Question

Two corners of an isosceles triangle are at `(1 ,3 )` and `(5 ,8 )`. If the triangle's area is `6`, what are the lengths of the triangle's sides?

Answer 1

possible triangle with sides:`sqrt(92537)=3.70974`,`sqrt(92537)=3.70974`,`sqrt41=6.40312`
possible triangle with sides:`sqrt413`,`" "sqrt41`,`" "1.89495`

Explanation:

Distance between the two points
`=sqrt((5-1)^2+(8-3)^2)=sqrt41`
Let this be the base `b=sqrt41`
Formula for Area of triangle:

Area`=1/2bh`
`6=1/2sqrt41*h`

`h=12/sqrt41`

The two equal sides can now be computed:
Let x be one of the sides
`x=sqrt(h^2+(b/2)^2)`
`x=sqrt((12/sqrt41)^2+((sqrt41)/2)^2)`
`x=1/82sqrt(92537)`
`x=3.70974`
First triangle with sides:
`x=3.70974`
`x=3.70974`
`b=sqrt41=6.40312`

For the second triangle:
the two equal sides are:
`x=sqrt41`
`x=sqrt41`
and the 3rd side `b`
compute the vertex angle of the isosceles triangle first:
Area`=1/2sqrt41*sqrt41*sin theta`
`6=1/2sqrt41*sqrt41*sin theta`
`sin theta=12/41`
`theta=17.0186^@`
Using cosine law
`b=sqrt(x^2+x^2-2*x*x*cos theta)`
`b=sqrt(sqrt41^2+sqrt41^2-2*sqrt41*sqrt41*cos(17.0186^@))`
`b=1.89495" "`units

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