Question

Two corners of an isosceles triangle are at `(8 ,5 )` and `(6 ,2 )`. If the triangle's area is `4`, what are the lengths of the triangle's sides?

Answer 1

Lengths of triangle's sides are `3.61(2dp) , 2.86(dp), 2.86(dp)` unit.

Explanation:

Length of base of isoceles triangle is `b=sqrt((x_1-x_2)^2+(y_1-y_2)^2) = sqrt((8-6)^2+(5-2)^2)= sqrt(4+9)=sqrt 13=3.61(2dp)`

Area of isoceles triangle is `A_t = 1/2 *b*h or 4 = 1/2*sqrt13*h or h = 8/sqrt 13 =2.22(2dp)`. Where `h` be the altitude of triangle.

Legs of isoceles triangle are `l_1=l_2= sqrt( h^2 +(b/2)^2) =sqrt( 2.22^2 +(3.61/2)^2) = 2.86(2dp)`unit

Lengths of triangle's sides are `3.61(2dp) , 2.86(dp),2.86(dp)` unit. [Ans]