# 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.

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Exe-18.1,

Exe-18.2 ,

Exe-18.3,

Exe-18.4

Objective Type Questions,

Mental Maths,

Multiple Choice Questions ,(MCQ)

Value Based Questions

HOTS

### 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.

Let the length = 9x and the breadth = 5x
Area = l × b ⇒ 14580 = 9x × 5x
⇒ 45x2 = 14580
∴ x2 = $\frac{14580}{45}$ = 324 ⇒ x = $\sqrt{324}$
⇒ x = $\sqrt{18 \times 18}$
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?

Area of rectangle = (16 × 9) m2 = 144 m2
Area of square = Area of rectangle (given)
∴ (side)2 = 144
Side = $\sqrt{144}=\sqrt{12 \times 12}$ = 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.

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

∴ Area of parallelogram = b × h = 24 × 12 = 288 cm2
Let d cm be the distance between the shortest sides.
∴ Area of parallelogram = (18 × d) cm2
⇒ 18 × d = 288
⇒ d = $\frac{288}{18}$ = 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 m2.

Side of square plot = 24 m
Length of house (l) = 18 m

Now area of square plot = (24)2 m2 = 24 × 24 = 576 m2
and area of hosue = 18 × 12 = 216 m2
Remaining area of the garden = 576 – 216 = 360 m2
Cost of developing the garden = ₹50 per m2
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 m2? (If required you can split the tiles in whatever way you want to fill up the comers).

Base of the parallelogram-shaped flooring tile = 18 cm
and height = 6 cm

∴ Area of one tile = Base × Height = 18 × 6 = 108 cm2
Area of floor = 540 m2
∴ Number of tiles = $\frac{\text { Total area }}{\text { Area of one tile }}$
$\frac{540 \times 100 \times 100}{108}$ = 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?

(a) Diameter of semicircle = 2.8 cm
∴ Perimeter = πr + 2r
$\frac{22}{7}$ × 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.

Radius of outer circle (R) = 21 cm.
radius of inner circle (r) = r cm.
Area of shaded portion = 770 cm2
⇒ π (R2 – r2) = 770
⇒ $\frac{22}{7}$ (212 – r2) = 770
⇒ 441 – r2 = 770 × $\frac{7}{22}$ = 35 × 7 = 245
⇒ r2 = 441 – 245
⇒ r2 = 196
⇒ r2 = 196
⇒ r = $\sqrt{196}=\sqrt{14 \times 14}$
⇒ 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.

Area of the square = 121 cm2
∴ Side = $\sqrt{121}=\sqrt{11 \times 11}$ = 11 cm
Perimeter = 4 a = 4 × 11= 44 cm
Now, circumference of the circle = 44 cm
∴ Radius = $\frac{44 \times 7}{2 \times 22}$ = 7cm
and area of the circle = πr2 = $\frac{22}{7}$(7)2
$\frac{22}{7}$ × 7 × 7 = 154 cm2

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

(i) Base = 3 cm, height = 4 cm.
Area = $\frac{1}{2}$ × base × height
$\frac{1}{2}$ × 3 × 4 = 6 cm2
(ii) By pythagoras theorem,
BC2 = AB2 + AC2
∴ BC2 = (3)2 + (4)2
= 9 + 16 = 25
⇒ BC = $\sqrt{25}$ cm = 5 cm
(iii) Now, Base = BC = 5 cm., h = AD = ?
Area = $\frac{1}{2}$ × b × h
⇒ 6 = $\frac{1}{2}$ × 5 × h
[∵ Area = 6 cm2 as in part (i)]
⇒ h = $\frac{12}{5}$ = 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.

Length of rectangular garden = 80 m
Width of crossing path 2.5 m

Area of length wise path
= 80 × 2.5 = 200 m2
= 40 × 2.5 = 100 m2
(i) Total area of both paths
= 200 + 100 – 2.5 × 2.5 m2
= 300 – 6.25 = 293.75 m2
= Area of garden – Area of paths
= 80 × 40 – 293.75 m2
= 3200 – 293.75 m2
= 2906.25 m2

Question 11.
In the given figure, ABCD is a rectangle. Find the area of the shaded region.

In the given figure.

Length of rectangle = 18 cm
∴ Area = l × b = 18 × 12 cm2 = 216 cm2
Area of triangle I = $\frac{1}{2}$ × 12 × 10 = 60 cm2
Area of triangle III = $\frac{1}{2}$ × 18 × 7 = 63 cm2
= Area of rectangle – Area of 3 triangles
= 216 – (60 + 63 + 20)
= 216 – 143 cm2
= 73 cm2

Question 12.
In the adjoining figure, ABCD is a square grassy lawn of area 729 m2. A path of uniform width runs all around it. If the area of the path is 295 m2, find
(i) the length of the boundary of the square field enclosing the lawn and the path.
(ii) the width of the path.

