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 cm^{2}

**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,

= 6 cm^{2}

**(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,

So we get

= 7 × 5 × 3 × 2

= 210 cm^{2}

**(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

So we get

= 2.4 × 2.4 × 6

= 34.56 cm^{2}

**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,

So , = 14 × 6 × 4

= 336 cm^{2}

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,

= 525 × 450 × 2

It is given that, 1 hectare = 10000 m^{2}

= (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.**

**Answer : **Given that,

ABC is a right angled triangle

BC = 12 cm and AB = 13 cm

Use the Pythagoras theorem

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

13^{2} = AC^{2} + 12^{2}

By calculation

AC^{2} = 13^{2} – 12^{2}

So , AC^{2} = 169 – 144 = 25

AC = √25 = 5 cm

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

= ½ × 12 × 5

= 30 cm^{2}

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 m^{2}

**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

Side^{2} = (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

, 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 cm^{2}

In triangle ABD

Using Pythagoras Theorem

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

Here BD = 12/2 = 6 cm

12^{2} = AD^{2} + 6^{2}

By calculation

144 =AD^{2} + 36

AD^{2} = 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

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,

= 12 × 4 × 2

= 96 cm^{2}

##### (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 ,

= 1500 × 1.732

= 2598 m^{2}

**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

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

BC^{2} = 4^{2} + 4^{2}

By calculation

BC^{2} = 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

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,

We can write

= 9 × 3.873

= 34.857

= 34.86 cm^{2}

**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,

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

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

60 = ½ × 5x (3x – 1)

By calculation

120 = 5x (3x – 1)

⇒ 120 = 15x^{2} – 5x

It can be written ,

15x^{2} – 5x – 120 = 0

Taking out the common terms

5 (3x^{2} – x – 24) = 0

⇒ 3x^{2} – x – 24 = 0

⇒ 3x^{2} – 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

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

AC^{2} = 15^{2} + 8^{2}

By calculation

AC^{2} = 225 + 64 = 289

⇒ AC^{2} = 17^{2}

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

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 = 2x^{2} + 3x + 1

We can write ,

0 = 2x^{2} + 3x + 1 – 120

⇒ 0 = 2x^{2} + 3x – 119

So , 2x^{2} + 3x – 119 = 0

⇒ 2x^{2} + 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

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

AC^{2} = 15^{2} + 8^{2}

⇒ AC^{2} = 225 + 64

⇒ AC^{2} = 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

x^{2} + (35 – x)^{2} = 25^{2}

By calculation

x^{2} + 1225 + x^{2} – 70x = 625

⇒ 2x^{2} – 70x + 1225 – 625 = 0

⇒ 2x^{2} – 70x + 600 = 0

Dividing by 2

x^{2} – 35x + 300 = 0

⇒ x^{2} – 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 cm^{2}

**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 cm^{2}

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

In right △ABD

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

13^{2} = (x/2)^{2} + AD^{2}

By calculation

169 = x^{2}/4 + AD^{2}

⇒ AD^{2} = 169 – x^{2}/4 **…(1)**

We know that, Area = 60 cm^{2}

AD = (area × 2)/base

AD = (60 × 2)/ x = 120/x **…(2)**

Using both the equations

169 – x^{2}/4 = (120/x)^{2}

By calculation

(676 – x^{2})/4 = 14400/x^{2}

By cross multiplication

676x^{2} – x^{4} = 57600

We can write,

x^{4} – 676x^{2} + 57600 = 0

⇒ x^{4} – 576x^{2} – 100x^{2} + 57600 = 0

Taking out the common terms

x^{2} (x^{2} – 576) – 100 (x^{2} – 576) = 0

⇒ (x^{2} – 576) (x^{2} – 100) = 0

Here, x^{2} – 576 = 0 where x^{2} = 576

So, x = 24

x^{2} – 100 = 0 where x^{2} = 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 m^{2} = Rs 60000

Here the cost of cultivating the field of Rs 25 for 100 m^{2}

So, the cost of cultivating the field of Rs 1 = 100/25 m^{2}

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

= 4 ×60000

= 240000 m^{2}

So, the area of field = 240000 m^{2}

½ × base × height = 240000 **…(1)**

Height of triangular field = h m^{2}

Base of triangular field = 3h m^{2}

Substituting the values in equation (1)

½ × 3h × h = 240000

By calculation

½ × 3h^{2} = 240000

⇒ h^{2} = (240000 × 2)/3

⇒ h^{2} = 80000 ×2

⇒ h^{2} = 160000

So , h = √160000 = 400

Here, the height of triangular field = 400 m

Base of triangular field = 3 ×400 = 1200 m^{2}

**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.\**

**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

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

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?**

**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

**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.**

**(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.**

** 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 × 10^{2}

= √3/4 × 100

So we get

= √3× 25

= 1.73 × 25

= 43.3 cm^{2}

In right angled triangle BDC

∠D = 90^{0}

BC = 10 cm

CD = 6 cm

Using Pythagoras Theorem

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

BD^{2} + 6^{2} = 10^{2}

BD^{2} + 36 = 100

So we get

BD^{2} = 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 cm^{2}

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

= 43.3 – 24

= 19.3 cm^{2}

##### (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 = 90^{0}

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

= ½ × 3 × 3

= 9/2 cm^{2}

So triangle BDG is an isosceles right triangle

DG^{2} + BG^{2} = BD^{2}

DG^{2} + DG^{2} = 4^{2}

By calculation

2DG^{2} = 16

DG^{2} = 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 cm^{2}

Area of isosceles right triangle EFC = 4 cm^{2}

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

Taking LCM

= (9 + 8 + 8)/ 2

= 25/2

= 12.5 cm^{2}

— : 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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