You assume with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 13 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 5:12:13. Or in other words: if \(a^2+b^2=13^2\) DOES NOT mean that \(a=5\) and \(b=12\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=13^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=5\) and \(b=12\).

For example: \(a=1\) and \(b=\sqrt{168}\) or \(a=2\) and \(b=\sqrt{165}\) ...

Hope it's clear.

BACK TO THE ORIGINAL 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

Since ABC is a right triangle and BC is hypotenuse then the area is \(\frac{1}{2}*AB*AC\).

\(AB + AC = 15\) --> square it: \(AB^2 + 2*AB*AC + AC^2=225\).

Also, since BC is the hypotenuse, then \(AB^2 + AC^2=BC^2=169\).

Substitute the second equation in the first: \(169+ 2*AB*AC=225\) --> \(2*AB*AC=56\) --> \(AB*AC=28\).

The area = \(\frac{1}{2}*AB*AC=14\).

Answer: C.
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Hope you guys like this method:

Refer diagram & form the equations

We require to find area of right triangle

\(= \frac{1}{2} * x * (15-x)\)

\(= \frac{1}{2} * (15x - x^2)\) .................... (1)

Using Pythagoras theorem,

\(13^2 = x^2 + (15-x)^2\)

\(169 = x^2 + 225 - 30x + x^2\)

(IMPORTANT: Don't try to solve it; just re-arrange it per requirement)

\(30x - 2x^2 = 56\)

\(15x - x^2 = 28\)..................... (2)

Placing 28 obtained from equation (2) in equation (1)

Area of right triangle \(= \frac{1}{2} * 28\)

Answer = 14 = C

Area of the triangle is 1/2*B*H = 1/2*AB*AC

Squaring on Both Sides

(AB+AC)^2 = AB^2 + AC^2 + 2AB.AC = 225

AB^2+AC^2 is 13^2

So on solving the equation we get

2AB.AC=56

1/2AB.AC=14

EDIT : Oh damn Bunuel beat me to it

How the hell did you know to square that? I tried using the pythag theorem for A^2+B^2=13^2, and since A+B=15, since A=15-B, subbing that back into the pythag theorem. Ended up with b^2-15b+28...which only has ugly solutions.

Just try to remember the relationship between area of a triangle with its sides involves The Pythagorean (or Pythagoras') Theorem a^2+b^2=c^2 and the equation (a+B)^2= a^2 + b^2 + 2ab---- in MGMAT series, they mention about this relationship too. GOOD LUCK

That is exactly what I did and it made sense right up to the last part.

We know that x+y = 15, we also know that x*y=28. There are only a few possibilities for what x and y could be:

x=1 y=28 (which we can rule out right off the bat)
x=2 y=14
x=4 y=7

None of which add up to 15. I see how you arrived at the answer: x*y (i.e. the base times the height) = 28 so you multiply by 1/2 to get the area of a triangle but shouldn't one of those sets of x, y also = 15?

Why do you assume that the lengths of the sides are integers?
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That's a good point - all that is relevant is that it is a right angle and we know that the two leg lengths, whatever they are individually, multiply to 14.

Also if you want to go from : We know that x+y = 15, we also know that \(x*y=28\)--->>>
\(y=28/x\)
\(x+28/x= 15\)
\(x^2-15x+28=0\)
\(x1= (15+\sqrt{113} )/2\) or
\(x2=(15-\sqrt{113} )/2\)
So\(Y1=(15-\sqrt{113} )/2\)
\(Y2=(15-\sqrt{113} )/2\)

--NOTICE that the results is not integers or multiply to 14 or st
So X1, Y1 and X2, Y2 just switch with each other so it's very reasonable coz' no assumption of which one is greater.
And \(X1.Y1/2=X2.Y2/2= 28/2=14\).

Hope it help!!

Would make 1 small note to "remember the relationship between the area of a RIGHT triangle with its sides involves the PT a^2+b^2=c^2 and the equation (a+b)^2=a^2+b^2" ... this relationship won't help you much in problems involving non-right triangles

x^2+ (15-x)^2=13^2
x^2-15x+28=13^2
product of two roots, AB.AC=28
Area=14

GUYS YOU DO NOT NEED TO SOLVE FOR THE SIDE LENGTHS OF THE TRIANGLE!!!

The first reply illustrated this perfectly.

All right, so let the two legs be "a" and "b". The area is ab/2. We also know that a^2+b^2=13^2, or 169.

Since a + b = 15, you can square this equation and it will stay the same: a^2+b^2+2ab = 225 (this is called "FOIL")

a^2+b^2 is known to be 169, so substitute that in to get 169+2ab=225, so 2ab=56.

We want ab/2. Divide 2ab=56 by 4 on both sides to get ab/2 = 14, which is the answer. Pick C.

I don't understand how 169 gets plugged in as (AB)^2? Help pls

Ok this problem is beginning to become more clear- but kudos if someone can explain further how we knew/ why to square AB + BC

I think the difficulty in this question lies in the fact that you cannot use the Pythagorean Theorem to solve for sides AB and BA, or alternatively the individual values of X and Y. The objective of this question, as posted, is to solve for the product of X and Y. It isn't necessary to know the individual values of X and Y in order to solve the question- the area of right triangle is

Area = (1/2) Base * H

Yet in this problem the product of sides AB and BA, or again X and Y, can be substituted for (Base * Height).

The only thing I'd like to know about this problem is, well, what are the values of sides AB and BA? The factors of 28 are 1, 2, 4 , 7 , 14, and 28
28, 14, 2 and 1 are out because 28 and 14 are larger numbers then the hypotenuse- though
7^2 + 4^2 does not equal 13^2?

Why do you assume that AB and AC must be integers? If you solve \(AB + AC = 15\) and \(AB^2 + AC^2=169\) you'll get that \(AB=\frac{15}{2} - \frac{\sqrt{113}}{2}\) and \(AC=\frac{15}{2} + \frac{\sqrt{113}}{2}\) or vise-versa.
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By the way your doubt is already addressed in the very first paragraph of the very post you are quoting as well as in several other posts in this thread.
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