Two corners of an isosceles triangle are at $(2 ,9 )$ and $(1 ,3 )$ . If the triangle's area is $9$ , what are the lengths of the triangle's sides?

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Answer 1 · 2017-12-08T10:53:02

Measure of the three sides are (6.0828, 4.2435, 4.2435)

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Length $a = sqrt((2-1)^2 + (9-3)^2) = sqrt 37 = 6.0828$

Area of $Delta = 9$
$:. h = (Area) / (a/2) = 9 / (6.0828/2) = 9 / 3.0414 = 2.9592$ #

$side b = sqrt((a/2)^2 + h^2) = sqrt((3.0414)^2 + (2.9592)^2)$
$b = 4.2435$

Since the triangle is isosceles, third side is also $= b = 4.2435$

Measure of the three sides are (6.0828, 4.2435, 4.2435)

Two corners of an isosceles triangle are at (2 ,9 )(2 ,9 ) and (1 ,3 )(1 ,3 ). If the triangle's area is 9 9 , what are the lengths of the triangle's sides?