Question

Two corners of an isosceles triangle are at `(7 ,2 )` and `(3 ,6 )`. If the triangle's area is `6`, what are the lengths of the triangle's sides?

Answer 1

The lengths of sides are: `a=5/2sqrt2=3.5355339` and `b=5/2sqrt2=3.5355339` and `c=4sqrt2=5.6568542`

Explanation:

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

Also Let corners `A(7, 2)` and `B(3, 6)`

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

`sqrt((x_c-3)^2+(y_c-6)^2)=sqrt((x_c-7)^2+(y_c-2)^2)`

simplify to obtain
`x_c-y_c=1" " "`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((7,3,x_c,7),(2,6,y_c,2))=`

`Area=1/2*(42+3y_c+2x_c-6-6x_c-7y_c)`

`Area=6` this is given

We now have the equation

`6=1/2*(42+3y_c+2x_c-6-6x_c-7y_c)`

`12=-4x_c-4y_c+36`

`x_c+y_c=6" " "`second equation

Solving simultaneously the system
`x_c-y_c=1`
`x_c+y_c=6`
`x_c=7/2` and `y_c=5/2`

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((3-7/2)^2+(6-5/2)^2)`

`a=b=5/2sqrt(2)=3.5355339" " "`units

compute side `c`:

`c=sqrt((x_a-x_b)^2+(y_a-y_b)^2)`

`c=sqrt((7-3)^2+(2-6)^2)`

`c=sqrt(2(16))`

`c=4sqrt2=5.6568542`