ML Aggarwal Mensuration Chapter Test Class 9 ICSE Maths Solutions

ML Aggarwal Mensuration Chapter Test Class 9 ICSE Maths Solutions Ch-16. Step by Step Solutions of Ch-Test Questions on Mensuration of ML Aggarwal for ICSE Class 9th Mathematics. Visit official website CISCE for detail information about ICSE Board Class-9.

ML Aggarwal Mensuration Chapter Test Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-16 Mensuration
Topics Solution of Ch-Test Questions
Academic Session 2024-2025

Chapter Test Solutions of ML Aggarwal for ICSE Class-9 Ch-16, Mensuration

Question 1.

(a) Calculate the area of the shaded region.
(a) Calculate the area of the shaded region
(b) If the sides of a square are lengthened by 3 cm, the area becomes 121 cm2. Find the perimeter of the original square.

Answer :

(a) According to figure,

OA is perpendicular to BC

AC = 15 cm, AO = 12 cm, BO = 5 cm, BC = 14 cm

OC = BC – BO = 14 – 5 = 9 cm

Area of right △AOC = ½ × base × altitude

= ½ × 9 × 12

= 54 cm2

(b) Consider the side of original square = x cm

So the length of given square = (x + 3) cm

Area = side × side

121 = (x + 3) (x + 3)

It can be written as

112 = (x + 3)2

11 = x + 3

⇒ x = 11 – 3

⇒ x = 8 cm

Question 2.

(a) Find the area enclosed by the figure (i) given below. All measurements are in centimetres:
(b) Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimetres.
(c) Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in metres.
(a) Find the area enclosed by the figure (i) given below. All measurements are in centimetres

Answer :

(a)

Find the area enclosed by the figure (i) given below. All measurements are in centimetres.

Area of figure (i) = Area of ABCD – Area of both triangles

= (9 ×9) – (½ ×5×6)×2

= 81 – (15 ×2)

= 81 – 30

= 51 cm2

(b) In △ABD

 Find the area of the quadrilateral ABCD shown in figure (ii) given below. All measurements are in centimetres.

Using Pythagoras theorem

BD2 = AB2 + AD2

BD2 = 62 + 82

⇒ BD2 = 36 + 64

BD2 = 100

⇒ BD = 10 cm

In △BCD

Using Pythagoras theorem

BC2 = BD2 + CD2

262 = 102 + CD2

⇒ 676 = 100 + CD2

CD2 = 676 – 100 = 576

⇒ CD = √576

⇒ CD = 24 cm

Here,

Area of the given figure = Area of △ABD + Area of △BCD

Area of the given figure = (½ × base × height) + (½ × base × height)

Area of the given figure = (½ ×AB ×AD) + (½ ×CD ×BD)

Area of the given figure = (½ ×6 ×8) + (½ ×24 ×10)

Area of the given figure = (3×8) + (12×10)

= 24 + 120

= 144 cm2

(c)

Calculate the area of the shaded region shown in figure (iii) given below. All measurements are in metres.

Area of the figure (iii) = Area of ABCD – (Area of 1st part + Area of 2nd part + Area of 3rd part)

It can be written as

= (AB×BC) – [(1/2 × base × height) + (1/2 × base × height) + ½(sum of parallel side × height)]

= (12×12) – [(1/2 ×5×5) + (½ ×5×7) + {½(7+3)×12]}

= 144 – [25/2 + 35/2 + (10×6)]

= 144 – (60/2 + 60)

= 144 – (30 + 60)

= 144 – 90

= 54 m2

Hence, the required area of given figure = 54 m2.

Question 3.  Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.

Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.

Answer :

Consider M as the midpoint of AB

AM = MB = 1 cm

Asifa cut an aeroplane from a coloured chart paper (as shown in the adjoining figure). Find the total area of the chart paper used, correct to 1 decimal place.

Now join MN and CN

△AMN, △NCB and △MNC are equilateral triangles having 1 cm side each

Area of △GHF

ml class 9 Mensuration img 84

Here

= ¾ × 3.316

= 3 × 0.829

= 2.487

= 2.48 cm2

Area of rectangle II (MCFH) = l × b

= 6.5 × 1

= 6.5 cm2

Area of △ III + IV = 2×½ ×6×1.5 = 9 cm2

Area of three equilateral triangles formed trapezium III = 3 × √3/4 × 12

= ¾ ×1.732

= 3 ×0.433

= 1.299

= 1.3 cm2

Total area = 2.48 + 6.50 + 9 + 1.30

= 19.28

= 19.3 cm2

Question 4. If the area of a circle is 78.5 cm2, find its circumference. (Take π = 3.14)

Answer :

Area of a circle = 78.5 cm2

Consider r as the radius

r2 = Area/π

r2 = 78.50/3.14

⇒ r2 = 25 = 52

r = 5 cm

Here,

Circumference = 2 πr

= 2 × 3.14 × 5

= 31.4 cm

Question 5. From a square cardboard, a circle of biggest area was cut out. If the area of the circle is 154 cm2, calculate the original area of the cardboard.

