ML Aggarwal Mensuration Exe-16.1 Class 9 ICSE Maths Solutions

ML Aggarwal Mensuration Exe-16.1 Class 9 ICSE Maths Solutions Ch-16. Step by Step Solutions of 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 Exe-16.1 Class 9 ICSE Maths Solutions

Board ICSE
Subject Maths
Class 9th
Chapter-16 Mensuration
Topics Solution of Exe-16.1 Questions
Academic Session 2024-2025

Solution of Exe-16.1 Questions

ML Aggarwal Mensuration Exe-16.1 Class 9 ICSE Maths Solutions Ch-16

Question 1. Find the area of a triangle whose base is 6 cm and corresponding height is 4 cm.

Answer : Base of triangle = 6 cm

Height of triangle = 4 cm

We know that, Area of triangle = ½ × base × height

Substituting the values

= ½ × 6 × 4

By further calculation

= 6 × 2

= 12 cm2

Question 2. Find the area of a triangle whose sides are

(i) 3 cm, 4 cm and 5 cm
(ii) 29 cm, 20 cm and 21 cm
(iii) 12 cm, 9.6 cm and 7.2 cm

Answer :

(i) Consider a = 3 cm, b = 4 cm and c = 5 cm

We know that, S = Semi perimeter = (a + b + c)/ 2

= (3 + 4 + 5)/ 2

= 12/2

= 6 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 2

= 6 cm2

(ii) Consider a = 29 cm, b = 20 cm and c = 21 cm

We know that, S = Semi perimeter = (a + b + c)/ 2

= (29 + 20 + 21)/ 2

= 70/2

= 35 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 3

So we get

= 7 × 5 × 3 × 2

= 210 cm2

(iii) Consider a = 12 cm, b = 9.6 cm and c = 7.2 cm

We know that, S = Semi perimeter = (a + b + c)/ 2

= (12 + 9.6 + 7.2)/ 2

= 28.8/2

= 14.4 cm

Here

ml class 9 Mensuration img 1

ml class 9 Mensuration img 4

So we get

= 2.4 × 2.4 × 6

= 34.56 cm2

Question 3. Find the area of a triangle whose sides are 34 cm, 20 cm and 42 cm. Hence, find the length of the altitude corresponding to the shortest side.

Answer :  Consider 34 cm, 20 cm and 42 cm as the sides of triangle

a = 34 cm, b = 20 cm and c = 42 cm

We know that, S = Semi perimeter = (a + b + c)/ 2

= (34 + 20 + 42)/ 2

= 96/2

= 48 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 5

So ,  = 14 × 6 × 4

= 336 cm2

Here the shortest side of the triangle is 20 cm

Consider h cm as the corresponding altitude

Area of triangle = ½ × base × height

336 = ½ × 20 × h

By calculation

h = (336 × 2)/ 20

So ,  h = 336/10

h = 33.6 cm

Hence, the required altitude of the triangle is 33.6 cm.

Question 4. The sides of a triangular field are 975m, 1050 m and 1125 m. If this field is sold at the rate of Rs. 1000 per hectare, find its selling price. [1 hectare = 10000 m²].

Answer :   Given that,  a = 975 m, b = 1050 m and c = 1125 m

We know that, S = Semi perimeter = (a + b + c)/ 2

= (975 + 1050 + 1125)/ 2

= 3150/2

= 1575 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 6

= 525 × 450 × 2

It is given that, 1 hectare = 10000 m2

= (525 × 900)/ 10000

By calculation

= (525 × 9)/ 100

= 4725/100

= 47.25 hectares

We know that, Selling price of 1 hectare field = Rs 1000

Selling price of 47.25 hectare field = 1000 × 47.25 = Rs 47250

Question 5. The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter.

The base of a right angled triangle is 12 cm and its hypotenuse is 13 cm long. Find its area and the perimeter

Answer :  Given that,

ABC is a right angled triangle

BC = 12 cm and AB = 13 cm

Use the Pythagoras theorem

AB2 = AC2 + BC2

132 = AC2 + 122

By calculation

AC2 = 132 – 122

So ,  AC2 = 169 – 144 = 25

AC = √25 = 5 cm

We know that, Area of triangle ABC = ½ × base × height

= ½ × 12 × 5

= 30 cm2

Similarly

Perimeter of triangle ABC = AB + BC + CA

= 13 + 12 + 5

= 30 cm

Question 6. Find the area of an equilateral triangle whose side is 8 m. Given your answer correct to two decimal places.

