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 `14` and the triangle has an area of `42`. What are the lengths of sides A and B?

Answer 1

`A= sqrt(6^2+7^2) => A= 5sqrt(3) = B`

Explanation:

Given: Area of an isosceles triangle, `A_Delta =42; C = 14`
Required: Length of sides `A` and `B`:
Solution: `A_Delta = 1/2 C*h; C = 14; h= height`
`42=1/2 (14*h); h=6` Now the height forms a right angle triangle with hypotenuse of `A` and sides of `h` and `C/2`. So you can use the Pythagoras Theorem:
`A^2= h^2+(C/2)^2` you know `h` and `C/2=7` solve of A
`A= sqrt(6^2+7^2) => A= 5sqrt(3) = B`