Question

Two corners of an isosceles triangle are at `(6 ,4 )` and `(4 ,1 )`. If the triangle's area is `8`, what are the lengths of the triangle's sides?

Answer 1

the lengths are `a=sqrt(15509)/26` and `b=sqrt(15509)/26` and `c=sqrt13`

Also `a=4.7898129` and `b=4.7898129` and `c=3.60555127`

Explanation:

First we let `C(x, y)` be the unknown 3rd corner of the triangle.

Also Let corners `A(4, 1)` and `B(6, 4)`

We set the equation using sides by distance formula
`a=b`

`sqrt((x_c-6)^2+(y_c-4)^2)=sqrt((x_c-4)^2+(y_c-1)^2)`

simplify to obtain
`4x_c+6y_c=35" " "`first equation

Use now the matrix formula for Area:

`Area=1/2((x_a,x_b,x_c,x_a),(y_a,y_b,y_c,y_a))=`

`=1/2(x_ay_b+x_by_c+x_cy_a-x_by_a-x_cy_b-x_ay_c)`

`Area=1/2((6,4,x_c,6),(4,1,y_c,4))=`

`Area=1/2*(6+4y_c+4x_c-16-x_c-6y_c)`

`Area=8` this is given

We now have the equation

`8=1/2*(6+4y_c+4x_c-16-x_c-6y_c)`

`16=3x_c-2y_c-10`

`3x_c-2y_c=26" " "`second equation

Solving simultaneously the system
`4x_c+6y_c=35`
`3x_c-2y_c=26`
`x_c=113/13` and `y_c=1/26`

We can now solve for the lengths of sides `a` and `b`
`a=b=sqrt((x_b-x_c)^2+(y_b-y_c)^2)`

`a=b=sqrt((6-113/13)^2+(4-1/26)^2)`

`a=b=sqrt(15509)/26=4.7898129" " "`units