Area of square ABCD = 729 m2
Side = $\sqrt{729}=\sqrt{27 \times 27}$ = 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 m2
Area of PQRS – Area of ABCD = Area of path
∴ (27 + 2x)2 m2 – 729 m2 = 295 m2
⇒ 729 + 4x2 + 108x – 729 = 295
⇒ 4x2 + 108x – 295 = 0
⇒ $x=\frac{-108 \pm \sqrt{(108)^{2}-4 \times(4) \times(-295)}}{8}$

∴ 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

(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, BC2 = OC2 + OB2
132 = 52 + OB2
⇒ OB2 = 169 – 25 = 144
⇒ OB = $\sqrt{144}=\sqrt{12 \times 12}$ = 12 cm
∴ Diagonal BD = 2 × OB = 2 × 12 = 24 cm
(ii) Area of rhombus = $\frac{1}{2}$ × d1 × d2
$\frac{1}{2}$ × 10 × 24 = 120 cm2

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.

Here, two parallel sides of trapezium are AD and BC
and distance between them is 14 m.
∴ Area of trapezium = $\frac{1}{2}$ (1·5 + 8) × 14
$\frac{1}{2}$ × 9·5 × 14 = 66 × 5 m2

Question 3.
The area of a trapezium is 360 m2, the distance between two parallel sides is 20 m and one of the parallel side is 25 m. Find the other parallel side.

Area of a trapezium = 360 m2
Distance between two parallel lines = 20 m
One parallel side = 25 m

Let Second parallel side = 11 m
∴ Area = $\frac{1}{2}$ (25 + x) × 20
⇒ 360 = $\frac{1}{2}$(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.

Side of rhombus = 6.5 cm
and altitude = 5 cm
Area of a rhombus = Side × Altitude = 6.5 × 5 = 32.5 cm2
One diagonal = 13 cm

Length of other diagonal = $\frac{\text { Area } \times 2}{\text { One diagonal }}$ = $\frac{32.5 \times 2}{13}$ = 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.

(i) Area of trapezium ACDE
$\frac{1}{2}$(AC + DE) × h
$\frac{1}{2}$(13 + 7) × 6.5 = 65 m2
(ii) Area of parallelogram ABDE = b × h = $\frac{1}{2}$ × 6 × 6·5 = 15·5 m2
[∵ 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.

Base of triangle = 24.8 cm
and altitude = 16.5 cm
Area = $\frac{1}{2}$ base × altitude
$\frac{1}{2}$ × 24.8 × 16.5 cm2 = 204.6 cm2
Now, Area of ∆ = Area of rhombus
∴ Area of rhombus = 204.6 cm2
Length of one diagonal = 22 cm
Area of rhombus = $\frac{1}{2}$(First diagonal × Second diagonal)
∴ Second diagonal = $\frac{\text { Area } \times 2}{\text { First diagonal }}$
$\frac{204.6 \times 2}{22}$ = 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.

Perimeter of a trapezium = 52 cm
Length of each non-parallel side = 10 cm
Altitude DL = 8 cm

In right ∆DAL (By Pythagoras Theorem)
DA2 = DL2 + AL2
⇒ (10)2 = (8)2 + AL2
⇒ 100 = 64 + AL2
⇒ AL2 = 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 = $\frac{1}{2}$ (sum of parallel sides) × altitude
$\frac{1}{2}$ × 32 × 8 = 128 cm2

Question 8.
The area of a trapezium is 540 cm2. If the ratio of parallel sides is 7 : 5 and the distance between them is 18 cm, find the lengths of parallel sides.

Let, the two parallel sides of trapezium are 7x and 5x.
Height = 18 cm
⇒ Area of trapezium = $\frac{1}{2}$ [(Sum of ||gm sides) × height]
540 = $\frac{1}{2}$(7x + 5x) × 18
∴ 540 = $\frac{1}{2}$ × 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.

(i) Area of trapezium ABCD

$\frac{1}{2}$ (Sum of opposite ||gm sides) × height
$\frac{1}{2}$ [(AB + CD) × (AF + FD)]
$\frac{1}{2}$[(AB + CD) × (AF + FD)
$\frac{1}{2}$[(5 + 3) × (5 + 4)]
$\frac{1}{2}$ (5 + 3) × 9 = 36 cm2
Area of rectangle GAFE = Length × Breadth
= 2 × 5 = 10 cm2
Total area of the figure
= Area of trapezium ABCD + Area of rectangle GAFE
= (36 + 10) cm2
= 46 cm2
(iii) Area of rectangle ABCD
Area of given figure = Area of rect. ABCD
+ Area of ||gm BIHJ + Area of rectangle EFGH.
= 9 × 2 = 18 cm2

Area of rectangle EFGH = Length × Breadth
= (EJ + JH) × EF
= (7 + 2) × 2
= 9 × 2 = 18 cm2
Area of parallelogram BIHJ = 2 × 5 = 10 cm2
[∵ Distance between BI and HJ = 9 – 2 – 2 = 5 cm]
Total area of the figure = (18 + 18 + 10) cm2 = 46 cm2

Question 10.
(ii) the area of trapezium ABCD
(iii) the area of triangle BCD

(i) In right angled ∆ ABD (By Pythagoras Theorem)
⇒ AD2 = BD2 – AB2 = (41)2 – (40)= 1681 – 1600 = 81
∴ AD = $\sqrt{81}$ = 9 cm.
(ii) Area of trapezium ABCD
$\frac{1}{2}$ (Sum of opposite ||gm lines) × height
$\frac{1}{2}$ (AB + CD) × AD
$\frac{1}{2}$ (40+ 15) × 9 = 247·5 cm2
(iii) Area of triangle BCD = Area of trapezium ABCD – Area of ∆ ABD
= (247·5 – $\frac{1}{2}$ × 40 × 9 ) cm2
= (247·5 – 180) cm2 = 67·5 cm2

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.

