Question

The perimeter of an isosceles triangle is 42 cm and its base is (3/2) times each of the equal sides. Find the length of each side of the triangle, area of the triangle and the height of the triangle.

Answer 1

Let the equal sides of isosceles triangle is x

Base of the triangle = \(\frac{3x}2\)

Since the perimeter of a triangle is given by:

a+b+c = perimeter

x + x +\(\frac{3x}2\) = 42

{2x+2x+3x)\(\frac{1}2\) = 42

x = 12 cm

Therefore sides of triangle are:a = x = 12,b = x = 12,c = \(\frac{3x}2\) = 18

Let 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]

a = 12, b = 12, c = 18

S = \(\frac{a+b+c}2\)= \(\frac{12+12+18}2\)= 21

A = \(\sqrt{21(21-12)(21-12)(21-18)}\)

A = \(\sqrt{21\times 9\times 9\times3}\) = 71.43 cm²

Area of triangle =\(\frac{1}2\)(Base x Altitude)

71.43 =\(\frac{1}2\)(18 x Altitude)

Altitude = 7.93 cm