Two corners of an isosceles triangle are at $(6, 4)$ $(6 ,4 )$ and $(9, 7)$ $(9 ,7 )$ . If the triangle's area is $36$ $36$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-09T18:37:30

The lengths of the sides are $= 4.24$ $=4.24$ , $17.1$ $17.1$ and $17.1$ $17.1$

The length of the base is

$b = \sqrt{{(9 - 6)}^{2} + {(7 - 4)}^{2}} = \sqrt{3^{2} + 3^{2}} = 3 \sqrt{2}$ $b=sqrt((9-6)^2+(7-4)^2)=sqrt(3^2+3^2)=3sqrt2$

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

The area is

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

$\frac{1}{2} \cdot 3 \sqrt{2} \cdot h = 36$ $1/2*3sqrt2*h=36$

$h = \frac{36 \cdot 2}{3 \sqrt{2}} = \frac{24}{\sqrt{2}} = 12 \sqrt{2}$ $h=(36*2)/(3sqrt2)=24/sqrt2=12sqrt2$

Let the lengths of the second and third sides of the triangle be $= c$ $=c$

Then,

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

$c^{2} = {(12 \sqrt{2})}^{2} + {(3 \frac{\sqrt{2}}{2})}^{2}$ $c^2=(12sqrt2)^2+(3sqrt2/2)^2$

$c^{2} = 288 + \frac{9}{2} = \frac{587}{2}$ $c^2=288+9/2=587/2$

$c = \sqrt{\frac{585}{2}} = 17.1$ $c=sqrt(585/2)=17.1$

Two corners of an isosceles triangle are at (6 ,4 )(6,4)(6 ,4 ) and (9 ,7 )(9,7)(9 ,7 ). If the triangle's area is 36 3636 , what are the lengths of the triangle's sides?