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 = $\frac{32-24}{2}=\frac{8}{2}$ = 4 cm
i. e., height = 4 cm
Now area of each portion of length side
$\frac{1}{2}$ (24 + 32) × 4
$\frac{1}{2}$ × 56 × 4 = 112 cm2
and area of each portion of breadth side
$\frac{1}{2}$(20 + 28) × 4
$\frac{1}{2}$ × 48 × 4 = 96 cm2
∴ Area each sections =112 cm2, 96 cm2, 112 cm2, 96 cm2

Question 12.
In the given quadrilateral ABCD, ∠BAD = 90° and ∠BDC = 90°. All measurements are in centimetres. Find the area of the quadrilateral ABCD.

In right angled triangle ABD, (By Pythagoras Theorem)
BD2 = AB2 + AD= (6)2 + (8)2 = 36 + 64 = 100 cm2
BD = $\sqrt{100 \mathrm{cm}^{2}}$
∴ BD = 10 cm.
Now, Area of A ABD = $\frac{1}{2}$ × b × h
$\frac{1}{2}$ × 6 × 8 = 24cm2 …(i)
In ∆ BDC, BD = 10 cm.,
BC = 26 cm., DC = ?
By Pythagoras theorem,
BC2 = BD2 + DC2
(26)2 = (10)2 + DC2
676 – 100 = DC2
⇒ DC = $\sqrt{576}$ = 24 cm.
Now, Area of ∆ BDC = $\frac{1}{2}$ × b × h
$\frac{1}{2}$ × 24 × 10 = 12 cm2 …(ii)
Add (i) and (ii), we get
Area of ∆ABD + Area of ∆BDC = (24 + 120) cm2
Area of quadrilateral ABCD = 144 cm2

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.

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
$\frac{1}{2}$(GD + FE) × FL
$\frac{1}{2}$(15 + 8) × 6
= 23 × 3 cm2 = 69 cm2
Area of trapezium FEDG = Area of trapezium ABCH = 69 cm2
and area of rectangle HCDG
= HC × CD = 15 × 8 = 120 cm2
Total area = Area of trapezium FEDG + Area of trapezium ABCH
+ Area of rectangle HCDG.
Total area = 69 + 69 + 120 = 258 cm2

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?

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
$\frac{1}{2}$(AE + DM) × AM
$\frac{1}{2}$(32 + 18 ) × 9
$\frac{50 \times 9}{2}$ = 225m2
According to Rahul’s the figure is divided into shapes
one square and on isoscles triangle.

Area of square ABCE = (Side)= (18)2 = 324 m2
and area of isosceles ∆EDC
$\frac{1}{2}$ × EC × DC
$\frac{1}{2}$ × 18 × 14 = 126 m2
∴ Total area = 225 × 2 = 450 m2
Third way to find out the area of given figure is as follow:

Here, DL ⊥ ED, DL = 14 m
Area of ∆DEC = $\frac{1}{2}$ × EC × LD
$\frac{1}{2}$ × 18 × 14 = 126 m2
Area of ∆AEB = $\frac{1}{2}$ × AB × AE
$\frac{1}{2}$ × 18 × 18= 162 m2
Area of ∆BEC = $\frac{1}{2}$ × BC × EC
$\frac{1}{2}$ × 18 × 18 = 162 m2
Now, area of pentagon ABCDE = Area ∆DEC
+ Area of ∆AEB + Area of ∆BEC
= (126 + 162 + 162) m2 = 450 m2

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.

Area of rectangle ABCD = Length × Breadth
= 18 × 10 = 180 cm2
In right angled ∆ BEC,
BC2 = CE2 + BE2 (By Pythagoras theorem)
(10)2 = 82 + BE2
∴ BE2 = 100 – 64 = 36
⇒ BE = $\sqrt{36}$ ⇒ BE = 6 cm.
∴ Area of rt. ∆ BEC = $\frac{1}{2}$ × 6 × 8 = 24cm2
Area of pentagon ABECD = Area of rectangle – area of ∆
= (180 – 24) cm2 = 156 cm2

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.

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 = $\frac{1}{2}$ AD × GE
$\frac{1}{2}$ × 8 × 2.5 = 10 cm2
Area of ∆ABF = $\frac{1}{2}$ AF × BF
$\frac{1}{2}$ × 3 × 2 = 3 cm2
Area of ∆CDH = $\frac{1}{2}$ × HD × CH
$\frac{1}{2}$(AD – AH) × 3
$\frac{1}{2}$(8 – 6) × 3
$\frac{1}{2}$ × 2 × 3 = 3 cm2
Area of trapezium BFHC
$\frac{1}{2}$(BF + CH) × FH
$\frac{1}{2}$ (2 + 3) × (AH – AF)
$\frac{1}{2}$ × 5 × (6 – 3)
$\frac{1}{2}$ × 5 × 3 = 7.5 cm2
∴ 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 cm2

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.