Answer :  Given that,

Side of equilateral triangle = 8 m

We know that, Area of equilateral triangle = √3/4 (side)2

= √3/4 × 8 × 8

By calculation

= √3 × 2 × 8

= 1.73 × 16

= 27.71 m2

Question 7. If the area of an equilateral triangle is 81√3 cm² find its. perimeter.

Answer :  Area of equilateral triangle = √3/4 (side)2

81 √3 = √3/4 (side)2

By calculation

Side2 = (81 √3 × 4)/ √3

So,  (side)2 = 81 × 4

side = √(81 × 4)

side = 9 × 2 = 18 cm

So the perimeter of equilateral triangle = 3 × side

= 3 × 18    = 54 cm

Question 8. If the perimeter of an equilateral triangle is 36 cm, calculate its area and height.

Answer :  The perimeter of an equilateral triangle = 3 × side

ml class 9 Mensuration img 7

, 36 = 3 × side

By calculation

side = 36/3 = 12 cm

So AB = BC = CA = 12 cm

Here, Area of equilateral triangle = √3/4 (side)2

= √3/4 (12)2

By calculation

= √3/4 × 12 × 12

So ,  = √3 × 3 × 12

= 1.73 × 36

= 62.4 cm2

In triangle ABD

Using Pythagoras Theorem

AB2 = AD2 + BD2

Here BD = 12/2 = 6 cm

122 = AD2 + 62

By calculation

144 =AD2 + 36

AD2 = 144 – 36 = 108

So we get

AD = √108 = 10.4

Hence, the required height is 10.4 cm.

Question 9. (i) If the length of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 48 cm, find its area.  (ii) The sides of a triangular plot are in the ratio 3: 5:7 and its perimeter is 300 m. Find its area

Answer :  

(i) Consider ABC as the triangle

(i) If the length of the sides of a triangle are in the ratio 3:4:5 and its perimeter is 48 cm, find its area.

Ratio of the sides are 3x, 4x and 5x

Take a = 3x cm, b = 4x cm and c = 5x cm

a + b + c = 48

3x + 4x + 5x = 48

12x = 48

So , x = 48/12 = 4

Here,  a = 3x = 3 × 4 = 12 cm

b = 4x = 4 × 4 = 16 cm

c = 5x = 5 × 4 = 20 cm

We know that, S = Semi perimeter = (a + b + c)/ 2

= (12 + 16 + 20)/ 2

= 48/2

= 24 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 9

ml class 9 Mensuration img 10

= 12 × 4 × 2

= 96 cm2

(ii) given Sides of a triangle are in the ratio = 3: 5: 7

Perimeter = 300 m

We know that, First side = (300 × 3)/ sum of ration

Substituting the values

= (300 × 3)/ (3 + 5 + 7)

By calculation

= (300 × 3)/ 15

= 60 m

Second side = (300 × 5)/ 15 = 100 m

Third side = (300 × 7)/ 15 = 140 m

Here, S = perimeter/2 = 300/2 = 150 m

So ,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 11

= 1500 × 1.732

= 2598 m2

Question 10. ABC is a triangle in which AB = AC = 4 cm and ∠ A = 90°. Calculate the area of ∆ABC. Also find the length of perpendicular from A to BC.

Answer :  AB = AC = 4 cm

Using the Pythagoras theorem

BC2 = AB2 + AC2

BC2 = 42 + 42

By calculation

BC2 = 16 + 16 = 32

BC = √32 = 4√2 cm

We know that, Area of △ABC = ½ × BC × h

Substituting the values

8 = ½ × 4√2 × h

By calculation

h = (8 × 2)/ 4√2

We can write,

h = (2 × 2)/ √2 × √2/√2

So ,  h = 4√2/2 = 2 × √2

h = 2 × 1.41 = 2.82 cm

Question 11. Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.

