GMATPrepNow wrote:

Right triangle ABC has sides with length x, y and z. If triangle ABC has perimeter 17, and x² + y² + z² = 98, then what is the area of triangle ABC?

A) 12.75

B) 13.25

C) 14

D) 14.5

E) 15.25

* kudos for all correct solutions

Let z = length of hypotenuse

Let x and y = lengths of the sides that make up the triangle’s right angle

NOTE: since triangle area = (base)(height)/2, we know that the

area of triangle ABC = [color=red]xy/2Given: x² + y² + z² = 98

The Pythagorean Theorem tells us that

x² + y² = z²So, we can write: x² + y² + (

x² + y²) = 98

Rewrite: 2(x² + y²) = 98

Or we can write: x² + y² = 49

Since we already know that

x² + y² = z², we can conclude that z² = 49, which means z =

7If the perimeter = 17, we can write: x + y + z = 17

Since z =

7, we can write: x + y +

7 = 17

Simplify to get: x + y = 10

IMPORTANT: If we square both sides of the above equation, some nice things happen.

(x + y) ² = 10²

Expand: x² + 2xy + y² = 100

Rearrange to get: (x² + y²) + 2xy = 100

Replace x² + y² with 49 to get: 49 + 2xy = 100

Simplify: 2xy = 51

NOTE: Our goal is to determine the value of

xy/2So, take 2xy = 51 and divide both sides by 4 to get: xy/2 = 51/4 = 12.75

Answer:

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Brent Hanneson – Founder of gmatprepnow.com