Question

An isosceles triangle has sides A, B, and C, such that sides A and B have the same length. Side C has a length of `3` and the triangle has an area of `42`. What are the lengths of sides A and B?

Answer 1

If we presume that C is the base of the triangle:

A = 28.04
B = 28.04

Explanation:

First we need to find the height or `h` of the triangle. We do this by using simple/basic algebra. In this case side C = `b`

= `(b*h)/2`
= `(3*h)/2 = 42`
= `(3*h) = 84`
= `h = 84/3`
= `28`

Then we use one of the most powerful tools in geometry to decode the side lengths of A and B, the Pythagoras Theorem. In this case, as A and B meet at the vertex and a perpendicular drawn from this point on base bisects it. Hence, it forms a right angled triangle, whose two legs are `h=28` and `3/2` (say `a` and `b`) and one of the equal side forms hypotenuse or `c`. And then we have

`a^2 + b^2 = c^2`
= `(3/2)^2 + 28^2 = c^2`
= `2.25 + 784 = c^2`
= `786.25 = c^2`
= `sqrt 786.25`
`approx` `28.04`

All the best!