Brent@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
Answer:
A
Source:
www.gmatprepnow.com
Difficulty level: 700+
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 =
xy/2
Given: x² + y² + z² = 98
The Pythagorean Theorem tells us that
x² + y² = z²
So, 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 =
7
If 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/2
So, take 2xy = 51 and divide both sides by 4 to get: xy/2 = 51/4 = 12.75
Answer:
A
