The perimeter of a triangular field is 144 m and the ratio of the side

1.	The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.
Answer» Sides of triangle are in ratio: 3 : 4 : 5 a = 3x, b = 4x, c = 5x Since the perimeter of a triangle is given by: a+b+c = perimeter 3x+4x+5x = 144 x = \(\frac{144}{12}\)= 12 x = 12 Therefore sides of the triangle are: a = 3x = 3 x 12 = 36, b = 4x = 4 x 12 = 48, c = 5x = 5 x 12 = 60 When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by: \(A = \sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}2\)[Heron’s Formula] s = \(\frac{a+b+c}2\) = \(\frac{36+48+60}2\) = 72 \(A = \sqrt{72(72-36)(72-48)(72-60)}\) A = \(\sqrt{72\times36\times24\times12}\) = 864 cm²

The perimeter of a triangular field is 144 m and the ratio of the sides is 3 : 4 : 5. Find the area of the field.

Answer»

Sides of triangle are in ratio: 3 : 4 : 5

a = 3x, b = 4x, c = 5x

Since the perimeter of a triangle is given by:

a+b+c = perimeter

3x+4x+5x = 144

x = \(\frac{144}{12}\)= 12

x = 12

Therefore sides of the triangle are:

a = 3x = 3 x 12 = 36,

b = 4x = 4 x 12 = 48,

c = 5x = 5 x 12 = 60

When a, b and c are the sides of triangle and s is the semi-perimeter, then its area is given by:

\(A = \sqrt{s(s-a)(s-b)(s-c)}\) where s = \(\frac{a+b+c}2\)[Heron’s Formula]

s = \(\frac{a+b+c}2\) = \(\frac{36+48+60}2\) = 72

\(A = \sqrt{72(72-36)(72-48)(72-60)}\)

A = \(\sqrt{72\times36\times24\times12}\) = 864 cm²