**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?**

Solution:

Each side of a triangle = a

Perimeter of a triangle = 3a

s = ^{3a}/_{2}

Area of the signal traingel signal board = √[s(s-a)(s-b)(s-c)] *[sign board is an equilateral triangle a = b = c ]*

= √[s(s-a)(s-a)(s-a)

= (s – a) √[s(s-a)]

= (^{3a}/_{2} – a) √[^{3a}/_{2}(^{3a}/_{2} – a)]

= ^{a}/_{2} √(^{3a^2}/_{4})

=^{a}/_{2 }. ^{a}/_{2} .√3

= ^{a^2}/_{4} . √3

Thus, the area of the signal sign board = ^{a^2}/_{4}. √3 sq. units

Therefore, perimeter = 180cm (given)

⇒each side of the triangle = ^{180}/_{3 }= 60 cm

⇒ area of the triangle = ^{(60)^2}/_{4} x √3 cm^{2} = 900√3 cm^{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 (see Fig. 12.9). The advertisements yield an earning of Rs 5000 per m**^{2}per year. A company hired one of its walls for 3 months. How much rent did it pay?

Solution:

Given sides of the triangular walls are 122m, 22m , 120m i.e., a = 122m , b = 22m , c = 120m

Therefore, s = ^{3a}/_{2 }= ^{122+22+120}/_{2} = 132 m

Area of the triangular side wall = √[s(s-a)(s-b)(s-c)

= √[132(132 – 122)(132 – 22)(132 – 120)] m^{2}

= √[132(10)(110)(12) m^{2}

= 1320 m^{2}

Given: The advertisements yield an earning of Rs 5000 per m^{2} per year. i.e., the advertisement earning of 1m^{2} of the wall for one year is Rs. 5000

⸫ advertisement earning of 1m^{2} wall for 1month = Rs. ^{5000}/_{12}

⸫ advertisement earning of 1m^{2} of the wall for 3 months = Rs. ^{5000}/_{12} x 3

⸫ advertisement earning of complete wall for 3 months = Rs. ^{5000}/_{12} x 3 x 1320 = Rs. 15, 50, 000

**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” (see Fig. 12.10 ). If the sides of the wall are 15 m, 11 m and 6 m, find the area painted in colour. Fig. 12.10**

Solution:

The sides of the walls are 15m , 11m and 6m

Therefore, s = ^{15+11+6}/_{2} = ^{32}/_{2} = 16m

Area of the triangular walls = √[s(s-a)(s-b)(s-c)] *[a= 15m, b = 11m, c = 6m]*

=√[16(16-15)(16-11)(16-6)]

=√[16x1x5x10]

=4×5√2

= 20√2

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

Solution:

Perimeter = 42m

Two sides of the triangle are 18cm and 10 cm , i.e., a = 18cm, b = 10cm

Perimeter = a + b + c

42 = 18 + 10 + c

c = 42 – 18 – 10 = 14m

Therefore, s = ^{a+b+c}/_{2} = ^{18+10+14}/_{2} = ^{42}/_{2 }= 21cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

= √[21(21-18)(21-10)(21-14)] cm^{2}

=√[21x3x11x7] cm^{2}

=√[7x3x3x11x7] cm^{2}

= 21√11cm^{2}

**Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540cm. Find it’s area.**

Solution:

Sides of the triangle are in the ratio 12:17:25 i.e., a = 12x , b = 17x, c = 25x

Perimeter = 540cm = 12x + 17x + 25x

54x = 540

x = ^{540}/_{54} = 10cm

Thus, a = 120 , b = 170 and c = 250

s =^{a+b+c}/_{2} = ^{540}/_{2} = 270 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

= √[270(270-120)(270-170)(270-250)

=√[270x150x100x20]

= √30x9x30x5x10x10x5x4]

=30x3x5x10x2

= 9000cm^{2}

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

Solution:

Perimeter of an isosceles triangle = 30 cm

Perimeter = a+b+c = 30cm

Each of the equal side is 12cm i.e., a = b = 12 cm

12+12+c = 30

12+12+c = 60

24 + c = 30

c = 30 – 24 = 6 cm

s = ^{30}/_{2} = 15 cm

Area of the triangle = √[s(s-a)(s-b)(s-c)

=√[15(15-12)(15-12)(15-6)]

=√[15x3x3x9]

=√[3x5x9x9]

=9√15 cm^{2}

For next exercise – Heron’s Formula – Exercise 12.2 – Class IX

## One thought on “Heron's Formula – Exercise 12.1 – Class IX”

Comments are closed.