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 = $\frac{1}{2}$ PV + QV
$\frac{1}{2}$ × 3 × 5 = $\frac{15}{2}$ =7.5 cm2
Area of ∆RXS = $\frac{1}{2}$ × S × R
= 3 × 6 = 9 cm2
Area ∆PUW = $\frac{1}{2}$ × PW × UW
$\frac{1}{2}$ × 5 × 4 = 10 cm2
Area ∆YTS = $\frac{1}{2}$ × YS × TY
$\frac{1}{2}$ × 2 × 2 = 2 cm2
Area of trapezium ∆VX R
$\frac{1}{2}$(QV + RX) × VX
$\frac{1}{2}$ (5 + 6) × 5 = $\frac{1}{2}$ × 11 × 5 cm2
$\frac{55}{7}$ = 27.5 cm2
Area of trapezium WUTY
$\frac{1}{2}$(UW + TY) × WY
$\frac{1}{2}$(4 + 2) × 4 = $\frac{1}{2}$ × 6 × 4 = 12 cm2
Now area of the figure = 7.5 + 9 + 10 + 2 + 27.5 + 12 cm2 = 68 cm2

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

Question 1.
The volume of a cube is 343 cm3, find the length of an edge of cube.

Volume of a cube = 343 cm3
Let a be the edge of cube, then
V = a3 = 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 cm3 and base area is 26 cm2.

Volume of a cuboid = 312 cm3
Base area = l × b = 26 cm2
∴ Height= $\frac{\text { Volume }}{\text { Base area }}=\frac{312}{26}$ = 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 m3?

Length of a godown (l) = 55 m
Height (h) = 30 m
Volume = lbh = 55 × 45 × 30 m3
Volume of one box = 1.25 m3
Number of box = $\frac{74250}{1.25}$ = 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.

Here l = 1·4 m = 140 cm, b = 90 cm, h = 70 cm
Volume of rectangular pit = l × b × h
= (140 × 90 × 70) cm3 = 882000 cm3
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?

Let edge of a cube = x
Then volume = x3
If the edge is trippled, then
Edge = 3x
Now, volume = (3x)3 = 27x3
∴ 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.

Radius of the milk cylindrical tank = 1.4 m and height (h) = 8 m
∴ Volume of milk in the tank = πr2h
$\frac{22}{7}$ × 1.4 × 1.4 × 8 m3
= 49.28 m3
= 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.

Thickness of the wood used in a closed box = 2 cm
External length of box (L) = 84 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 cm3
= 403200 – 340800 cm3
= 62400 cm3

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.

Ratio in diameters of two cylindrical jars = 3 : 4
But their volume is same.
Let h1 and h2 be the heights of the two jars respectively.
Let radius of the first jar (r1) = $\frac{3 x}{2}$
and radius of the second jar (r2) = $\frac{4 x}{2}$
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?

Let radius of a cylinder = r
and height = h
Volume = πr2h
Its radius is halved and height is doubled, then

∴ Ratio in the volumes of the new cylinder to old one
$\frac{\pi r^{2} h}{2} : \pi r^{2} h$ = 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.

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.

Speed of water flow through cylindrical pipe = 5 m/sec.
Internal diameter of the pipe = 7 cm
∴ Radius (r) = $\frac{7}{2}$cm
∴ Length of water flow in 1 minutes (h)
= 5 × 60 = 300 m
∴ Volume of water = πr2h

= 1155 litres
Now volume of water = 1155000 cm3
Volume of rectangular tank of size
= 4m × 3m × 2.31m
= 27.72 m3
Speed of water 4 m/sec.
Radius of pipe = $\frac{7}{2}$ 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.

Radius of one cylinder (r1) = 15 cm
and height (h1) = 40 cm
and radius of second cylinder (r2) = 20 cm
and height (h2) = 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 m3 .

Height of pole (h) = 7 m
Diameter = 20 cm

Weight of wood = 225 kg per m3
Total weight = 225 × $\frac{22}{100}=\frac{99}{2}$ = 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.

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) = $\frac{14}{2}$ = 7 cm
and height (h) = 30 cm
Volume of cuboid = 30 × 14 × 14 = 5880 cm3
Volume of cylinder = πr2h
$\frac{22}{7}$ × 7 × 7 × 30 = 4620 cm3
and wastage of wood = 5880 – 4620 = 1260 cm3

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

Question 1.
The surface area of a cube is 384 cm2. Find
(i) the length of an edge
(ii) volume of the cube.

Surface area of a cube = 384 cm2

(ii) Volume = (Edge)3 = (8)= 8 × 8 × 8 cm3 = 512 cm3

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.

Radius of a solid cylinder (r) = 5 cm
Height (h) = 10 cm
Total surface area = 2πrh + 2πr2
= 2rπ(h + r)
= 2π × 5(10 + 5)
= π × 10 × 15 = 150π cm2

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.

A cuboid shaped aquarium,
Length (l) = 70 cm
and height (h) = 35 cm

Area of base = 70 × 28 cm3 = 1960 cm3
Area of side face = (28 × 35) × 2 cm= 1960 cm2
Area of back face = 70 × 35 cm2 = 2450 cm
∴. Total area = 1960 + 1960 + 2450 = 6370 cm2
∴ Area of paper required = 6370 cm2

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 m2. What will be the cost of white washing if the ceiling of the hall is also white washed?

