ML Aggarwal Solutions for Class 9 Maths Chapter 16 ...

ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration

Exercise 16.1

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

It is given that Base of triangle = 6 cm Height of triangle = 4 cm

We know that Area of triangle = ? ? base ? height Substituting the values = ? ? 6 ? 4 By further calculation = 6 ? 2 = 12 cm2

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 Solution:

(i) Consider a = 3 cm, b = 4 cm and c = 5 cm We know that S = Semi perimeter = (a + b + c)/ 2 Substituting the values = (3 + 4 + 5)/ 2 = 12/2 = 6 cm

Here

= 6 cm2

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

S = Semi perimeter = (a + b + c)/ 2 Substituting the values = (29 + 20 + 21)/ 2 = 70/2 = 35 cm

Here

ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration

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 Substituting the values = (12 + 9.6 + 7.2)/ 2 = 28.8/2 = 14.4 cm

Here

ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration

So we get = 2.4 ? 2.4 ? 6 = 34.56 cm2 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. Solution: 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 Substituting the values = (34 + 20 + 42)/ 2 = 96/2 = 48 cm Here

So we get

= 14 ? 6 ? 4 = 336 cm2

ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration

Here the shortest side of the triangle is 20 cm Consider h cm as the corresponding altitude Area of triangle = ? ? base ? height Substituting the values 336 = ? ? 20 ? h By further calculation h = (336 ? 2)/ 20 So we get h = 336/10 h = 33.6 cm

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

4. The sides of a triangular field are 975 m, 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)

Solution:

It is given that a = 975 m, b = 1050 m and c = 1125 m We know that S = Semi perimeter = (a + b + c)/ 2 Substituting the values = (975 + 1050 + 1125)/ 2 = 3150/2 = 1575 cm

Here

So we get = 525 ? 450 ? 2 It is given that 1 hectare = 10000 m2 = (525 ? 900)/ 10000 By further calculation = (525 ? 9)/ 100 = 4725/100 = 47.25 hectares

ML Aggarwal Solutions for Class 9 Maths Chapter 16 Mensuration

We know that Selling price of 1 hectare field = Rs 1000 Selling price of 47.25 hectare field = 1000 ? 47.25 = Rs 47250

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

Solution:

It is given that ABC is a right angled triangle BC = 12 cm and AB = 13 cm

Using the Pythagoras theorem AB2 = AC2 + BC2 Substituting the values 132 = AC2 + 122 By further calculation AC2 = 132 - 122 So we get AC2 = 169 ? 144 = 25 AC = 25 = 5 cm

We know that Area of triangle ABC = ? ? base ? height Substituting the values = ? ? 12 ? 5 = 30 cm2

Similarly Perimeter of triangle ABC = AB + BC + CA

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