Answer :  The isosceles triangles is

Find the area of an isosceles triangle whose equal sides are 12 cm each and the perimeter is 30 cm.

Here AB = AC = 12cm

Perimeter = 30 cm

So, BC = 30 – (12 + 12) = 30 – 24 = 6 cm

We know that, S = Semi perimeter = (a + b + c)/ 2

Substituting the values

= 30/ 2

= 15 cm

Here,

ml class 9 Mensuration img 1

ml class 9 Mensuration img 13

We can write

= 9 × 3.873

= 34.857

= 34.86 cm2

Question 12. Find the area of an isosceles triangle whose base is 6 cm and perimeter is 16 cm.

Answer :  in an isosceles triangle, any two sides are equal

let each of the equal side be x
perimeter=16cm
base=6cm

therefore,
x+x+6=16
2x=16-6
2x=10
x = 10/2
x=5

area of triangle =½×base×height

since we need the height,  draw a perpendicular from the base which will divide this triangle into two right angled triangles with each base = to 6/2=3cm
hypotenuse is the side of isosceles triangle which is 5cm
let the height be x

Using Pythagoras theorem,
ml class 9 Mensuration img 14

h=4
height is 4cm

area of triangle= ½×base×height
= ½ 6 x 4
= 12 cm²
is the area

Question 13. The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm².

Answer :  Consider ABC as a right angled triangle

AB = 5x cm and BC = (3x – 1) cm

The sides of a right angled triangle containing the right angle are 5x cm and (3x – 1) cm. Calculate the length of the hypotenuse of the triangle if its area is 60 cm2.

We know that, Area of △ABC = ½ × AB × BC

60 = ½ × 5x (3x – 1)

By calculation

120 = 5x (3x – 1)

⇒ 120 = 15x2 – 5x

It can be written ,

15x2 – 5x – 120 = 0

Taking out the common terms

5 (3x2 – x – 24) = 0

⇒ 3x2 – x – 24 = 0

⇒ 3x2 – 9x + 8x – 24 = 0

Taking out the common terms

3x (x – 3) + 8 (x – 3) = 0

⇒ (3x + 8) (x – 3) = 0

Here,  3x + 8 = 0 or x – 3 = 0

We can write ,

3x = -8 or x = 3

⇒ x = -8/3 or x = 3

x = -8/3 is not possible

So x = 3

AB = 5 × 3 = 15 cm

BC = (3 × 3 – 1) = 9 – 1 = 8 cm

In right angled △ABC

Using Pythagoras theorem

AC2 = AB2 + BC2

AC2 = 152 + 82

By calculation

AC2 = 225 + 64 = 289

⇒ AC2 = 172

So, AC = 17 cm

hypotenuse of the right angled triangle is 17 cm.

Question 14. In ∆ ABC, ∠B = 90°, AB = (2A + 1) cm and BC = (A + 1). cm. If the area of the ∆ ABC is 60 cm², find its perimeter.

Answer :    given that

AB = (2x + 1) cm

BC = (x + 1) cm

 In △ABC, ∠B = 90°, AB = (2A + 1) cm and BC = (A + 1) cm. If the area of the △ABC is 60 cm2, find its perimeter.

We know that, Area of △ABC = ½ × AB × BC

60 = ½ ×(2x + 1) (x + 1)

By cross multiplication

60 ×2 = (2x + 1) (x + 1)

By calculation

120 = 2x2 + 3x + 1

We can write ,

0 = 2x2 + 3x + 1 – 120

⇒ 0 = 2x2 + 3x – 119

So ,  2x2 + 3x – 119 = 0

⇒ 2x2 + 17x – 14x – 119 = 0

Taking out the common terms

x (2x + 17) – 7 (2x + 17) = 0

⇒ (x – 7) (2x + 17) = 0

Here,  x – 7 = 0 or 2x + 17 = 0

⇒ x = 7 or 2x = – 17

⇒ x = 7 or x = -17/2

AB = (2x + 1) = 2 × 7 + 1

⇒ AB = 14 + 1 = 15 cm

⇒ BC = (x + 1) = 7 + 1 = 8 cm

In right angled △ABC

Using Pythagoras Theorem

AC2 = AB2 + BC2

AC2 = 152 + 82

⇒ AC2 = 225 + 64

⇒ AC2 = 289

So ,  AC = 17 cm

So the perimeter = AB + BC + AC

= 15 + 8 + 17

= 40 cm

Question 15. If the perimeter of a right angled triangle is 60 cm and its hypotenuse is 25 cm, find its area.

