Two corners of an isosceles triangle are at $(5, 6)$ $(5 ,6 )$ and $(4, 8)$ $(4 ,8 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

Question

1375 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-09T11:07:29

The lengths of the sides are $= 2.24, 32.21, 32.21$ $=2.24, 32.21,32.21$

The length of the base is

$b = \sqrt{{(4 - 5)}^{2} + {(8 - 6)}^{2}} = \sqrt{1 + 4} = \sqrt{5}$ $b=sqrt((4-5)^2+(8-6)^2)=sqrt(1+4)=sqrt5$

The area of the triangle is

$A = \frac{1}{2} \cdot b \cdot h = 36$ $A=1/2*b*h=36$

So,

The altiude is $h = 36 \cdot \frac{2}{b} = \frac{72}{\sqrt{5}}$ $h=36*2/b=72/sqrt5$

We apply Pythagoras' theorem

The length of the side is

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

$= \sqrt{(\frac{5}{4} + \frac{72^{2}}{5})}$ $=sqrt((5/4+72^2/5))$

$= \sqrt{1038.05}$ $=sqrt(1038.05)$

$= 32.21$ $=32.21$

Two corners of an isosceles triangle are at (5 ,6 )(5,6)(5 ,6 ) and (4 ,8 )(4,8)(4 ,8 ). If the triangle's area is 36 3636 , what are the lengths of the triangle's sides?