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.**

**(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 cm^{2}

**(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

11^{2} = (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.**

**Answer :**

**(a)**

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

= (9 ×9) – (½ ×5×6)×2

= 81 – (15 ×2)

= 81 – 30

= 51 cm^{2}

**(b)** In △ABD

Using Pythagoras theorem

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

BD^{2} = 6^{2} + 8^{2}

⇒ BD^{2} = 36 + 64

BD^{2} = 100

⇒ BD = 10 cm

In △BCD

Using Pythagoras theorem

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

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

⇒ 676 = 100 + CD^{2}

CD^{2} = 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 cm^{2}

**(c)**

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

Hence, the required area of given figure = 54 m^{2}.

**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.**

**Answer :**

Consider M as the midpoint of AB

AM = MB = 1 cm

Now join MN and CN

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

Area of △GHF

Here

= ¾ × 3.316

= 3 × 0.829

= 2.487

= 2.48 cm^{2}

Area of rectangle II (MCFH) = l × b

= 6.5 × 1

= 6.5 cm^{2}

Area of △ III + IV = 2×½ ×6×1.5 = 9 cm^{2}

Area of three equilateral triangles formed trapezium III = 3 × √3/4 × 1^{2}

= ¾ ×1.732

= 3 ×0.433

= 1.299

= 1.3 cm^{2}

Total area = 2.48 + 6.50 + 9 + 1.30

= 19.28

= 19.3 cm^{2}

**Question 4. ****If the area of a circle is 78.5 cm**^{2}, find its circumference. (Take π = 3.14)

^{2}, find its circumference. (Take π = 3.14)

**Answer :**

Area of a circle = 78.5 cm^{2}

Consider r as the radius

r^{2} = Area/π

r^{2} = 78.50/3.14

⇒ r^{2} = 25 = 5^{2}

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 cm^{2}

Consider r as the radius

πr^{2} = 154

⇒ 22/7 r^{2} = 154

r^{2} = (154×7)/22 = 49 = 7^{2}

⇒ r = 7 cm

Side of square = 7 × 2 = 14 cm

So the area of the original cardboard = a^{2}

= 14^{2}

= 196 cm^{2}

**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

Area = l × b

= 200 × 150

= 30000 cm^{2}

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 = πr^{2}

= 3.14 × 5 × 5 cm^{2}

Area of 300 circles = 300 × 314/100 × 25 = 23550 cm^{2}

So,

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

= 30000 – 23550

= 6450 cm^{2}

**(b) **Diameter of semicircular protractor = 14 cm

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 (R^{2} – r^{2}) = 22/7 ×[(17.5)^{2} – (14)^{2}]

= 22/7 (17.5 + 14) ×(17.5 – 14)

= 22/7 × 31.5 × 3.5

= 246.5 m^{2}

Rate of paving the road = ₹ 60 per m^{2}

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.**

**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 (R^{2} – r^{2})]

= 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 m^{2}

**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)**

**Answer :**

In a semicircle ∠ACB = 90°

△ ABC is a right-angled triangle

Using Pythagoras theorem

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

= 12^{2} + 16^{2}

= 144 + 256

= 400

AB^{2} = (20)^{2}

⇒ AB = 20 cm

Radius of semicircle = 20/2 = 10 cm

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

= ½ πr^{2} – (AC × BC)/ 2

= ½ × 3.142 (10)^{2} – (12 × 16)/2

= 314.2/2 – 96

= 157.1 – 96

= 61.1 cm^{2}

**(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).**

**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

OB = 2 + 1 = 3 cm

AB = 1 cm

Using Pythagoras theorem

OA = √OB^{2} + AB^{2}

OA = √(3)^{2} + (1)^{2}

= √9 + 1

= √10

So the radius of the circle = √10 cm

Area of the circle = πr^{2}

= 3.14 × (√10)^{2}

= 3.14 × 10

= 31.4 cm^{2}

Area of 5 square of side 2 cm each = 2^{2} × 5

= 4 × 5

= 20 cm^{2}

So the area of shaded portion = 31.4 – 20 = 11.4 cm^{2}

**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**

**cm2, find the radius of the quadrants. Take π = .**

**Answer :**

**(a) **Radius of quadrant = 7 cm

OA = 3 cm, OB = 4 cm

AX = 7 – 3 = 4 cm

BY = 7 – 4 = 3 cm

AB^{2} = OA^{2} + OB^{2}

= 3^{2} + 4^{2}

= 9 + 16

= 25

AB = √25 = 5 cm

**(i) **Area of shaded portion = ¼ πr^{2} – ½ OA × OB

= ¼ × 22/7 × 7^{2} – ½ × 3 × 4

= ¼ × 22/7 × 49 – 6

= 77/2 – 6

= 65/2

= 32.5 cm^{2}

**(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 = a^{2} – 4 × [¼ π(a/2)^{2}]

= a^{2} – (4 × ¼ π× a^{2}/4)

