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 m2.
Solution 6
Solution 7
Solution 8
Solution 9
Solution 10
Solution 11
Heron's Formula Exercise Ex. 17.2
Solution 1
For 
ABC
AC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So,ABC is a right angle triangle, right angled at point B.
Area ofABC
ABCAC2 = AB2 + BC2
(5)2 = (3)2 + (4)2
So,
ABC is a right angle triangle, right angled at point B.Area of
ABCFor ADC
Perimeter = 2s = AC + CD + DA = (5 + 4 + 5) cm = 14 cm
s = 7 cm
By Heron's formula
Area of triangle
ADCPerimeter = 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) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)
ABC + Area of ACD= (6 + 9.166) cm2 = 15.166 cm2 = 15.2 cm2 (approximately)
Solution 2
Solution 3
Solution 4
Let us join BD.
In
BCD applying Pythagoras theorem
BD2 = BC2 + CD2
= (12)2 + (5)2
= 144 + 25
BD2 = 169
BD = 13 m
In
BCD applying Pythagoras theoremBD2 = BC2 + CD2
= (12)2 + (5)2
= 144 + 25
BD2 = 169
BD = 13 m
Area of BCD
BCD
For ABD
ABD
By Heron's formula 
Area of triangle 
Area of park = Area of ABD + Area of BCD
= 35.496 + 30 m2
ABD + Area of BCD= 35.496 + 30 m2
= 65.496 m2
= 65. 5 m2 (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