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 m2
= 216 m2
Area of 4 windows of size = 2 × 1.5 = 2 × 1.5 × 4 = 12 m2
Area of 2 door of size = 1.5 × 2.5 = 2 × 1.5 × 2.5 = 7.5 m2
∴ Area of remaining hall = 216 – (12 + 7.5) = 216 – 19.5 m2 = 196.5 m2
Area of ceiling = l × b = 15 × 12= 180 m2
Cost of white washing the walls at the rate of ₹5 per m2
= 196.5 × 5 = ₹982.50
Area of ceiling = l × b = 15 × 12= 180 m2
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.

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 m2
Area of four walls = 2 (50 + 30) × 2·5 = 160 × 2·5 = 400 m2
Area to be cemented = 1500 m2 + 400 m2 = 1900 m2
Cost of cementing 1m2 = ₹27
Cost of cementing 1900m= ₹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.

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 m2
Cost of painting 1 m2 = ₹8·40
Cost of painting 1062 m2 = ₹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.

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 = $\frac{20 \times 3}{4}$ = 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 cm2

Question 8.
The volume of a cuboid is 448 cm3. 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.

Volume of a cuboid = 448 cm3
Height = 7 cm
∴ Area of base = $\frac{448}{7}$ = 64 cm2
∵ Base is a square.
(i) ∴ Side of square base = $\sqrt{64}$ = 8 cm
(ii) Surface area of the cuboid = 2 [lb + bh + hl]
= 2[8 × 8 + 8 × 7 + 7 × 8] cm2
= 2[64 + 56 + 56]
= 2 × 176 = 352 cm2

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 cm2, find the volume of the solid.

Ratio in length, breadth and height of a rectangular solid = 5 : 4 : 2
Total surface area =1216 cm2
Let Length = 5x,
and height = 2x
Total surface area = 2[5x × 4x + 4x × 2x + 2x × 5x]
= 2[20x2 + 8x2 + 10x2 ]
= 2 × 38x2 = 76x2
∴ 94x2 = 1216
⇒ x2 = $\frac{1216}{76}$ = 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 cm3

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.

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 m2
Area of 2 doors and 3 windows
= (2 × 1·1 × 2 + 3 × 1·5 × 1·4)
= (44 + 6·3) m2 = 10·7 m2
Area of ceiling = l × b = 6 × 5 = 30 m2
Total area for white washing
= (77 – 10·7 + 30) m2 = 96·3 m2
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?

(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) cm3 = 28800 cm3.
Volume of cube = (side)= (4)2 = 64 cm2
Number of cubes recasting from cubiod = $\frac{28800}{64}$ = 450
(ii) Surface area of 1 cube = 6 × a= 6 × 16 = 96 cm2
Surface area of 450 cubes = 96 × 450 = 43200 cm2
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?

Let a cm be the edge of new cube
∴ According to question,
a3 = 33 + 43 + 53 = 27 + 64 + 125 = 216 cm3
a = $\sqrt[3]{216}$
⇒ a = 6 cm.
Surface area of new cube = 6 × (side)= 6 × (6)2 = 216 cm2
Cost of coating the surface of new cube = ₹3·50 × 216 = ₹156

Question 13.
The curved surface area of a hollow cylinder is 4375 cm2, it is cut along its height and formed a rectangular sheet of width 35 cm. Find the perimeter of the rectangular sheet.

Curved surface area of a hollow cylinder = 4375 cm2
By cutting it from the height,
it becomes a rectangular sheet whose width = 35 cm

∴Height of cylinder = 35 cm
∴ Length of sheet = $\frac{\text { Area }}{\text { Height }}$
$\frac{4375}{35}$ = 125 cm
Now perimeter of the sheet = 2(l + b)
= 2 × (125 + 35)
= 2 × 160 = 320 cm

Question 14.
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.

Diameter of a road roller = 0.7 m = 70 cm
∴ Radius (r) = $\frac{70}{2}$ cm = 35 cm = $\frac{35}{100}$m
and width (h) = 1.2 m
Curved surface area = 2πrh
$2 \times \frac{22}{7} \times \frac{35}{100} \times 1.2 \mathrm{m}^{2}$
$\frac{264}{100} \mathrm{m}^{2}$
Area of polyground = 120 m × 44 m
= 120 × 44 m2
= 5280 m2
$\frac{5280}{264}$ × 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) = $\frac{14}{2}$ = 7 cm
and height (h) = 20 cm
Width of lable = 2 0 – (2 + 2) cm = 20 – 4 = 16 cm
∴ Area of lable = 2πrh = 2 × $\frac{22}{7}$ × 7 × 16 = 704 cm2

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 cm2. Find the height and the volume of the cylinder.

Sum of height and radius of a cylinder = 37 cm
Total surface area = 1628 cm2
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 = πr2h = $\frac{22}{7}$ × 7 × 7 × 30 cm= 4620 cm3

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 cm3.

Ratio between curved surface and total surface area of a cylinder = 1 : 2
Total surface area = 616 cm2

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.

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 cm3

### 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 = $\frac{1}{2}$ × ……….. × distance between parallel sides.
(iii) Area of a rhombus = $\frac{1}{2}$ × 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 = …………… cm3
(ix) 1 m3 = ………… 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 ………..

(i) Area of a parallelogram = base × height.
(ii) Area of a trapezium = $\frac{1}{2}$ × (sum of parallel sides)
× distance between parallel sides.
(iii) Area of a rhombus = $\frac{1}{2}$ × 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 cm3
(ix) 1 m3 = 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 = πr2.
(iv) Volume of a cube = 6 × (side)2
(v) 1 m3 = 100000 cm3
(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.