Answer :

Area of circle cut out from the square board = 154 cm2

Consider r as the radius

πr2 = 154

⇒ 22/7 r2 = 154

r2 = (154×7)/22 = 49 = 72

⇒ r = 7 cm

Side of square = 7 × 2 = 14 cm

So the area of the original cardboard = a2

= 142

= 196 cm2

Question 6.

(a) From a sheet of paper of dimensions = 2m x 1.5m, how many circles can you cut of radius 5cm. Also find the area of the paper wasted. Take π = 3.14.
(b) If the diameter of a semicircular protractor is 14cm, then find its perimeter.

Answer :

(a) Length of sheet of paper = 2 m = 200 cm

Breadth of sheet = 1.5 m = 150 cm

From a sheet of paper of dimensions 2 m × 1.5 m, how many circles of radius 5 cm can be cut? Also find the area of the paper wasted. Take π = 3.14.

Area = l × b

= 200 × 150

= 30000 cm2

Radius of circle = 5 cm

Number of circles in lengthwise = 200/(5×2) = 20

Number of circles in widthwise = 150/10 = 15

So,

the number of circles = 20 × 15 = 300

Here

Area of one circle = πr2

= 3.14 × 5 × 5 cm2

Area of 300 circles = 300 × 314/100 × 25 = 23550 cm2

So,

the area of remaining portion = area of square – area of 300 circles

= 30000 – 23550

= 6450 cm2

(b) Diameter of semicircular protractor = 14 cm
If the diameter of a semi-circular protractor is 14 cm, then find its perimeter.

Perimeter = ½ πd + d

= (½× 22/7 ×14) + 14

= 22 + 14

= 36 cm


Mensuration Exercise-Chapter Test

ML Aggarwal Class 9 ICSE Maths Solutions

Page 400

Question 7. A road 3.5 m wide surrounds a circular park whose circumference is 88 m. Find the cost of paving the road at the rate of Rs. 60 per square metre.

Answer :

Width of the road = 3.5 m

Circumference of the circular park = 88 m

Consider r as the radius of the park

2 πr = 88

2 × 22/7 ×r = 88

r = (88×7)/(2×22) = 14 m

Here

Outer radius (R) = 14 + 3.5 = 17.5 m

Area of the path = 22/7 × (17.5+14)×(17.5–14)

π (R2 – r2) = 22/7 ×[(17.5)2 – (14)2]

= 22/7 (17.5 + 14) ×(17.5 – 14)

= 22/7 × 31.5 × 3.5

= 246.5 m2

Rate of paving the road = ₹ 60 per m2

So ,

the total cost = 60 × 346.5 = ₹ 20790

Question 8. The adjoining sketch shows a running tract 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.

The adjoining sketch shows a running tract 3.5 m wide all around which consists of two straight paths and two semicircular rings. Find the area of the track.
Answer :

Width of track = 3.5 m

Inner length of rectangular base = 140 m

Width = 42 m

Outer length of rectangular base = 140 + (2×3.5)

= 140 + 7 = 147 m

Width = 42 + (2×3.5)

= 42 + 7 = 49 m

Radius of inner semicircle (r) = 42/2 = 21 m

Outer radius (R) = 21 + 3.5 = 24.5 m

Here

Area of track = [2 ×(140×3.5)] + [2× ½ π (R2 – r2)]

= 2 (490) + 22/7 [(24.5)2 – (21)2]

= 980 + 22/7 (24.5 + 21) (24.5 – 21)

= 980 + (22/7 × 45.5×3.5)

= 980 + 500.5

= 1480.5 m2

Question 9. In the adjoining figure, O is the centre of a circular arc and AOB is a line segment. Find the perimeter and the area of the shaded region correct to one decimal place. (Take π = 3.142)

In the adjoining figure, O is the centre of a circular arc and AOB is a line segment. Find the perimeter and the area of the shaded region correct to one decimal place

Answer :

In a semicircle ∠ACB = 90°

△ ABC is a right-angled triangle

Using Pythagoras theorem

AB2 = AC2 + BC2

= 122 + 162

= 144 + 256

= 400

AB2 = (20)2

⇒ AB = 20 cm

Radius of semicircle = 20/2 = 10 cm

(i) Area of shaded portion = Area of semicircle – Area of △ ABC

= ½ πr2 – (AC × BC)/ 2

= ½ × 3.142 (10)2 – (12 × 16)/2

= 314.2/2 – 96

= 157.1 – 96

= 61.1 cm2

(ii) Here, Perimeter of shaded portion = circumference of semicircle + AC + BC

= πr + 12 + 16

= (3.142×10) + 28

= 31.42 + 28

= 59.42 cm

= 59.4 cm

Question 10. 

