Question

Two corners of an isosceles triangle are at `(2 ,6 )` and `(4 ,8 )`. If the triangle's area is `36`, what are the lengths of the triangle's sides?

Answer 1

The length of the sides are `=sqrt8, sqrt650, sqrt650`

Explanation:

The length of side `A=sqrt((8-6)^2+(4-2)^2)=sqrt8`

Let the height of the triangle be `=h`

The area of the triangle is

`1/2*sqrt8*h=36`

The altitude of the triangle is `h=(36*2)/sqrt8=36/sqrt2`

The mid-point of `A` is `(6/2,14/2)=(3,7)`

The gradient of `A` is `=(8-6)/(4-2)=1`

The gradient of the altitude is `=-1`

The equation of the altitude is

`y-7=-1(x-3)`

`y=-x+3+7=-x+10`

The circle with equation

`(x-3)^2+(y-7)^2=36^2/2=648`

The intersection of this circle with the altitude will give the third corner.

`(x-3)^2+(-x+10-7)^2=648`

`x^2-6x+9+x^2-6x+9=648`

`2x^2-12x-630=0`

`x^2-6x-315=0`

We solve this quadratic equation

`x=(6+-sqrt(6^2+4*1*315))/(2)`

`=(6+-36)/2`

`x_1=42/2=21`

`x_2=-30/2=-15`

The points are `(21,-11)` and `(-15,-25)`

The length of `2` sides are `=sqrt((2-21)^2+(6+11)^2)=sqrt650`

graph{(y+x-10)((x-2)^2+(y-6)^2-0.1)((x-4)^2+(y-8)^2-0.1)((x-3)^2+(y-7)^2-648)=0 [-52.4, 51.64, -21.64, 30.4]}