Two corners of an isosceles triangle are at $(2 ,4 )$ and $(3 ,8 )$ . If the triangle's area is $18$ , 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 $=sqrt[(3-2)^2+(8-4)^2]=sqrt17$

Next, find the height ...

Triangle Area = $(1/2) xx("base")xx("height")$

$18=(1/2)xxsqrt17xx("height")$

height $=36/sqrt17$

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

$(height)^2+[(1/2)(base)]^2=(side)^2$

$(36/sqrt17)^2+[(1/2)(sqrt17)]^2=(side)^2$

Sides $=sqrt(5473/68)~~8.97$

In summary, the isosceles triangle has two equal sides of length $~~8.97$ and a base length of $sqrt17$

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 18 18 , what are the lengths of the triangle's sides?