(a) In the figure (1) given below, the radius is 3.5 cm. Find the perimeter of the quarter of the circle.
(b) In the figure (ii) given below, there are five squares each of side 2 cm.
(i) Find the radius of the circle.
(ii) Find the area of the shaded region. (Take π= 3.14).

(a) In the figure (1) given below, the radius is 3.5 cm. Find the perimeter of the quarter of the circle

Answer :

(a) Radius of quadrant = 3.5 cm

Perimeter = 2r + (¼×2 πr)

= 2r + (½×πr)

= (2×3.5) + (½× 22/7 ×3.5)

= 7 + 5.5

= 12.5 cm

(b) According to the  figure
(b) In the figure (ii) given below, there are five squares each of side 2 cm.

OB = 2 + 1 = 3 cm

AB = 1 cm

Using Pythagoras theorem

OA = √OB2 + AB2

OA = √(3)2 + (1)2

= √9 + 1

= √10

So the radius of the circle = √10 cm

Area of the circle = πr2

= 3.14 × (√10)2

= 3.14 × 10

= 31.4 cm2

Area of 5 square of side 2 cm each = 22 × 5

= 4 × 5

= 20 cm2

So the area of shaded portion = 31.4 – 20 = 11.4 cm2

Question 11.

(a) In the figure (i) given below, a piece of cardboard in the shape of a quadrant of a circle of radius 7 cm is bounded by perpendicular radii OX and OY. Points A and B lie on OX and OY respectively such that OA = 3 cm and OB = 4 cm. The triangular part OAB is removed. Calculate the area and the perimeter of the remaining piece.

(b) In the figure (ii) given below, ABCD is a square. Points A, B, C and D are centres of quadrants of circles of the same radius. If the area of the shaded portion is 21\frac { 3 }{ 7 }
cm2, find the radius of the quadrants. Take π = \frac { 22 }{ 7 }.

Answer :

(a) Radius of quadrant = 7 cm

OA = 3 cm, OB = 4 cm

AX = 7 – 3 = 4 cm

BY = 7 – 4 = 3 cm

AB2 = OA2 + OB2

= 32 + 42

= 9 + 16

= 25

AB = √25 = 5 cm

(i) Area of shaded portion = ¼ πr2 – ½ OA × OB

= ¼ × 22/7 × 72 – ½ × 3 × 4

= ¼ × 22/7 × 49 – 6

= 77/2 – 6

= 65/2

= 32.5 cm2

(ii) Perimeter of shaded portion = ¼ × 2 πr + AX + BY + AB

= ½ × 22/7 × 7 + 4 + 3 + 5

= 11 + 12

= 23 cm

(b) ABCD is a square with centres A, B, C and D quadrants drawn.

Consider a as the side of the square

Radius of each quadrant = a/2

Here,

Area of shaded portion = a2 – 4 × [¼ π(a/2)2]

= a2 – (4 × ¼ π× a2/4)

= a2 – (22/7 × a2/4)

= a2 – 11a2/14

= 3a2/14

Here

Area of shaded portion = 21 3/7 = 150/7 cm2

By equating both we get

3a2/14

= 150/7

a2 = 150/7 × 14/3

⇒ a2 = 100

= 102

⇒ a = 10 cm

So the radius of each quadrant = a/2

= 10/2

= 5 cm


Mensuration Exercise-Chapter Test

ML Aggarwal Class 9 ICSE Maths Solutions

Page 401

Question 12. In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircle are drawn on AB, BC and CA as diameter. Show that the sum of areas of semi circles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.

. In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircles are drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.

Answer :

ABC is a right angled triangle right angled at B

Using Pythagoras theorem

AC2 = AB2 + BC2 …(i)

. In the adjoining figure, ABC is a right angled triangle right angled at B. Semicircles are drawn on AB, BC and CA as diameter. Show that the sum of areas of semicircles drawn on AB and BC as diameter is equal to the area of the semicircle drawn on CA as diameter.

