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## OG 13 128

tagged by: Brent@GMATPrepNow

This topic has 3 expert replies and 1 member reply
oquiella Master | Next Rank: 500 Posts
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#### OG 13 128

Tue Dec 22, 2015 3:34 pm
Elapsed Time: 00:00
• Lap #[LAPCOUNT] ([LAPTIME])
The hypotenuse of a right triangle is 10cm. What is the perrimeter, in centimeters, of the triangle?

1. The area of the triangle is 25 square cetimeters
2. The 2 legs of the triagle are of equal length.

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Marty Murray Legendary Member
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Tue Dec 22, 2015 9:35 pm
Quote:
The hypotenuse of a right triangle is 10cm. What is the perimeter, in centimeters, of the triangle?

1. The area of the triangle is 25 square centimeters
2. The 2 legs of the triangle are of equal length.
The perimeter of the triangle = the total of the lengths of the three sides of the triangle, a + b + c. Since we already know the length of the hypotenuse, c, we just need either a and b, or a + b to calculate the perimeter.

Statement 1:

Since we know that c = 10, we know that a² + b² = 100

The area is 25, and the triangle is a right triangle. So a and b can be the height and base.

So, we know that ab/2 = 25.

So ab = 50.

We can can combine what we have to get the following.

a² + 2ab + b² = 200 --> (a + b)² = 200

So a + b = √200 and we have the answer to the rephrased question.

NOTICE, that while we don't have a or b, we know that a + b + c is √200 + 10.

Sufficient.

Statement 2:

From Statement 2 we know that a = b.

So we can use a² + a² = c² = 100 to calculate the lengths of a and b and answer the rephrased question.

Sufficient.

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GMATGuruNY GMAT Instructor
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Wed Dec 23, 2015 5:02 am
oquiella wrote:
The hypotenuse of a right triangle is 10cm. What is the perrimeter, in centimeters, of the triangle?

1. The area of the triangle is 25 square cetimeters
2. The 2 legs of the triangle are of equal length.

Statement 2:
A right triangle with 2 equal legs is an ISOSCELES right triangle.
In an isosceles right triangle, the sides are in the following ratio:
x : x : x√2.

The prompt indicates that the hypotenuse has a length of 10.
Since x√2 = 10, we get:
x = 10/√2.

Thus, the triangle has a base and height of 10/√2 and a hypotenuse of 10.
Since the lengths of all 3 sides are known, the perimeter can be determined.
SUFFICIENT.

Statement 1:
Since the two statements cannot contradict each other, it must be possible that the triangle implied by statement 2 also satisfies statement 1.
Thus, an area of 25 MUST be yielded by a right triangle with a base and height of 10/√2 and a hypotenuse of 10.
Proof:
If b = 10/√2 and h = 10/√2, then A = (1/2)bh = (1/2)(10/√2)(10/√2) = 25.

Check whether a DIFFERENT right triangle with a hypotenuse of 10 can also yield an area of 25.
A common right triangle with a hypotenuse of 10 is a 6-8-10 triangle.
The area of a 6-8-10 triangle = (1/2)(6)(8) = 24.
Doesn't work.
Implication:
For a right triangle with a hypotenuse of 10 to yield an area of 25, it MUST have a base and height of 10/√2.

Since the lengths of all 3 sides are known, the perimeter can be determined.
SUFFICIENT.

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Last edited by GMATGuruNY on Thu Dec 24, 2015 5:38 am; edited 1 time in total

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### GMAT/MBA Expert

Rich.C@EMPOWERgmat.com Elite Legendary Member
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Wed Dec 23, 2015 9:30 am
Hi oquiella,

One of the interesting things about the GMAT is that most questions can be approached in a few different ways, so it is beneficial to be a "strong thinker" on Test Day. Knowing how to do math is sometimes what's needed (although sometimes the math can be done in several different ways); in other situations, the ability to figure out patterns is what's needed.

Whoever wrote this explanation is taking a real heavy "math" approach to the problem, which is fine. I think you'll find this next approach to be a bit easier…

The prompt tells us that the hypotenuse of a right triangle is 10. The question asks us for the perimeter, so we're going to need the lengths of the other two sides.

Since we have a right triangle, we can use the Pythagorean Theorem:

X^2 + Y^2 = 10^2

**Important: triangles can't have "negative" sides, so both X and Y MUST be positive.**

We have 2 variables, but only 1 equation, so we can't figure out the exact values of X and Y

Fact 1: Triangle area = 25

Since Area = (1/2)(B)(H) we know that….

(1/2)(X)(Y) = 25
(X)(Y) = 50

Now we have ANOTHER equation. Combined with what we were given in the beginning, we have…

Two variables AND two UNIQUE equations, which means we have a "system" of equations and we CAN solve for X and Y. No more math is needed.
Fact 1 is SUFFICIENT.

Fact 2: The two legs of the triangle are equal.

Now we know that X = Y.

This gives us a SECOND equation to work with. Just as in Fact 1, we have a "system" and we can solve for X and Y.
Fact 2 is SUFFICIENT

Be on the lookout for "system" questions. You'll see a couple on Test Day and they're built around some useful math "patterns" that you can use to avoid doing work.

GMAT assassins aren't born, they're made,
Rich

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### GMAT/MBA Expert

Brent@GMATPrepNow GMAT Instructor
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Wed Dec 23, 2015 3:53 pm
Quote:
The hypotenuse of a right triangle is 10 cm. What is the perimeter, in centimeters, of the triangle?

(1) The area of the triangle is 25 square centimeters.
(2) The 2 legs of the triangle are of equal length
Plugging in numbers might be challenging, because we'd need to find values that satisfy BOTH the given information (hypotenuse = 10) AND the information in the statements.

IMPORTANT: For geometry DS questions, we are typically checking to see whether the statements "lock" a particular angle or length into having just one value. This concept is discussed in much greater detail in our free video: http://www.gmatprepnow.com/module/gmat-data-sufficiency?id=1103

So, for this question, if a statement FORCES our right triangle into having ONE AND ONLY ONE shape and size, then that statement is sufficient. Moreover, we NEED NOT find the actual perimeter of the triangle. We need only recognize that we could find its perimeter (finding the perimeter will just waste time).

Okay, onto the question....

Target question: What is the perimeter of the right triangle?

Given: The hypotenuse of the triangle has length 10 cm.

Statement 1: The area of the triangle is 25 square centimeters.
Let's let x = length of one leg
Also, let y = length of other leg
So, if the area is 25, we can write (1/2)xy = 25 [since area = (1/2)(base)(height)]
Multiply both sides by 2 to get xy = 50
Multiply both sides by 2 again to get 2xy = 100 [you'll soon see why I performed this step]

Now let's deal with the given information (hypotenuse has length 10)
The Pythagorean Theorem tells us that x² + y² = 10²
In other words, x² + y² = 100

We now have two equations:
2xy = 100
x² + y² = 100

Since both equations are set equal to 100, we can write: 2xy = x² + y²

Rearrange this to get x² - 2xy + y² = 0
Factor to get (x - y)(x - y) = 0
This means that x =y, which means that the two legs of our right triangle HAVE EQUAL LENGTH.

So, the two legs of our right triangle have equal length AND the hypotenuse has length 10.
There is only one such right triangle in the universe, so statement 1 FORCES our right triangle into having ONE AND ONLY ONE shape and size.

This means that statement 1 is SUFFICIENT

Statement 2: The 2 legs of the triangle are of equal length
We already covered this scenario in statement 1.
So, statement 2 is also SUFFICIENT

Cheers,
Brent

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