9th Maths

Heron's Formula

Main Formulae

Area of triangle `=1/2xxhe\ig\htxxba\se`

Area of equilateral triangle with side 'a' `=(sqrt3)/4\a^2`

Heron’s formula for area of triangle `=sqrt(s(s-a)(s-b)(s-c))`

Where; a, b and c are sides of triangle and;

Semiperimeter: `s=(a+b+c)/2`


Exercise 12.1

Question 1: A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side ‘a’. Find the area of the signal board, using Heron’s formula. If its perimeter is 180 cm, what will be the area of the signal board?

Answer: Given; perimeter = 180 cm
Hence, s = 180/2 = 90 cm
Side = 180/3 = 60 cm

Area `=sqrt(s(s-a)(s-b)(s-c))`
`=sqrt(90(90-60)^3)`
`=sqrt(90xx30^3)`
`=sqrt(30^4xx3)`
`=30^2sqrt3=900sqrt3`

Question 2: The triangular side walls of a flyover have been used for advertisements. The sides of the walls are 122 m, 22 m and 120 m. The advertisements yield an earning of Rs 5000 per m2 per year. A company hired one of its walls for 3 months. How much rent did it pay?

Answer: Given; a = 122 m, b = 22 m, and c = 120 m
Rate = Rs. 5000 per square meter per year
`s = (a + b + c)/2 = (122 + 22 + 120)/2

= 264/2 = 132`

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

`=sqrt(132(132-122)(132-22)(132-120))`

`=sqrt(132xx10xx110xx12)`

`=sqrt(11xx12xx10xx11xx10xx12)`

`=sqrt(11^2xx12^2xx10^2)`

`=11xx12xx10=1320 sq m`

`Co\st=Ar\ea\xx\Ra\te\xx\Ti\me`
`=1320xx5000xx3/12=Rs. 1,650,000`


Question 3: There is a slide in a park. One of its side walls has been painted in some colour with a message “KEEP THE PARK GREEN AND CLEAN”. If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour.

Answer: Given; a = 15, b = 11 and c = 6
`S = (a + b + c)/2 = (15 + 11 + 6)/2`

`= 32/2 = 16`

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

`=sqrt(16(16-15)(16-11)(16-10))`

`=sqrt(16xx1xx5xx10)`

`=sqrt(4xx4xx5xx5xx2)`

`=4xx5sqrt2=20sqrt2 sq m`

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`



Ex 12.1

Ex 12.2 Part 1

Ex 12.2 Part 2