Area of semicircle on AC as diameter = ½ π (AC/2)2

= ½ π × AC2/4

= πAC2/8

Area of semicircle on AB as diameter = ½ π (AB/2)2

= ½ π × AB2/4

= πAB2/8

Area of semicircle on BC as diameter = ½ π (BC/2)2

= ½ π × BC2/4

= πBC2/8

πAB2/8 + πBC2/8 = π/8 (AB2 + BC2)

From equation (i)

= π/8 (AC2)

= πAC2/8

Hence, it is proved.

Question 13.  The length of minute hand of a clock is 14 cm. Find the area swept by the minute hand in 15 minutes.

Answer :

Radius of hand = 14 cm

So ,

the area swept in 15 minutes = πr2 × 15/60

= 22/7 × 14 × 14 × ¼

= 154 cm2

Question 14. Find the radius of a circle if a 90° arc has a length of 3.5 n cm. Hence, find the area of sector formed by this arc.

Answer :

Length of arc of the sector of a circle = 3.5 π cm

Angle at the centre = 900

Radius of the arc = 3.5 π/2 π × 360/90

= (3.5×4)/2

= 7 cm

Area of the sector = πr2 × 90°/360°

By calculation

= 22/7 × 7 × 7 × ¼

= 77/2

= 38.5 cm2

Question 15. A cube whose each edge is 28 cm long has a circle of maximum radius on each of its face painted red. Find the total area of the unpainted surface of the cube.

Answer :

Edge of cube = 28 cm

Surface area = 6 a2

= 6 × (28)2

= 6 × 28 × 28

= 4704 cm2

Diameter of each circle = 28 cm

So the radius = 28/2 = 14 cm

Area of each circle = πr2

= 22/7 × 14 × 14

= 616 cm2

Area of such 6 circles drawn on 6 faces of cube = 616 × 6 = 3696 cm2

Area of remaining portion of the cube = 4704 – 3696 = 1008 cm2

Question 16.  Can a pole 6.5 m long fit into the body of a truck with internal dimensions of 3.5m, 3 m and 4m?

Answer :

No, a pole 6.5 m long cannot fit into the body of a truck with internal dimensions of 3.5 m, 3 m and 4 m

Length of pole = 6.5 m

Internal dimensions of truck are 3.5 m, 3 m and 4 m

So the pole cannot fit into the body of truck with given dimensions.

Question 17. A car has a petrol tank 40 cm long, 28 cm wide and 25 cm deep. If the fuel 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) cm3

Here 1000 cm3 = 1 litre

= (40×28×25)/1000 litre

Average of car = 13.5 km per litre

Distance travelled by car = (40×28×25)/1000 × 13.5

Multiply and divide by 10

= (40 × 25) × 28/1000 × 135/10

= (1×28)/1 × 135/10

= (14× 135)/5

= 14 × 27

= 378 km

Hence,

The car can travel 378 km with a full tank of petrol.

Question 18. An aquarium took 96 minutes to completely fill with water. Water was filling the aquarium at a rate of 25 litres every 2 minutes. Given that the aquarium was 2 m long and 80 cm wide, compute the height of the aquarium.

Answer :

Water filled in 2 minutes = 25 litres

Water filled in 1 minute = 25/2 litres

Water filled in 96 minutes = 25/2 × 96

= 25 × 48

= 1200 litres

So ,

the capacity of aquarium = 1200 litres …(1)

Length of aquarium = 2m = 2 × 100 = 200 cm

Breadth of aquarium = 80 cm

Consider h cm as the height of aquarium

So the capacity of aquarium = 200 × 80 × h cm3

= (200×80×h)/1000 litre

= 1/5 × 80 × h litre

= 16 h litre …(2)

Using equation (1) and (2)

16h = 1200

h = 1200 / 16

⇒ h = 75 cm

Hence, height of aquarium = 75 cm.

Question 19. The lateral surface area of a cuboid is 224 cm2. Its height is 7 cm and the base is a square. Find :

(i) a side of the square, and
(ii) the volume of the cuboid.

Answer :

Lateral surface area of a cuboid = 224 cm2

Height of cuboid = 7 cm

Base is square

x cm as the breadth of cuboid (Since the base is square both length and breadth are same)

Here

Lateral surface area = 2 (l + b) × h

224 = 2 (x + x) × 7

224 = 2 × 2x × 7

⇒ 224 = 28x

28x = 224

⇒ x = 224/28 = 8 cm

(i) Side of the square = 8 cm
(ii) Volume of the cuboid = l × b × h

= 8 × 8 × 7

= 448 cm3

Question 20. If the volume of a cube is V m3, its surface area is S m2 and the length of a diagonal is d metres, prove that 6√3 V = S d.

