Question

Find the base of an isosceles triangle whose area is 12 cm sq and one of the equal side is 5 cm

Answer 1

Following are the available values:

Base: ?

Area: 12 cm sq

side 1: 5 cm

The formula of area is :

1/2 X base X height

- let the base be 2b and height be h.

- bXh= 12

- h= 12/b

- use Pythagoras theorem:

b²+h²= 5²

so b²+ 12²/b²= 25

so next your equation will be: b( power 4) - 25b²+144=0

So next (b + 4)(b – 4)(b + 3)(b – 3) = 0

Neglecting the negative values, b could be 3 giving h = 4

or b could be 4 giving h = 3

This means the whole base could be 6 and the height is 4

OR the whole base could be 8 and the height is 3

Answer 2

Cut your isosceles triangle in two equals right-angled triangles, and consider one of them.

(A) Its area is 12/2 = 6cm2.

(B) Its hypothenuse length is 5 cm.

Let us call x and y the other sides, x being half the base of the initial triangle, and y the height from the base . We have :

(A) x*y/2 = 6, so x*y = 12 = 3 * 4

(B) x^2 + y^2= 5^2 = 25 from Pythagore.

There are many ways to solve this simple quadratic system,

one of them is to recognize the “canonical” Pythagore example 3^2 + 4^2 = 5^2,

leading to the two solutions (x=3 and y=4), or (x=4 and y=3).

The isosceles triangle base length is 2*x, so it is 6 or 8.