(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 = πr2. 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 m3 = 100000 cm3 False
Correct:
1 m3 = 1000000 cm3
(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 cm2. If its base is 10 cm, then its height is
(a) 5 cm
(b) 6 cm
(c) 7 cm
(d) 8 cm

Area of a triangle = 30 cm2
Base = 10 cm
Area = $\frac{1}{2}$ × Base × Height
Height = $\frac{A \times 2}{B}=\frac{30 \times 2}{10}$ =6 cm (b)

Question 4.
If the perimeter of a square is 80 cm, then its area is
(a) 800 cm2
(b) 600 cm2
(c) 400 cm2
(d) 200 cm2

Perimeter of a square = 80 cm
Perimeter of square = 4(Side)
∴ Side = $\frac{\text { Perimeter }}{4}$
∴ Side = $\frac{80}{4}$ = 20 cm
Area = (side)2 = (20)2 = 400 cm2 (c)

Question 5.
Area of a parallelogram is 48 cm2. If its height is 6 cm then its base is
(a) 8 cm
(b) 4 cm
(c) 16 cm
(d) none of these

Area of parallelogram = 48 cm2
Height = 6 cm
Area of ||gm = Base × Height
Base = $\frac{\mathrm{A}}{h}=\frac{48}{6}$ = 8 cm (a)

Question 6.
If d is the diameter of a circle, then its area is
(a) πd2
(b) $\frac{\pi d^{2}}{2}$
(c) $\frac{\pi d^{2}}{4}$
(d) 2πd2

d is the diameter a circle
∴ Radius = r =$\frac{d}{2}$
Area = πr2 = π$\left(\frac{d}{2}\right)^{2}$ = π$\frac{d^{2}}{4}$ (c)

Question 7.
If the area of a trapezium is 64 cm2 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

Area of trapezium = 64 cm2
Distance between parallelogram (h) = 8 cm
Area of trapezium = $\frac{1}{2}$ × (Sum of ||gm sides) × h
Sum of parallel lines = $\frac{\mathrm{A} \times 2}{h}$
$\frac{64 \times 2}{8}$ = 16 cm (d)

Question 8.
Area of a rhombus whose diagonals are 8 cm and 6 cm is
(a) 48 cm2
(b) 24 cm2
(c) 12 cm2
(d) 96 cm2

Area of rhombus =$\frac{d_{1} \times d_{2}}{2}=\frac{8 \times 6}{2}$
$\frac{48}{2}$ = 24 cm2 (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

Area of rhombus1 = $\frac{1}{2}$ × d1 × d2
Now the diagonals are doubled
Area of rhombus = $\frac{1}{2}$ × 2d1 × 2d2 = 2d1d2
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 cm2
(b) 200 cm2
(c) 50 cm2
(d) none of these

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 = $\frac{1}{2}$ (4 + 6) × 10 cm2
$\frac{1}{2}$ × 10 × 10 = 50 cm2 (c)

Question 11.
Area of a rhombus is 90 cm2. 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

Area of a rhombus = 90 cm2
Length of one diagonal = 10 cm
Area of rhombus = $\frac{1}{2}$ × d1 × d2
∴ Length of second diagonal = $\frac{A \times 2}{d_{1}}$
$\frac{90 \times 2}{10}$ = 18cm (a)

Question 12.
If the volume of a cube is 729 cm3, then its surface area is
(a) 486 cm2
(b) 324 cm2
(c) 162 cm2
(d) none of these

Volume of a cube = 729 cm3
V = (Side)3
∴ Side (edge) = $\sqrt[3]{729}=\sqrt[3]{9 \times 9 \times 9}$ = 9 cm
Then surface area = 6 × (side)2
= 6 × (9)2 = 6 × 81 cm2
= 486 cm2 (a)

Question 13.
If the lateral surface area of a cube is 100 cm2, then its volume is
(a) 25 cm3
(b) 125 cm3
(c) 625 cm3
(d) none of these

Lateral surface area of a cube = 4(Edge)2 = 100 cm2
∴ 4 × (edge)2 = 100
⇒ (edge)2 = $\frac{100}{4}$ = 25 = (5)2
∴ Edge of cube = 5 cm
Volume = (edge)= (5)3 = 125 cm3 (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

Let original side of a cube = x
Then volume = x3
If edge is doubled i.e. edge = 2x
Then volume = (2x)3 = 8x3
∴ Ratio between new cube and original cube
= 8x3 : 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 m2 is
(a) ₹3186
(b) ₹3618
(c) ₹3168
(d) none of these

Dimensions of a room = 10m× 12 × 9m
Area of 4 walls = 2(l + b)h
= 2(10 + 12) × 9 = 2 × 22 × 9 m2
= 396 cm2
Cost of painting = ₹8 per m2
∴ Total cost = 396 × 8 = ₹3168 (c)

Question 16.
Volume of a cylinder is 1848 cm2. 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

Volume of a cylinder = 1848 cm2
Diameter of base = 14 cm
∴ Radius (r) = $\frac{14}{2}$ = 7 cm
V = πr2h
∴ Height = $\frac{\mathrm{V}}{\pi r^{2}}$
$\frac{1848 \times 7}{22 \times 7 \times 7}$ = 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

and height = h
Then volume = πr2h
If radius is doubled i. e. 2 r and height is halved
i.e. $\frac{h}{2}$, then
Volume = π(2r)2 × $\frac{h}{2}$
= π × 4r2 × $\frac{h}{2}$ = 2πr2h
∴ 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 m2 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?