= a^{2} – (22/7 × a^{2}/4)

= a^{2} – 11a^{2}/14

= 3a^{2}/14

Here

Area of shaded portion = 21 3/7 = 150/7 cm^{2}

By equating both we get

3a^{2}/14

= 150/7

a^{2} = 150/7 × 14/3

⇒ a^{2} = 100

= 10^{2}

⇒ 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.**

**Answer :**

ABC is a right angled triangle right angled at B

Using Pythagoras theorem

AC^{2} = AB^{2} + BC^{2}** …(i)**

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

= ½ π × AC^{2}/4

= πAC^{2}/8

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

= ½ π × AB^{2}/4

= πAB^{2}/8

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

= ½ π × BC^{2}/4

= πBC^{2}/8

πAB^{2}/8 + πBC^{2}/8 = π/8 (AB^{2} + BC^{2})

From equation (i)

= π/8 (AC^{2})

= πAC^{2}/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 = πr^{2} × 15/60

= 22/7 × 14 × 14 × ¼

= 154 cm^{2}

**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 = 90^{0}

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

= (3.5×4)/2

= 7 cm

Area of the sector = πr^{2} × 90°/360°

By calculation

= 22/7 × 7 × 7 × ¼

= 77/2

= 38.5 cm^{2}

**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 a^{2}

= 6 × (28)^{2}

= 6 × 28 × 28

= 4704 cm^{2}

Diameter of each circle = 28 cm

So the radius = 28/2 = 14 cm

Area of each circle = πr^{2}

= 22/7 × 14 × 14

= 616 cm^{2}

Area of such 6 circles drawn on 6 faces of cube = 616 × 6 = 3696 cm^{2}

Area of remaining portion of the cube = 4704 – 3696 = 1008 cm^{2}

**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) cm^{3}

Here 1000 cm^{3} = 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 cm^{3}

= (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 cm^{2}

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 cm^{3}

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

^{3}, 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 = a^{3} and S = 6a^{2}

Diagonal (d) = √3. a

Sd = 6a^{2} × √3a = 6√3a^{3}

Here, V = a^{3}

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).**

**Answer :**

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

Volume of part (3) = 50×40×12 = 24000 cm^{3}

Volume of part (1) = 50×40×(16 + 24)

= 50×40×40

= 80000 cm^{3}

Volume of part (2) = 50×40×24 = 48000 cm^{3}

So the total volume = 24000 + 8000 + 48000 = 153000 cm^{3}

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

= 2(50×12 + 50×40 + 50×24) cm^{2} + (12×40 + 28×40 + 16×40 + 24×40) cm^{2} + 3(50×40) cm^{2}

= 2 (600 + 2000 + 1200) cm^{2} + (480 + 1120 + 640 + 960) cm^{2} + (3×2000) cm^{2}

= 2 (3800) + 3200 + 6000 cm^{2}

= 7600 + 3200 + 6000

= 16800 cm^{2}

**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 cm^{3}

**(ii)** volume of box (Internal) = 564000 cm^{3}

External volume of box = 98 cm × 84 cm × 77 cm = 633864 cm^{3}

So,

the volume of wood used in making the box = 633864 – 564000 = 69864 cm^{3}

**(iii)** Weight of 1 cm^{3} wood = 0.8 gm

So the weight of 69864 cm^{3} 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) cm^{3}

Volume of the cube having edge 4 cm = 4 × 4 × 4 = 64 cm^{3}

**(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 cm^{2}

So the total surface area of 450 cubes = 450 × 96 = 43200 cm^{2}

Cost of silver coating of the surface for 1 cm^{2} = ₹ 1.25

Cost of silver coating the surface for 43200 cm^{2} = 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 = edge^{3}

= (3 cm)^{3}

= 3 × 3 × 3

= 27 cm^{3}

Volume of second cube = edge^{3}

= (4 cm)^{3}

= 4 × 4 × 4

= 64 cm^{3}

Volume of third cube = edge^{3}

= (5 cm)^{3}

= 5 × 5 × 5

= 125 cm^{3}

So the total volume = 27 + 64 + 125

= 216 cm^{3}

Make new cube whose volume = 216 cm^{3}

Edge^{3} = 216 cm^{3}

Edge^{3} = (6 cm)^{3}

Edge = 6 cm

Surface area of new cube = 6 (edge)^{2}

= 6 (6)^{2}

= 6 × 6 × 6

= 216 cm^{2}

Given

Cost of coating the surface for 1 cm^{2} = ₹ 3.50

So,

the cost of coating the surface for 216 cm^{2} = 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**

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