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

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Answer 1 · 2016-12-03T04:02:58

The sides are $a = 4.25, b = sqrt(40), c = 4.25$

Let side $b = sqrt((4 - 2)^2 + (3 - 9)^2)$

$b = sqrt((2)^2 + (-6)^2)$

$b = sqrt(4 + 36)$

$b = sqrt(40)$

We can find the height of the triangle, using $A = 1/2bh$

$9 = 1/2sqrt(40)h$

$h = 18/sqrt(40)$

We do not know whether b is one of the sides that are equal.

If b is NOT one of the sides that are equal, then the height bisects the base and the following equation is true:

$a^2 = c^2 = h^2 + (b/2)^2$

$a^2 = c^2 = 324/40 + 40/4$

$a^2 = c^2 = 8.1 + 10$

$a^2 = c^2 = 18.1$

$a = c ~~ 4.25$

$s = (sqrt(40) + 2(4.25))/2$

$s ~~ 7.4$

$A = sqrt(s(s - a)(s - b)(s - c))$

$A = sqrt(7.4(3.2)(1.07)(3.2))$

$A ~~ 9$

This is consistent with the given area, therefore, side b is NOT one of the equal sides.

The sides are $a = 4.25, b = sqrt(40), c = 4.25$

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