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In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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23 Nov 2013, 11:01

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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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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23 Nov 2013, 18:41

1

This post received KUDOS

Bunuel wrote:

ronr34 wrote:

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

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.

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.

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

Show Tags

23 Nov 2013, 22:49

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

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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12 Dec 2013, 14:11

Bunuel wrote:

ronr34 wrote:

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

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.

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?

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

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.

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?
_________________

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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13 Dec 2013, 06:19

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.

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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13 Dec 2013, 10:30

WholeLottaLove wrote:

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 \(X1.Y1/2=X2.Y2/2= 28/2=14\).

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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31 Mar 2014, 22:33

AccipiterQ wrote:

dreambig1990 wrote:

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

ah ok, thanks! I appreciate it

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

Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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25 Apr 2015, 18:43

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Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
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Re: In right triangle ABC, BC is the hypotenuse. If BC is 13 and [#permalink]

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15 Apr 2017, 17:59

Bunuel wrote:

ronr34 wrote:

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

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.

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?

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

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.

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.
_________________

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

I have a question regarding this. If Bc is the hypotenuse, and the triangle is a right triangle, that means that the other two side must be 5:12:13.... How can it be stated that the other sides sum to be 15?

Please explain. thanks

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 \(\frac{1}{2}*AB*AC\).

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.

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?

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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