Two corners of an isosceles triangle are at $(2, 4)$ $(2 ,4 )$ and $(3, 8)$ $(3 ,8 )$ . If the triangle's area is $18$ $18$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-02-08T00:22:33

First find the length of the base, then solve for the height using the area of 18.

Using the distance formula ...

length of base $= \sqrt{{(3 - 2)}^{2} + {(8 - 4)}^{2}} = \sqrt{17}$ $=sqrt[(3-2)^2+(8-4)^2]=sqrt17$

Next, find the height ...

Triangle Area = $(\frac{1}{2}) \times (base) \times (height)$ $(1/2) xx("base")xx("height")$

$18 = (\frac{1}{2}) \times \sqrt{17} \times (height)$ $18=(1/2)xxsqrt17xx("height")$

height $= \frac{36}{\sqrt{17}}$ $=36/sqrt17$

Finally, use Pythagorean theorem to find the length of the two equal sides ...

${(h e i g h t)}^{2} + {[(\frac{1}{2}) (b a s e)]}^{2} = {(s i d e)}^{2}$ $(height)^2+[(1/2)(base)]^2=(side)^2$

${(\frac{36}{\sqrt{17}})}^{2} + {[(\frac{1}{2}) (\sqrt{17})]}^{2} = {(s i d e)}^{2}$ $(36/sqrt17)^2+[(1/2)(sqrt17)]^2=(side)^2$

Sides $= \sqrt{\frac{5473}{68}} \approx 8.97$ $=sqrt(5473/68)~~8.97$

In summary, the isosceles triangle has two equal sides of length $\approx 8.97$ $~~8.97$ and a base length of $\sqrt{17}$ $sqrt17$

Hope that helped

Two corners of an isosceles triangle are at (2,4)(2 ,4 ) and (3,8)(3 ,8 ). If the triangle's area is 1818 , what are the lengths of the triangle's sides?