Answer  :

Volume of cube = (V) = (Side)3

Consider a as the side of cube

V = a3 and S = 6a2

Diagonal (d) = √3. a

Sd = 6a2 × √3a = 6√3a3

Here, V = a3

Sd = 6√3V

Hence, 6√3V = Sd.

Question 21. The adjoining figure shows a victory stand, each face is rectangular. All measurement are in centimetres. Find its volume and surface area (the bottom of the stand is open).

The adjoining figure shows a victory stand, each face is rectangular. All measurement are in centimetres. Find its volume and surface area (the bottom of the stand is open)

Answer :

Three parts are indicated as 3, 1 and 2 in this figure

Volume of part (3) = 50×40×12 = 24000 cm3

Volume of part (1) = 50×40×(16 + 24)

= 50×40×40

= 80000 cm3

Volume of part (2) = 50×40×24 = 48000 cm3

So the total volume = 24000 + 8000 + 48000 = 153000 cm3

Total surface area = Area of front and back + Area of vertical faces + Area of top faces

= 2(50×12 + 50×40 + 50×24) cm2 + (12×40 + 28×40 + 16×40 + 24×40) cm2 + 3(50×40) cm2

= 2 (600 + 2000 + 1200) cm2 + (480 + 1120 + 640 + 960) cm2 + (3×2000) cm2

= 2 (3800) + 3200 + 6000 cm2

= 7600 + 3200 + 6000

= 16800 cm2

Question 22. The external dimensions of an open rectangular wooden box are 98 cm by 84 cm by 77 cm. If the wood is 2 cm thick all around, find :

(i) the capacity of the box
(ii) the volume of the wood used in making the box, and
(iii) the weight of the box in kilograms correct to one decimal place, given that 1 cm3 of wood weighs 0.8 gm.

Answer :

Open rectangular wooden box (external) = 98 cm, 84 cm and 77 cm

Thickness = 2 cm

So,

the internal dimensions of open rectangular wooden box = (98 – 2×2) cm, (84 – 2×2) cm and (77 – 2) cm

= (98 – 4) cm, (84 – 4) cm, 75 cm

= 94 cm, 80 cm, 75 cm

(i)  Capacity of the box = 94 cm × 80 cm × 75 cm

= 564000 cm3

(ii) volume of box (Internal) = 564000 cm3

External volume of box = 98 cm × 84 cm × 77 cm = 633864 cm3

So,

the volume of wood used in making the box = 633864 – 564000 = 69864 cm3

(iii) Weight of 1 cm3 wood = 0.8 gm

So the weight of 69864 cm3 wood = 0.8 × 69864 gm

= (0.8 × 69864)/1000 kg

= 55891.2/1000 kg

= 55.9 kg

(correct to one decimal place)

Question 23. 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 Rs. 1.25 per square centimetre ?

Answer :

Dimensions of the cuboidal block = 36 cm, 32 cm and 0.25 m

= 36 cm ×32 cm ×(0.25×100) cm

= (36 × 32 × 25) cm3

Volume of the cube having edge 4 cm = 4 × 4 × 4 = 64 cm3

(i) Number of cubes = Volume of cuboidal block/ Volume of one cube

= (36×32×25)/64

= (36×25)/2

= 18 × 25

= 250

(ii) Total surface area of one cube = 6(a)2

= 6 (4)2

= 6 × 4 × 4

= 96 cm2

So the total surface area of 450 cubes = 450 × 96 = 43200 cm2

Cost of silver coating  of the surface for 1 cm2 = ₹ 1.25

Cost of silver coating the surface for 43200 cm2 = 43200 × 1.25

= ₹ 54000

Question 24. 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 Rs. 3.50 per square centimetre.

Answer :

Volume of first cube = edge3

= (3 cm)3

= 3 × 3 × 3

= 27 cm3

Volume of second cube = edge3

= (4 cm)3

= 4 × 4 × 4

= 64 cm3

Volume of third cube = edge3

= (5 cm)3

= 5 × 5 × 5

= 125 cm3

So the total volume = 27 + 64 + 125

= 216 cm3

Make new cube whose volume = 216 cm3

Edge3 = 216 cm3

Edge3 = (6 cm)3

Edge = 6 cm

Surface area of new cube = 6 (edge)2

= 6 (6)2

= 6 × 6 × 6

= 216 cm2

Given

Cost of coating the surface for 1 cm2 = ₹ 3.50

So,

the cost of coating the surface for 216 cm2 = 3.50 × 216

= ₹ 756

—  : End of ML Aggarwal Mensuration Chapter Test Class 9 ICSE Maths Solutions :–

Return to :-   ML Aggarawal Maths Solutions for ICSE  Class-9

Thanks

Please Share with Your Friends

Leave a Comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.