Answer :    Perimeter of a right angled triangle = 60 cm

Hypotenuse = 25 cm

Here, the sum of two sides = 60 – 25 = 35 cm

Consider base = x cm

Altitude = (35 – x) cm

Using the Pythagoras theorem

x2 + (35 – x)2 = 252

By calculation

x2 + 1225 + x2 – 70x = 625

⇒ 2x2 – 70x + 1225 – 625 = 0

⇒ 2x2 – 70x + 600 = 0

Dividing by 2

x2 – 35x + 300 = 0

⇒ x2 – 15x – 20x + 300 = 0

Taking out the common terms

x (x – 15) – 20 (x – 15) = 0

⇒ (x – 15) (x – 20) = 0

Here,  x – 15 = 0

So,  x = 15

x – 20 = 0

So ,  x = 20 cm

So, 15 cm and 20 cm are the sides of the triangle

Area = ½ × base × altitude

= ½ × 15 × 20

= 150 cm2

Question 16. The perimeter of an isosceles triangle is 40 cm. The base is two third of the sum of equal sides. Find the length of each side.

Answer :  Perimeter of an isosceles triangle = 40 cm

Consider x cm as each equal side

We know that, Base = 2/3 (2x) = 4/3 x

So according to the sum

2x + 4/3 x = 40

By further calculation

6x + 4x = 120

⇒ 10x = 120

By division

x = 120/10 = 12

Hence, the length of each equal side is 12 cm.

Question 17. If the area of an isosceles triangle is 60 cm2 and the length of each of its equal sides is 13 cm, find its base.

Answer :  Area of isosceles triangle = 60 cm2

Length of each equal side = 13 cm

Consider base BC = x cm

Construct AD perpendicular to BC which bisects BC at D

So BD = DC = x/2 cm

ml class 9 Mensuration img 17

In right △ABD

AB2 = BD2 + AD2

132 = (x/2)2 + AD2

By calculation

169 = x2/4 + AD2

⇒ AD2 = 169 – x2/4 …(1)

We know that, Area = 60 cm2

AD = (area × 2)/base

AD = (60 × 2)/ x = 120/x …(2)

Using both the equations

169 – x2/4 = (120/x)2

By calculation

(676 – x2)/4 = 14400/x2

By cross multiplication

676x2 – x4 = 57600

We can write,

x4 – 676x2 + 57600 = 0

⇒ x4 – 576x2 – 100x2 + 57600 = 0

Taking out the common terms

x2 (x2 – 576) – 100 (x2 – 576) = 0

⇒ (x2 – 576) (x2 – 100) = 0

Here,  x2 – 576 = 0 where x2 = 576

So, x = 24

x2 – 100 = 0 where x2 = 100

So, x = 10

Therefore, the base is 10 cm or 24 cm.

Question 18. The base of a triangular field is 3 times its height If the cost of cultivating the field at the rate of ₹25 per 100m2 is ₹60000, find its base and height.

Answer :  Cost of cultivating the field at the rate of Rs 25 per 100 m2 = Rs 60000

Here the cost of cultivating the field of Rs 25 for 100 m2

So,  the cost of cultivating the field of Rs 1 = 100/25 m2

Cost of cultivating the field of Rs 60000 = 100/25 ×60000

= 4 ×60000

= 240000 m2

So, the area of field = 240000 m2

½ × base × height = 240000 …(1)

Height of triangular field = h m2

Base of triangular field = 3h m2

Substituting the values in equation (1)

½ × 3h × h = 240000

By calculation

½ × 3h2 = 240000

⇒ h2 = (240000 × 2)/3

⇒ h2 = 80000 ×2

⇒ h2 = 160000

So ,  h = √160000 = 400

Here, the height of triangular field = 400 m

Base of triangular field = 3 ×400 = 1200 m2

Question 19. A triangular park ABC? has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardner Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of ₹20 per metre leaving a space 3 m wide for a gate on one side.\

A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardener Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gate on one side.

