Question

Two corners of an isosceles triangle are at `(8 ,5 )` and `(1 ,7 )`. If the triangle's area is `15`, what are the lengths of the triangle's sides?

Answer 1

see below.

Explanation:

Name the points `M(8,5) and N(1,7)`
By Distance formula,
`MN=sqrt((1-8)^2+(7-5)^2)=sqrt53`
Given Area `A=15`,
`MN` can be either one of the equal sides or the base of the isosceles triangle.

Case 1) : `MN` is one of the equal sides of the isosceles triangle.

`A=1/2a^2sinx`,
where `a` is one of the equal sides and `x` is the included angle between the two equal sides.
`=> 15=1/2sqrt53^2sinx`
`=> x=sin^-1((2*15)/sqrt53^2)=34.4774^@`
`=> MP` (the base) `= 2*MN*sin(x/2)`
`=2*sqrt53*sin(34.4774/2)=4.31`

Therefore, lengths of the triangle sides are : `sqrt53, sqrt53, 4.31`

Case 2) : MN is the base of the isosceles triangle.

`A=1/2bh`, where `b and h` are the base and the height of the triangle, respectively.
`=> 15=1/2*MN*h`
`=> h=(2*15)/sqrt53=30/sqrt53`
`=> MP=PN` (the equal side) `= sqrt(((MN)/2)^2+h^2)`
`= sqrt((sqrt53/2)^2+(30/sqrt53)^2)`
`=sqrt(6409/212)`

Therefore, lengths of the triangle's sides are `sqrt(6409/212), sqrt(6409/212), sqrt53`