# Class 9 RD SHARMA Solutions Maths Chapter 17 - Heron's Formula

Ex. 17.1

Ex. 17.2

17.24

17.25

## Heron's Formula Exercise Ex. 17.1

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

The sides of the triangular field are in the ratio 25:17:12.

Let the sides of triangle be 25x, 17x, and 12x.

Perimeter of this triangle = 540 m

25x + 17x + 12x = 540 m

54x = 540 m

x = 10 m

Sides of triangle will be 250 m, 170 m, and 120 m.

Let the sides of triangle be 25x, 17x, and 12x.

Perimeter of this triangle = 540 m

25x + 17x + 12x = 540 m

54x = 540 m

x = 10 m

Sides of triangle will be 250 m, 170 m, and 120 m.

Semi-perimeter (s) =

By Heron's formula:

So, area of the triangle is 9000 m

^{2}.### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10

### Solution 11

## Heron's Formula Exercise Ex. 17.2

### Solution 1

For ABC

AC

(5)

So, ABC is a right angle triangle, right angled at point B.

Area of ABC

AC

^{2}= AB^{2}+ BC^{2}(5)

^{2}= (3)^{2}+ (4)^{2}So, ABC is a right angle triangle, right angled at point B.

Area of ABC

For ADC

Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

s = 7 cm

By Heron's formula

Area of triangle

Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm

s = 7 cm

By Heron's formula

Area of triangle

Area of ABCD = Area of ABC + Area of ACD

= (6 + 9.166) cm

= (6 + 9.166) cm

^{2}= 15.166 cm^{2}= 15.2 cm^{2}(approximately)### Solution 2

### Solution 3

### Solution 4

Let us join BD.

In BCD applying Pythagoras theorem

BD

= (12)

= 144 + 25

BD

BD = 13 m

In BCD applying Pythagoras theorem

BD

^{2}= BC^{2}+ CD^{2}= (12)

^{2}+ (5)^{2}= 144 + 25

BD

^{2}= 169BD = 13 m

Area of BCD

For ABD

By Heron's formula

Area of triangle

Area of park = Area of ABD + Area of BCD

= 35.496 + 30 m

= 35.496 + 30 m

^{2} = 65.496 m

^{2} = 65. 5 m

^{2}(approximately)### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 09

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

## Heron's Formula Exercise 17.24

### Solution 1

### Solution 2

### Solution 3

### Solution 4

### Solution 5

## Heron's Formula Exercise 17.25

### Solution 6

### Solution 7

### Solution 21

### Solution 22

### Solution 23

### Solution 25

### Solution 8

### Solution 9

### Solution 24

### Solution 26