Two corners of an isosceles triangle are at $(5, 8)$ $(5 ,8 )$ and $(9, 2)$ $(9 ,2 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-07-16T12:49:46

The lengths of the sides are $= 10.6$ $=10.6$ , $10.6$ $10.6$ and $= 7.2$ $=7.2$

The length of the base is

$b = \sqrt{{(9 - 5)}^{2} + {(2 - 8)}^{2}} = \sqrt{16 + 36} = \sqrt{52} = 2 \sqrt{13} = 7.2$ $b=sqrt((9-5)^2+(2-8)^2)=sqrt(16+36)=sqrt52=2sqrt13=7.2$

Let the altitude of the triangle be $= h$ $=h$

Then

The area of the triangle is $A = \frac{1}{2} \cdot b \cdot h$ $A=1/2*b*h$

$h = 2 \frac{A}{b} = 2 \cdot \frac{36}{2 \sqrt{13}} = \frac{36}{\sqrt{13}}$ $h=2A/b=2*36/(2sqrt13)=36/sqrt13$

The sides of the triangle are

$= \sqrt{h^{2} + {(\frac{b}{2})}^{2}}$ $=sqrt(h^2+(b/2)^2)$

$= \sqrt{\frac{36^{2}}{13} + 13}$ $=sqrt(36^2/13+13)$

$= 10.6$ $=10.6$

Two corners of an isosceles triangle are at (5,8)(5 ,8 ) and (9,2)(9 ,2 ). If the triangle's area is 3636 , what are the lengths of the triangle's sides?