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I have been trying to solve the following problem for a while:

You are given a right triangle ABC (angle C is right). The perimeter ABC is 72. CK is the median, and CM is the height to the hypotenuse. CK - CM = 7. Find the area of the triangle.

You can find a diagram here.

Thank you in advance :)

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  • $\begingroup$ Hint: $m=\frac c2$. $\endgroup$ – Hagen von Eitzen May 27 '16 at 5:43
  • $\begingroup$ I know, I have used it while trying to solve it $\endgroup$ – Gale Staneva May 27 '16 at 5:44
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    $\begingroup$ You have $a+b+c=72$, $\frac{c}{2}-\frac{ab}{c}=7$, $a^2+b^2=c^2$, just solve the (system of) equations. $\endgroup$ – gamma May 27 '16 at 5:51
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Taking @Frank000's suggestion to give a full answer for those that wish to see one possible method.

Using the fact that $$\displaystyle \frac{c}{2} - \frac{ab}{c} = 7$$ we rearrange to find $$ c^2 -14c = 2ab. $$ Note also that $a+b=72-c$, squaring this gives $$ a^2 + 2ab + b^2 = 72^2 -144c +c^2,$$ since $a^2 + b^2 =c^2$ and we have an expression for $2ab$, $$ c^2 + c^2 - 14c = 72^2 -144c +c^2, $$ or rather $$ c^2 + 130c - 72^2 = 0, $$ the positive solution of which is $c=32$, giving $ a = \displaystyle \frac{288}{b}$, and substitution into the pythagorean theorem gives $$\frac{288^2}{b^2} + b^2 = 32^2,$$ which gives $a = 4(5 \pm 7)$, $b = 4(5 \mp 7)$.

This problem could probably also be done with multiple applications of Heron's formula, summing the area of the some of the interior triangles.

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  • $\begingroup$ Thank you so much for the full answer! $\endgroup$ – Gale Staneva May 27 '16 at 6:16

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