area =

(ii) ∵ Diagonals of rhombus bisect each other at right angles

∴ AO = OC = and BO = OD

AO = = 8 cm and BO = = 6 cm

Now in right ∆AOB

AB^{2} = AO^{2} + BO^{2} (Pythagoras Theorem)

= 8^{2} + 6^{2} = 64 + 36

= 100 = (10)^{2}

∴ AB = 10 cm

∴ Side of rhombus = 10 cm

Question 10.

The area of a parallelogram is 98 cm^{2}. If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.

#### Answer

Area of a parallelogram = 98 cm^{2}

One altitude = of its corresponding base

Let base = x cm

Then altitude = cm

∴ Area = Base × Altitude

98 = x × x

⇒ x^{2} = 98 × 2 = 196 = (14)^{2}

∴ x = 14

∴ Base = 14 cm and altitude = = 7 cm

Question 11.

Preeti is painting the walls and ceiling of a hall whose dimensions are 18 m × 15 m × 5 m. From each can of paint 120 m^{2} of area is painted. How many cans of paint does she need to paint the hall?

#### Answer

Length of a hall (l) = 18 m

Breadth (b) = 15m and

height (h) = 5 m

Area of 4-wall and ceiling = 2(l + b)h + lb

= 2(18 + 15) × 5 + 18 × 15 m^{2}

= 2 × 33 × 5 + 270

= 330 + 270 = 600 m^{2}

For paint of 120 m^{2} area, can of paint is required = 1

∴ Total number of cans required to paint the area of 600 m^{2} = = 5 cans

Question 12.

A rectangular paper is size 22 cm × 14 cm is rolled to form a cylinder of height 14 cm, find the volume of the cylinder. (Take π = )

#### Answer

Length of a rectangular paper = 22 cm

and breadth = 14 cm

By rolling it a cylinder is formed whose height is 14 cm

and circumference of the base = 22 cm

Circumference = 2πr

Question 13.

A closed rectangular wooden box has inner dimensions 90 cm by 80 cm by 70 cm. Compute its capacity and the area of the tin foil needed to line its inner surface.

#### Answer

Given that,

Inner length of rectangular box = 90 cm

Inner breadth of rectangular box = 80 cm

Inner height of rectangular box = 70 cm

Capacity of rectangular box = Volume of rectangular box

= l × b × h

= 90 cm × 80 cm × 70 cm = 504000 cm^{3}

Required area of tin foil = 2 (lb + bh + lh)

= 2(90 × 80 + 80 × 70 + 90 × 70) cm^{2}

= 2(7200 + 5600 + 6300) cm^{2}

= 2 × 19100 cm^{2} = 38200 cm^{2}

Question 14.

The lateral surface area of a cuboid is 224 cm^{2}. Its height is 7 cm and the base is a square. Find

(i) side of the square base

(ii) the volume of the cuboid.

#### Answer

Lateral surface area of a cuboid is 224 cm^{2}

Height (h) = 7 cm

(i) 2(l + b) × h = 224

⇒ 2(l + b) × 7 = 224

l + b = = 16 cm

But l = b (∵ The base of cuboid is a square)

∴ 2 × side = 16 cm

⇒ Side = = 8 cm

(ii) Volume of cuboid = lbh = 8 × 8 × 7 cm^{3} = 448 cm^{3}

Question 15.

The inner dimensions of a closed wooden box are 2 m by 1.2 m by 0.75 m. The thickness of the wood is 2.5 cm. Find the cost of wood required to make the box if 1 m^{3} of wood costs ₹5400.

#### Answer

Inner dimensions of wooden box are 2 m, 1.2 m, 0.75 m

Thickness of the wood = 2.5 cm

External dimensions of wooden box are

(2 + 2 × 0.025), (1.2 + 2 × 0.025), (0.75 + 2 × 0.025)

= (2 + 0.05), (1.2 + 0.05), (0.75 + 0.5)

= 2.05, 1.25, 0.80

Volume of solid

= External volume of box – Internal volume of box

= 2.05 × 1.25 × 0.80 m^{3} – 2 × 1.2 × 0.75m^{3}

= 2.05 – 1.80 = 0.25 m^{3}

Cost = ₹5400 for 1 m^{3}

Total cost = ₹5400 × 0.25 = ₹5400 ×

= ₹54 × 25 = ₹1350

Question 16.

A car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the ful consumption of the car averages 13.5 km per litre, how far can the car travel with a full tank of petrol?

#### Answer

Capacity of car tank = 40 cm × 28 cm × 25 cm = (40 × 28 × 25) cm^{3}

= litre (∵ 1000 cm^{3} = 1 litre)

Average of car = 13.5 km per litres

Then, distance travelled by car

Hence, the car can travel 378 km with a full tank of petrol.

Question 17.

The diameter of a garden roller is 1.4 m and it is 2 m long. How much area it will cover in 5 revolutions?

#### Answer

Diameter of a garden rolle = 1.4 m

∴ Radius (r) = = 0.7 m = 70 cm

and length (h) = 2m

∴ Curved surface area = 2πrh

= 2 × × 70 × 200 cm^{2}

= 88000 cm^{2}

∴ Area covered in 5 revolutions = m^{2} = 44 m^{2}

Question 18.

The capacity of an open cylindrical tank is 2079 m^{3} and the diameter of its base is 21m. Find the cost of plastering its inner surface at ₹40 per square metre.

#### Answer

Capacity of an open cylindrical tank = 2079 m^{3}

Diameter of base = 21 m

∴ Radius (r) = m

Let h be the height, then

Cost of plastering the surface = ₹40 × 742.5 = ₹29700

Question 19.

A solid right circular cylinder of height 1.21 m and diameter 28 cm is melted and recast into 7 equal solid cubes. Find the edge of each cube.

#### Answer

Height of solid right circular cylinder = 1.21 m = 121 cm

and diameter = 28 cm

∴ Radius (r) = = 14 cm

Volume of the metal used = πr^{2}h

= × 14 × 14 × 121 cm^{3}

= 74536 cm^{3}

∴ Volume of 7 solid cubes = 74536 cm^{3}

Volume of 1 cube = = 10648 cm^{3}

Question 20.

(i) How many cubic metres of soil must be dug out to make a well 20 m deep and 2 m in diameter?

(ii) If the inner curved surface of the well in part (i) above is to be plastered at the rate of ₹50 per m^{2}, find the cost of plastering.

#### Answer

(i) Depth of a well (h) = 20 m

and diameter = 2 m

Radius (r) = = 1 m

Volume of earth dug out = πr^{2}h

— End of **Mensuration Class-8 ML Aggarwal** Solutions :–

Return to **– ML Aggarwal Maths Solutions for ICSE Class -8**

Thanks