2008 AMC 10A Problem 18

Below is the professionally curated solution for Problem 18 of the 2008 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2008 AMC 10A solutions, or check the answer key.

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Concepts:Pythagorean Theoremsystem of equationsalgebraic manipulation

Difficulty rating: 1580

18.

A right triangle has perimeter 3232 and area 20.20. What is the length of its hypotenuse?

574\dfrac{57}{4}

594\dfrac{59}{4}

614\dfrac{61}{4}

634\dfrac{63}{4}

654\dfrac{65}{4}

Solution:

Let the legs be y,zy, z and the hypotenuse x.x. Then y2+z2=x2,y^2 + z^2 = x^2, y+z=32x,y + z = 32 - x, and yz=40.yz = 40.

Squaring the second equation, (32x)2=y2+z2+2yz=x2+80. (32 - x)^2 = y^2 + z^2 + 2yz = x^2 + 80.

This gives 102464x=80,1024 - 64x = 80, so x=594.x = \dfrac{59}{4}.

Thus, the correct answer is B.

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