Question

Two corners of an isosceles triangle are at `(2 ,6 )` and `(3 ,2 )`. If the triangle's area is `48`, what are the lengths of the triangle's sides?

Answer 1

The length of three sides of triangle are `4.12, 23.37 ,23.37` unit

Explanation:

The base of the isosceles triangle, `b=sqrt( (x_1-x_2)^2+(y_1-y_2)^2) = sqrt( (2-3)^2+(6-2)^2) =sqrt17=4.12(2dp)unit`

The area of an isosceles triangle is `A_t = 1/2 * b *h =1/2*4.12 *h ; A_t=48 :. h = (2* A_t )/b = (2*48)/4.12=96/4.12= 23.28(2dp) unit`. Where `h` is the altitude of triangle.

The legs of the isosceles triangle are `l_1=l_2= sqrt(h^2+(b/2)^2)=sqrt(23.28^2+(4.12/2)^2) =23.37(2dp) unit`

Hence the length of three sides of triangle are `4.12(2dp), 23.37(2dp) ,23.37(2dp)` unit [Ans]