Dimensions of a room = 10m × 12m × 10m
∴ Area of 4-walls = 2(l + b) × h
= 2(10 + 12) × 10 m2
= 2 × 22 × 10 = 440 m3
and area of cielings = l x b = 10 × 12 = 120 m2
Total area = 440 + 120 = 560 m2
Rate of painting = ₹10 per m2
Total changes for painting = ₹560 × 10 = ₹5600
Money gave to an orphanage = $\frac{1}{4}$ 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?

Parallel sides of a trapezium = 60 cm and 80 cm
and distance between then = 50 cm
∴ Area of trapezium = $\frac{1}{2}$ (sum of parallel sides) × height
$\frac{1}{2}$(60 + 80) × 50
$\frac{1}{2}$ × 140 × 50 cm3
= 3500 cm3
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 m2 is ₹924 and the cost of papering the walls at ₹3.30 m2 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.

Length of a room is 50% more than its breadth
Then length (l) = x + $\frac{150}{100}=\frac{3}{2} x \mathrm{m}$
Cost of carpeting the room at the rate of ₹38.50 = ₹924
Area of floor = ₹ $\frac{924}{38.50}$
$\frac{924 \times 100}{3850}$ = 24 m2
∴ l × b = 24 m2 …(i)
Cost of papering the walls at the rate of ₹33.30 per m2 = ₹3214.50
∴ Area of paper $\frac{214.50}{3.30}=\frac{21450}{330}=65 \mathrm{m}^{2}$
Area of one door of dimension 1 m × 2 m = 2 m2
and area of two windows of size
= 1 × 1.5 m = 1 × 1.5 × 2 = 3 m2
∴ Area of 4-walls = 65 + 2 + 3 = 70 m2
Now, l × b = 24 ⇒ $\frac{3}{2}$x × x = 24
⇒ x2 = $\frac{24 \times 2}{3}$ = 16 = (4)2
∴ x = 4
∴ Length = 4 × $\frac{3}{2}$ = 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.

Height of cylindrical shaped barrel (h) = 7 cm
Diameter = 5 mm
∴ Radius (r) = $\frac{5}{2}$ 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.

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.

From a sheet, squares of 6 cm sides are cut and cutout
Length of sheet = 32 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 cm3

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?

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 = $\frac{P}{2}-l=\frac{260}{2}$ – 75 = 130 – 75 = 55 m
Area of square field = (side)2 = (65)2 m= 4225 m2
and area of rectangular field = l × b = 75 × 55 m = 4125 m2
It is clear that area of square field is greater difference = 4225 – 4125 = 100 m2

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.

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 = $\frac{1}{2}$ (Sum of parallel sides) × height
$\frac{1}{2}$(1.5 + 2.5) × 0.8
$\frac{1}{2}$ × 4 × 0.8 m2 = 1.6 m2

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.

Length of a school hall (l) = 26 m
and breadth (b) = 22 m
∴ Area = l × b = 26 × 22 m2 = 572 m2
One student requires 1.1 sq. m area
∴ Number of students = $\frac{572}{1.1}$
$\frac{572 \times 10}{11}$ = 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.

Cost of fencing the square field at ₹7·80 per metre = ₹936.
∴ Total fence required = $\frac{936}{7 \cdot 80}$ = 120
⇒ Perimeter of the field = 120 m
⇒ 4 × Side = 120 m (∵ Field is square)
Side = $\frac{120}{4}$
∴ Side = 30 m
Hence, Area of square field
= (30)2 = 900 m2
= 900 × 2.50 = ₹2250

Question 5.
Find the area of the shaded portion in the following figures all measurements are given in cm.

(i) Outer length = 30 cm
Side of each rectangle of the corner (l)

$\frac{30-18}{2}=\frac{12}{2}$ = 6 cm
and b = 10 – 6 = $\frac{4}{2}$ = 2 cm
∴ Area of 4 comer = 6 × 2 × 4 = 48 cm2
and area of inner rectangle = 18 × 6 = 108 cm2
∴ Area of shaded portion = 108 + 48 = 156cm2
(ii) Area of rectangle I = 4 × 2 = 8 cm2
Area of rectangle II = 4 × 1 = 4 cm2
Area of rectangle III = 6 × l = 6 cm2
and area of square IV = 1 × 1 = 1 Cm2
∴ Total area of shaded portion = 8 + 4 + 6 + 1 = 19 cm2

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.

Area of trapezium = 160 cm2
Rate in length of parallel sides = 1 : 3
Smaller paralell side = 10 cm
Then length of greater side = $\frac{10 \times 3}{1}$ = 30 cm
Now, distance between them = h

Question 7.
The area of a trapezium is 729 cm2 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.

Area of a trapezium = 729 cm2
Distance between two parallel sides (Altitude) = 18 cm
∴ Sum of parallel sides = $\frac{\text { Area } \times 2}{\text { Altitude }}$
$\frac{729 \times 2}{18}$ = 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 = $\frac{84}{2}$ = 42
∴ Longer side = 42 cm
and shorter side = 42 – 3 = 39 cm

Question 8.

