Two corners of an isosceles triangle are at $(2 ,3 )$ and $(1 ,4 )$ . If the triangle's area is $64$ , what are the lengths of the triangle's sides?

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Answer 1 · 2016-10-21T20:13:49

The 3 sides are $90.5, 90.5, and sqrt(2)$

Let b = the length of the base from $(2,3)$ to $(1, 4)$

$b = sqrt((1 - 2)^2 + (4 - 3)^2)$

$b = sqrt(2)$

This cannot be one of the equal sides, because the maximum area of such a triangle would occur, when it is equilateral, and specifically:

$A = sqrt(3)/2$

This conflicts with our given area, $64 units^2$

We can use the Area to find the height of the triangle:

$Area = (1/2)bh$

$64 = 1/2sqrt(2)h$

$h = 64sqrt(2)$

The height forms a right triangle and bisects the base, therefore, we can use the Pythagorean theorem to find the hypotenuse:

$c^2 = (sqrt(2)/2)^2 + (64sqrt(2))^2$

$c^2 = 8192.25$

$c~~ 90.5$

Two corners of an isosceles triangle are at (2 ,3 )(2 ,3 ) and (1 ,4 )(1 ,4 ). If the triangle's area is 64 64 , what are the lengths of the triangle's sides?