In right triangle ABC, BC is the hypotenuse. If BC is 13 and : Problem Solving (PS)

HaseebR

Intern

Joined: 22 Nov 2013

Posts: 1

Own Kudos [?]: 1 [1]

Given Kudos: 0

Send PM

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

23 Nov 2013, 11:22

1

Kudos

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

AccipiterQ

Manager

Joined: 26 Sep 2013

Posts: 151

Own Kudos [?]: 598 [1]

Given Kudos: 40

Concentration: Finance, Economics

Schools: HBS '16 (D) Sloan '16 (D) Carroll '16 (WA$) Boston U '16 (A$) Northeastern '16 (A)

GMAT 1: 670 Q39 V41

GMAT 2: 730 Q49 V41

Send PM

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

23 Nov 2013, 18:41

1

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 $\frac{1}{2}*AB*AC$.

$AB + AC = 15$ --> 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.

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.

dreambig1990

Intern

Joined: 27 Feb 2013

Posts: 40

Own Kudos [?]: 6 [0]

Given Kudos: 77

Send PM

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

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

WholeLottaLove

Senior Manager

Joined: 13 May 2013

Posts: 314

Own Kudos [?]: 565 [0]

Given Kudos: 134

Send PM

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

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 $\frac{1}{2}*AB*AC$.

$AB + AC = 15$ --> 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.

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?

L

Bunuel

Math Expert

Joined: 02 Sep 2009

Posts: 92948

Own Kudos [?]: 619260 [0]

Given Kudos: 81609

Send PM

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

13 Dec 2013, 02:57

Expert Reply

WholeLottaLove wrote:

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 $\frac{1}{2}*AB*AC$.

$AB + AC = 15$ --> 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.

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?

WholeLottaLove

Senior Manager

Joined: 13 May 2013

Posts: 314

Own Kudos [?]: 565 [0]

Given Kudos: 134

Send PM

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

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.

dreambig1990

Intern

Joined: 27 Feb 2013

Posts: 40

Own Kudos [?]: 6 [0]

Given Kudos: 77

Send PM

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

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 $X1.Y1/2=X2.Y2/2= 28/2=14$.

Hope it help!!

meenanke

Intern

Joined: 26 Mar 2014

Posts: 4

Own Kudos [?]: 5 [0]

Given Kudos: 4

Send PM

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

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

jasimuddin

Intern

Joined: 16 Oct 2012

Status:My heart can feel, my brain can grasp, I'm indomitable.

Affiliations: Educator

Posts: 35

Own Kudos [?]: 23 [0]

Given Kudos: 51

Location: Bangladesh

WE:Social Work (Education)

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

24 Oct 2015, 02:02

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

ft029

Intern

Joined: 31 Oct 2015

Posts: 1

Own Kudos [?]: 1 [1]

Given Kudos: 0

Send PM

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

31 Oct 2015, 19:13

1

Bookmarks

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.

S

Nunuboy1994

Director

Joined: 12 Nov 2016

Posts: 569

Own Kudos [?]: 118 [0]

Given Kudos: 167

Location: United States

Schools: Yale '18

GMAT 1: 650 Q43 V37

GRE 1: Q157 V158

GPA: 2.66

Send PM

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

10 Mar 2017, 13:19

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

S

Nunuboy1994

Director

Joined: 12 Nov 2016

Posts: 569

Own Kudos [?]: 118 [0]

Given Kudos: 167

Location: United States

Schools: Yale '18

GMAT 1: 650 Q43 V37

GRE 1: Q157 V158

GPA: 2.66

Send PM

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

11 Mar 2017, 23:45

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

S

Nunuboy1994

Director

Joined: 12 Nov 2016

Posts: 569

Own Kudos [?]: 118 [0]

Given Kudos: 167

Location: United States

Schools: Yale '18

GMAT 1: 650 Q43 V37

GRE 1: Q157 V158

GPA: 2.66

Send PM

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

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 $\frac{1}{2}*AB*AC$.

$AB + AC = 15$ --> 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.

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?

L

Bunuel

Math Expert

Joined: 02 Sep 2009

Posts: 92948

Own Kudos [?]: 619260 [0]

Given Kudos: 81609

Send PM

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

16 Apr 2017, 01:18

Expert Reply

Nunuboy1994 wrote:

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 $\frac{1}{2}*AB*AC$.

$AB + AC = 15$ --> 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.

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.

L

Bunuel

Math Expert

Joined: 02 Sep 2009

Posts: 92948

Own Kudos [?]: 619260 [0]

Given Kudos: 81609

Send PM

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

16 Apr 2017, 01:23

Expert Reply

Nunuboy1994 wrote:

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 $\frac{1}{2}*AB*AC$.

$AB + AC = 15$ --> 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.

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.

B

Yashodh

Intern

Joined: 19 Dec 2021

Posts: 6

Own Kudos [?]: 3 [0]

Given Kudos: 138

GPA: 4

WE:Programming (Computer Software)

Send PM

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

06 Jan 2022, 01:16

area of triangle = (a+b+c) * r / 2
where , r is in-radii, c is hypotenuse

for a right triangle inradii = (a+b-c)/2 ( good formula to remember but only for right triangle)

so area of triangle is
= (a+b+c)(a+b-c)/4
= (15+13)(15-13)/4
= 28 * 2 / 4
= 14

another good formula to remember
area of triangle = (a*b*c)/(4R) here R is circumradii

if it is a right triangle R = c /2 , c is hypotenuse

bumpbot

Non-Human User

Joined: 09 Sep 2013

Posts: 32691

Own Kudos [?]: 822 [0]

Given Kudos: 0

Send PM

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

08 Mar 2023, 06:24

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.

gmatclubot

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

08 Mar 2023, 06:24

GMAT Problem Solving (PS) Questions

In right triangle ABC, BC is the hypotenuse. If BC is 13 and