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Area of a Triangle

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by freyesinsb » Wed Apr 24, 2013 1:30 pm
Looking for various approaches to this Algebra problem disguised as a Geometry question.

In right triangle ABC, BC is the hypotenuse. If BC is 13 and AB + AC = 15, what is the area of the triangle?

A. 2√7
B. 2√14
C. 14
D. 28
E. 56

Source: Veritas Prep

OA: C
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by rintoo22 » Wed Apr 24, 2013 2:01 pm
In right triangle ABC, BC is the hypotenuse. If BC is 13 and AB + AC = 15, what is the area of the triangle?

A. 2√7
B. 2√14
C. 14
D. 28
E. 56
Let AB = x and AC = y
Given AB + AC = 15 => x+y=15.......(1)
Hypotenuse = 13
Therefore x^2+y^2 = 169 .......(2)

Now solve eq 1
(x+y)^2=225
x^2 + y^2 + 2xy = 225
x^2 + y^2 = 225 - 2xy
We know from equation 2 that x^2+y^2 = 169
so 169 = 225 - 2xy
xy = 28

Area of the triangle 1/2 base * Height
1/2*28=14.....Answer c

Hope this helps
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by GMATGuruNY » Wed Apr 24, 2013 2:12 pm
freyesinsb wrote:Looking for various approaches to this Algebra problem disguised as a Geometry question.

In right triangle ABC, BC is the hypotenuse. If BC is 13 and AB + AC = 15, what is the area of the triangle?

A. 2√7
B. 2√14
C. 14
D. 28
E. 56

Source: Veritas Prep

OA: C
(AB + AC)² = 15²
AB² + 2(AB)(AC) + AC² = 225.
AB² + AC² + 2(AB)(AC) = 225.

Since ∆ABC is a right triangle:
AB² + AC² = BC²
AB² + AC² = 13²
AB² + AC² = 169.

Substituting AB² + AC² = 169 into AB² + AC² + 2(AB)(AC) = 225, we get:
169 + 2(AB)(AC) = 225
(AB)(AC) = 28.

Since A = (1/2)(AB)(AC), we get:
A = (1/2)(28) = 14.

The correct answer is C.
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by freyesinsb » Thu Apr 25, 2013 11:48 am
Thanks guys, that is the approach i took to solve. Is there any other way?
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by aaggar7 » Fri Apr 26, 2013 4:20 am
But the approach is simple and is definitely not time consuming..any good reason u r looking for an alternate approach
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