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# M32-06

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Math Expert
Joined: 02 Sep 2009
Posts: 50730

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04 Jul 2016, 04:39
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Difficulty:

35% (medium)

Question Stats:

67% (01:16) correct 33% (00:51) wrong based on 51 sessions

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If triangle ABC is right angled at vertex A, what is the area of triangle ABC?

(1) AB + AC = 8.

(2) The length of the largest side of the triangle is $$5 \sqrt 2$$

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Joined: 02 Sep 2009
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04 Jul 2016, 04:39
Official Solution:

If triangle ABC is right angled at vertex A, what is the area of triangle ABC?

The are of a right triangle is $$\frac{leg_1*leg_2}{2}=\frac{AB*AC}{2}$$. Basically to answer the qurestion we need to find the value of AB*AC.

(1) AB + AC = 8. This statement is clearly insufficient. We can have infinite number of combinations of the lengths.

(2) The length of the largest side of the triangle is $$5 \sqrt 2$$. The largest side of a right triangle is hypotenuse but knowing the length of the hypotenuse is not enough to get the area. Not sufficient.

(1)+(2) Square (1): $$(AB)^2 + 2(AB)(AC)+(AC)^2 = 8^2$$. From Pythagoras theorem we know that $$(AB)^2 +(AC)^2 = (BC)^2=(5 \sqrt 2)^2=50$$. Therefore, $$(AB)^2 + 2(AB)(AC)+(AC)^2 =50+2(AB)(AC)= 8^2$$. This leads to $$(AB)(AC)=7$$. Sufficient.

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23 Jul 2016, 16:17
Statement 2: (base)^2 + (height)^2 = (hypotenuse)^2
b^2 + h^2 = (5sqrt2)^2 = 50
only values of b & h that work are b = 1 and h = 7 or b = 7 and b = 1. (1 + 49 = 50). Both lead to the same area.

Ahh, nevermind. I incorrectly assumed b and h are integers. I'll post this anyway in case someone came to the same erroneous conclusion.
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Joined: 27 Apr 2015
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25 Oct 2016, 22:58
Bunuel wrote:
If triangle ABC is right angled at vertex A, what is the area of triangle ABC?

(1) AB + AC = 8.

(2) The length of the largest side of the triangle is $$5 \sqrt 2$$

1. Why is B not the answer? we can say it is an isosceles right-angled triangle.

2. I am not able to comprehend the solution, how do we arrive at 2(AB)(AC)

Thank you.
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Joined: 02 Sep 2009
Posts: 50730

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26 Oct 2016, 00:00
Bunuel wrote:
If triangle ABC is right angled at vertex A, what is the area of triangle ABC?

(1) AB + AC = 8.

(2) The length of the largest side of the triangle is $$5 \sqrt 2$$

1. Why is B not the answer? we can say it is an isosceles right-angled triangle.

2. I am not able to comprehend the solution, how do we arrive at 2(AB)(AC)

Thank you.

1. Knowing the length of the hypotenuse is not sufficient to get the area.

2:
Square (1): $$(AB)^2 + 2(AB)(AC)+(AC)^2 = 8^2$$.

From Pythagoras theorem we know that $$(AB)^2 +(AC)^2 = (BC)^2=(5 \sqrt 2)^2=50$$.

Substitute $$(AB)^2 +(AC)^2 = 50$$ into $$(AB)^2 + 2(AB)(AC)+(AC)^2 = 8^2$$ to get $$50+2(AB)(AC)= 8^2$$.

$$50+2(AB)(AC)= 8^2$$

$$(AB)(AC)=7$$. Sufficient.
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02 Dec 2016, 17:11
Hi Bunuel,

In these problems, is it wrong to assume that because triangle ABC is a right triangle and its hypotenuse is 5√2, then it's other two sides are 5 and 5? (its a 45,45,90 triangle).

