Question

Two corners of an isosceles triangle are at `(7 ,9 )` and `(5 ,3 )`. If the triangle's area is `64`, what are the lengths of the triangle's sides?

Answer 1

the sides are:`2sqrt10=6.32456`
and `sqrt(10490)/5=20.4841`
and `sqrt(10490)/5=20.4841`

Explanation:

To compute for the sides, we need to obtain the height `h` using the area`=64` and the base `b` that can be solved using the points (7, 9) and (5, 3)

We solve for the base first
`b=sqrt((7-5)^2+(9-3)^2)`
`b=sqrt(4+36)=sqrt(40)=2sqrt10`

Area`=1/2*b*h`
`64=1/2*2*sqrt10*h`
`h=64/sqrt10=(32sqrt10)/5`

the height h divides the triangle into 2 equal parts and it passes thru the midpoint of the base b. So , we have a right triangle formed
Let x and x be the two unknown equal sides

`x=sqrt((b/2)^2+h^2)=sqrt(((2sqrt10)/2)^2+((32sqrt10)/5)^2)`
`x=sqrt(10+(1024(10))/25)=sqrt((250+10240)/25)`
`x=sqrt((10490)/25)`
`x=sqrt(10490)/5`
`x=20.4841" "` units

God bless....I hope the explanation is useful.