Two corners of an isosceles triangle are at $(3, 9)$ $(3 ,9 )$ and $(2, 5)$ $(2 ,5 )$ . If the triangle's area is $4$ $4$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-06-10T10:18:04

The lengths of the triangle's sides are $2.83$ $2.83$ , $2.83$ $2.83$ and $4.12$ $4.12$

The length of the base is

$b = \sqrt{{(3 - 2)}^{2} + {(9 - 5)}^{2}} = \sqrt{1^{2} + 4^{2}} = \sqrt{17}$ $b=sqrt((3-2)^2+(9-5)^2)=sqrt(1^2+4^2)=sqrt17$

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

The area is

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

$\frac{1}{2} \cdot \sqrt{17} \cdot h = 4$ $1/2*sqrt17*h=4$

$h = \frac{4 \cdot 2}{\sqrt{17}} = \frac{8}{\sqrt{17}}$ $h=(4*2)/(sqrt17)=8/sqrt17$

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

Then,

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

$c^{2} = {(\frac{8}{\sqrt{17}})}^{2} + {(\frac{\sqrt{17}}{2})}^{2}$ $c^2=(8/sqrt17)^2+(sqrt17/2)^2$

$c^{2} = 3.76 + 4.25 = 8.01$ $c^2=3.76+4.25=8.01$

$c = \sqrt{8.01} = 2.83$ $c=sqrt(8.01)=2.83$

Two corners of an isosceles triangle are at (3,9)(3 ,9 ) and (2,5)(2 ,5 ). If the triangle's area is 44 , what are the lengths of the triangle's sides?