Two corners of an isosceles triangle are at $(4 ,2 )$ and $(1 ,5 )$ . If the triangle's area is $64$ , what are the lengths of the triangle's sides?

Question

1266 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-22T14:47:16

$color(blue)(a=b=sqrt(32930)/6 and c=3sqrt(2)$

Let $A=(4,2)$ and $B=(1,5)$

If $AB$ is the base of an isosceles triangle then $C=(x,y)$ is the vertex at the altitude.

Let The sides be $a,b,c$ , $a=b$

Let h be the height, bisecting AB and passing through point C:

Length $AB = sqrt((4-1)^2+(2-5)^2)=sqrt(18)=3sqrt(2)$

To find $h$ . We are given area equals 64:

$1/2AB*h=64$

$1/2(3sqrt(2))h=64=>h=(64sqrt(2))/3$

By Pythagoras' theorem:

$a=b=sqrt(((3sqrt(2))/2)^2+((64sqrt(2))/3)^2)=sqrt(32930)/6$

So the lengths of the sides are:

$color(blue)(a=b=sqrt(32930)/6 and c=3sqrt(2)$

Two corners of an isosceles triangle are at (4 ,2 )(4 ,2 ) and (1 ,5 )(1 ,5 ). If the triangle's area is 64 64 , what are the lengths of the triangle's sides?