Find the area of the polygon given in the figure:

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
$\frac{1}{2}$(EF + DH) × FH
$\frac{1}{2}$(24 + 14) × 30
$\frac{1}{2}$ × 38 × 30 = 570 m2
∴ Total area of the figure,
= Area of ∆ABC + area ∆AEF + area ∆DHC + area trapezium EFHD
= 480 + 192 + 98 + 570 = 1340 m2

Question 9.
The diagonals of a rhombus are 16 m and 12 m, find:
(i) its area
(ii) length of a side
(iii) perimeter.

Diagonals of a rhombus are d1 = 16 cm
and d2 = 12 cm

(i) Area = $\frac{d_{1} \times d_{2}}{2}=\frac{16 \times 12}{2}=96 \mathrm{cm}^{2}$
(ii) ∵ Diagonals of rhombus bisect each other at right angles
∴ AO = OC = and BO = OD
AO = $\frac{16}{2}$ = 8 cm and BO = $\frac{12}{2}$ = 6 cm
Now in right ∆AOB
AB2 = AO2 + BO2 (Pythagoras Theorem)
= 82 + 62 = 64 + 36
= 100 = (10)2
∴ AB = 10 cm
∴ Side of rhombus = 10 cm

Question 10.
The area of a parallelogram is 98 cm2. If one altitude is half the corresponding base, determine the base and the altitude of the parallelogram.

Area of a parallelogram = 98 cm2
One altitude = $\frac{1}{2}$ of its corresponding base
Let base = x cm

Then altitude = $\frac{x}{2}$cm
∴ Area = Base × Altitude
98 = x × $\frac{1}{2}$x
⇒ x2 = 98 × 2 = 196 = (14)2
∴ x = 14
∴ Base = 14 cm and altitude = $\frac{x}{2}=\frac{14}{2}$ = 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 m2 of area is painted. How many cans of paint does she need to paint the hall?

Length of a hall (l) = 18 m
height (h) = 5 m
Area of 4-wall and ceiling = 2(l + b)h + lb
= 2(18 + 15) × 5 + 18 × 15 m2
= 2 × 33 × 5 + 270
= 330 + 270 = 600 m2
For paint of 120 m2 area, can of paint is required = 1
∴ Total number of cans required to paint the area of 600 m2 = $\frac{600}{120}$ = 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 = $\frac{22}{7}$)

Length of a rectangular paper = 22 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.

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 cm3
Required area of tin foil = 2 (lb + bh + lh)
= 2(90 × 80 + 80 × 70 + 90 × 70) cm2
= 2(7200 + 5600 + 6300) cm2
= 2 × 19100 cm2 = 38200 cm2

Question 14.
The lateral surface area of a cuboid is 224 cm2. Its height is 7 cm and the base is a square. Find
(i) side of the square base
(ii) the volume of the cuboid.

Lateral surface area of a cuboid is 224 cm2
Height (h) = 7 cm
(i) 2(l + b) × h = 224
⇒ 2(l + b) × 7 = 224
l + b = $\frac{224}{7 \times 2}$ = 16 cm
But l = b (∵ The base of cuboid is a square)
∴ 2 × side = 16 cm
⇒ Side = $\frac{16}{2}$ = 8 cm
(ii) Volume of cuboid = lbh = 8 × 8 × 7 cm3 = 448 cm3

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 m3 of wood costs ₹5400.

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 m3 – 2 × 1.2 × 0.75m3
= 2.05 – 1.80 = 0.25 m3
Cost = ₹5400 for 1 m3
Total cost = ₹5400 × 0.25 = ₹5400 × $\frac{25}{100}$
= ₹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?

Capacity of car tank = 40 cm × 28 cm × 25 cm = (40 × 28 × 25) cm3
$\frac{40 \times 28 \times 25}{1000}$ litre (∵ 1000 cm3 = 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?

Diameter of a garden rolle = 1.4 m
∴ Radius (r) = $\frac{1.4}{2}$ = 0.7 m = 70 cm
and length (h) = 2m
∴ Curved surface area = 2πrh
= 2 × $\frac{22}{7}$ × 70 × 200 cm2
= 88000 cm2
∴ Area covered in 5 revolutions = $\frac{88000 \times 5}{10000}$ m2 = 44 m2

Question 18.
The capacity of an open cylindrical tank is 2079 m3 and the diameter of its base is 21m. Find the cost of plastering its inner surface at ₹40 per square metre.

Capacity of an open cylindrical tank = 2079 m3
Diameter of base = 21 m
∴ Radius (r) = $\frac{21}{2}$ 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.

Height of solid right circular cylinder = 1.21 m = 121 cm
and diameter = 28 cm
∴ Radius (r) = $\frac{28}{2}$ = 14 cm
Volume of the metal used = πr2h
$\frac{22}{2}$ × 14 × 14 × 121 cm3
= 74536 cm3
∴ Volume of 7 solid cubes = 74536 cm3
Volume of 1 cube = $\frac{74536}{7}$ = 10648 cm3

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 m2, find the cost of plastering.

(i) Depth of a well (h) = 20 m
and diameter = 2 m
Radius (r) = $\frac{2}{2}$ = 1 m
Volume of earth dug out = πr2h

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