Also, if I think that this (2 sides with a length of 5) could be an option, can I compare these results with the first statement?
Knowing that they don't add up to 8 could be a clue that those values can't be valid. Would this comparison be okay? (considering that statements don't contradict each other)

Thanks!
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Joined: 02 Sep 2009
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03 Dec 2016, 00:28
nelliegu wrote:
Hi Bunuel,

In these problems, is it wrong to assume that because triangle ABC is a right triangle and its hypotenuse is 5√2, then it's other two sides are 5 and 5? (its a 45,45,90 triangle).

Also, if I think that this (2 sides with a length of 5) could be an option, can I compare these results with the first statement?
Knowing that they don't add up to 8 could be a clue that those values can't be valid. Would this comparison be okay? (considering that statements don't contradict each other)

Thanks!

As a rule one shouldn't assume anything on the DS. Why should ABC be isosceles? Is there any single piece of information that supports this? NO!

You are right about the second observation though - if you wrongly assume that ABC is in fact isosceles you could quickly see that it cannot be true because it contradicts the first statement and the statements in DS never contradict each other.
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25 Mar 2017, 05:23
hi,

can someone pl. confirm if the following relation is true for a right angled trianle:

Median = 0.5 * hypotenuse
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25 Mar 2017, 05:28
Ashokshiva wrote:
hi,

can someone pl. confirm if the following relation is true for a right angled trianle:

Median = 0.5 * hypotenuse

Yes.

In a right triangle, the median drawn to the hypotenuse has the measure half the hypotenuse.
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05 Apr 2017, 02:45
Bunuel

I have a query on this question.
I reached to the conclusion that 1 and 2 are not sufficient independently.
However, while combining 1 & 2, I took the below options.

AB+BC=8 can have options many options of value like 1+7, 2+6, 3+5, 4+4.
and as (2) says that Hyp \sqrt{50} then it means that sides AB and BC have to have values whose sum = 8.

So, while solving it, I came across that AB and AC can have all different values. (are we not supposed to take multiple combinations for sum of 8?)
With different values of the sides, there shall be multiple options, and hence I concluded that C is not sufficient and that E is the answer.

Also, I have gone through the OA and OE given by you. That makes sense to my mind.
But I still can't figure where my reasoning went wrong?

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05 Apr 2017, 03:03
GMATAspirer09 wrote:
Bunuel

I have a query on this question.
I reached to the conclusion that 1 and 2 are not sufficient independently.
However, while combining 1 & 2, I took the below options.

AB+BC=8 can have options many options of value like 1+7, 2+6, 3+5, 4+4.
and as (2) says that Hyp \sqrt{50} then it means that sides AB and BC have to have values whose sum = 8.

So, while solving it, I came across that AB and AC can have all different values. (are we not supposed to take multiple combinations for sum of 8?)
With different values of the sides, there shall be multiple options, and hence I concluded that C is not sufficient and that E is the answer.

Also, I have gone through the OA and OE given by you. That makes sense to my mind.
But I still can't figure where my reasoning went wrong?

1. From x + y = 8 you cannot assume that the only solutions are the ones you've listed. We are not told that x and y are integers, thus x + y = 8 has infinitely many solutions.

2. We have two equations: x + y = 8 and x^2 + y^2 = 50. If you solve them in conventional way, you'll get that x = 1 and y = 7 OT x = 7 and y = 1. The area is xy/2, so in either way you get the same answer 7/2. In my solution above we were able to get the area without solving the equation, so we took the shortcut way to get the value of xy and not individual values of x and y.

Hope it helps.
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05 Dec 2017, 13:23
How did you know to go from "Substitute (AB)2+(AC)2=50 into (AB)2+2(AB)(AC)+(AC)2=8^2"? Thanks
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05 Dec 2017, 20:07
gkestler wrote:
How did you know to go from "Substitute (AB)2+(AC)2=50 into (AB)2+2(AB)(AC)+(AC)2=8^2"? Thanks

Substitute (AB)^2 +(AC)^2 = 50 into (AB)^2 + 2(AB)(AC)+(AC)^2 = 8^2 to get 50+2(AB)(AC)= 8^2.

Could you please tell me what is unclear there?
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Re: M32-06 &nbs [#permalink] 05 Dec 2017, 20:07
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# M32-06

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