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

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## Chapter 17 - Heron's Formula 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.

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 ## Chapter 17 - Heron's Formula 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 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  Area of ABCD = Area of 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 Area of BCD For ABD By Heron's formula Area of triangle  Area of park = Area of 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 ## Chapter 17 - Heron's Formula 17.24

Solution 1 Solution 2 Solution 3 Solution 4 Solution 5  ## Chapter 17 - Heron's Formula 17.25

Solution 6 Solution 7 Solution 21 Solution 22 Solution 23  Solution 25 Solution 8  Solution 9 Solution 24  Solution 26  ### STUDY RESOURCES

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