Class 9 Maths

Heron's Formula

Exercise 12.1

Part 2

Question 4: Find the area of a triangle two sides of which are 18cm and 10cm and the perimeter is 42cm.

Answer: Given; Perimeter = 42 cm, a = 18 cm, b = 10 cm and c = ?
Value of c = Perimeter – (a + b) = 42 – (18 + 10)
= 42 – 28 = 14 cm
`s = 42/2 = 21`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(21(21-18)(21-14)(21-10))`

`=sqrt(21xx3xx7xx11)`
`=sqrt(3^2xx7^2xx11)=21sqrt11 sq cm`

Question 5: Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find its area.

Answer: Given; Perimeter = 540 cm and a:b:c = 12:17:25
Sides can be calculated as follows:
`12x + 17x + 25x = 540`
Or, `54x = 540`
Or, `x = 10`
Hence, `a = 120, b = 170 and c = 250`
Now, `s = 540/2 = 270`

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(270(270-120)(270-170)(270-250))`

`=sqrt(270xx150xx100xx20)`

`=sqrt(10xx27xx10xx15xx10xx2)`

`=sqrt(10^4xx3^4)``=10^2xx3^2=900 sq cm`

Question 6: An isosceles triangle has perimeter 30 cm and each of the equal sides is 12 cm. Find the area of the triangle.

Answer: Given; perimeter = 30 cm, a = 12, b = 12 and c = ?
Value of c = 30 – (12 + 12) = 30 – 24 = 6 cm
S = 30/2 = 15

Area `=sqrt(s(s-a)(s-b)(s-c))`

`=sqrt(15(15-12)(15-12)(15-6))`

`=sqrt(15xx3xx3xx9)=9sqrt15 sq cm`