Answer :  ABC is a triangular park with sides 120 m, 80 m and 50 m.

Here the perimeter of triangle ABC = 120 + 80 + 50 = 250 m

A triangular park ABC has sides 120 m, 80 m and 50 m (as shown in the given figure). A gardener Dhania has to put a fence around it and also plant grass inside. How much area does she need to plant? Find the cost of fencing it with barbed wire at the rate of Rs 20 per metre leaving a space 3 m wide for a gate on one side.

Portion at which a gate is build = 3m

Remaining perimeter = 250 – 3 = 247 m

So the length of fence around it = 247 m

Rate of fencing = Rs 20 per m

Total cost of fencing = 20 × 247 = Rs 4940

We know that, S = Semi perimeter = (a + b + c)/2

= 250/2

= 125 cm

ml class 9 Mensuration img 1

ml class 9 Mensuration img 19

So,  = 375√15m²

Question 20. An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (shown in the given figure), each piece measuring 20 cm, 50 cm and 50 cm. How much cloth of each colour is required for the umbrella?

An umbrella is made by stitching 10 triangular pieces of cloth of two different colours
Answer:   An umbrella is made by stitching 10 triangular pieces of cloth of two different colors

20 cm, 50 cm, 50 cm are the measurement of each triangle

So we get

S = (20 + 50 + 50) / 2

S = 120/2 = 60

ml class 9 Mensuration img 1

ml class 9 Mensuration img 20

Question 21.

(a) In the figure (1) given below, ABC is an equilateral triangle with each side of length 10 cm. In ∆ BCD, ∠D = 90° and CD = 6 cm.
Find the area of the shaded region. Give your answer correct to one decimal place.
(a) In the figure (1) given below, ABC is an equilateral triangle with each side of length
(b) In the figure(ii) given, ABC is an isosceles right angled triangle and DEFG is a rectangle. If AD = AE = 3 cm and DB = EC = 4 cm, find the area of the shaded region.
(b) In the figure(ii) given, ABC is an isosceles right angled triangle and DEFG is a rectangle

 Answer :   (a) It is given that

ABC is an equilateral triangle of side = 10 cm

We know that, Area of equilateral triangle ABC = √3/4 × (side)2

= √3/4 × 102

= √3/4 × 100

So we get

= √3× 25

= 1.73 × 25

= 43.3 cm2

In right angled triangle BDC

∠D = 900

BC = 10 cm

CD = 6 cm

Using Pythagoras Theorem

BD2 + DC2 = BC2

BD2 + 62 = 102

BD2 + 36 = 100

So we get

BD2 = 100 – 36 = 64

BD = √64 = 8 cm

We know that, Area of triangle BDC = ½ × base × height

So we get

= ½ × BD × DC

= ½ × 8 × 6

= 4 × 6

= 24 cm2

Here the area of shaded portion = Area of triangle ABC – Area of triangle BDC

= 43.3 – 24

= 19.3 cm2

(b) It is given that

AD = AE = 3 cm

DB = EC = 4 cm

By addition we get

AD + DB = AE + EC = (3 + 4) cm

AB = AC = 7 cm

∠A = 900

We know that, Area of right triangle ADE = ½ × AD × AE

= ½ × 3 × 3

= 9/2 cm2

So triangle BDG is an isosceles right triangle

DG2 + BG2 = BD2

DG2 + DG2 = 42

By calculation

2DG2 = 16

DG2 = 16/2 = 8

DG = √8 cm

Area of triangle BDG = ½ × BG × DG

We can write,

= ½ × DG × DG

= ½ (√8)2

= ½ × 8

= 4 cm2

Area of isosceles right triangle EFC = 4 cm2

So the area of shaded portion = 9/2 + 4 + 4

Taking LCM

= (9 + 8 + 8)/ 2

= 25/2

= 12.5 cm2

—  : End of ML Aggarwal Mensuration Exe-16.1 Class 9 ICSE Maths Solutions Ch-16 :–

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

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