Two corners of an isosceles triangle are at $(1 ,6 )$ and $(2 ,9 )$ . If the triangle's area is $24$ , what are the lengths of the triangle's sides?

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Answer 1 · 2018-05-09T02:02:39

base $sqrt{10},$ common side $sqrt{2329/10}$

Archimedes' Theorem says the area $a$ is related to the squared sides $A, B$ and $C$ by

$16a^2 = 4AB-(C-A-B)^2$

$C=(2-1)^2+(9-6)^2 = 10$

For an isosceles triangle either $A=B$ or $B=C$ . Let's work out both. $A=B$ first.

$16 (24^2) = 4A^2 - (10-2A)^2$

$16(24^2) = -100 +40A$

$A = B = 1/40 ( 100+ 16(24^2)) = 2329/10$

$B=C$ next.

$16 (24)^2 = 4 A (10) - A^2$

$(A - 20)^2 = - 8816 quad$ has no real solutions

So we found the isosceles triangle with sides

base $sqrt{10},$ common side $sqrt{2329/10}$

Two corners of an isosceles triangle are at (1 ,6 )(1 ,6 ) and (2 ,9 )(2 ,9 ). If the triangle's area is 24 24 , what are the lengths of the triangle's sides?