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Can we plug in this and solve

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shibsriz@gmail.com Master | Next Rank: 500 Posts Default Avatar
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Can we plug in this and solve

Post Sun Apr 13, 2014 7:32 am
157.
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

Ans-D

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Post Sun Apr 13, 2014 8:44 am
shibsriz@gmail.com wrote:
157.
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

Answer = D

Cheers,
Brent

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Post Sun Apr 13, 2014 5:18 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.
Start with statement 2, which seems easier to evaluate.

Statement 2: The 2 legs of the triangle are of equal length
In an isosceles right triangle, the sides are in the following ratio: s : s : s√2.
Since the hypotenuse = s√2 = 10, s = 10/√2.
Thus, p = 10/√2 + 10/√2 + 10.
SUFFICIENT.

Statement 1: The area of the triangle is 25 square centimeters
Only ONE case satisfies statement 2 above: a triangle with sides of 10/√2, 10/√2, and 10.
Since the two statements cannot contradict each other, this case MUST also satisfy statement 1.
In other words, the area of a triangle with sides of 10/√2, 10/√2, and 10 must be 25:
(1/2)(10/√2)(10/√2) = 100/4 = 25.

Check whether this is the ONLY case that satisfies statement 1.
Another right triangle with a hypotenuse of 10 is a 6-8-10 triangle.
In a 6-8-10 triangle, the area = (1/2)(6)(8) = 24, which doesn't satisfy the constraint in statement 1 that the area = 25.

Implication:
Only ONE right triangle has a hypotenuse of 10 and an area of 25 -- the same triangle implied by statement 1:
10/√2, 10/√2, 10.
Thus, p = 10/√2 + 10/√2 + 10.
SUFFICIENT.

The correct answer is D.

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Post Sun Apr 13, 2014 11:26 pm
Hi shibsriz,

This Geometry question is actually based on an algebra concept: System math.

We're given the hypotenuse of a right triangle (10cm) and we're asked for the PERIMETER of the triangle. This means that we'll need to know the length and width of the triangle.

To start, we can use the Pythagorean Theorem: L^2 + W^2 = 10^2. This is one equation with 2 variables. Normally, this would be a much more complex idea, but since we're dealing with Geometry rules, there's no such thing as a "negative side length", so we know that both L and W must be positive. We'll need another formula to combine with this one if we want to answer this question.

Fact 1: The area of the triangle is 25 cm^2.

This Fact helps us to create the following formula: (1/2)(Base)(Height) = (1/2)(L)(W) = 25. so (L)(W) = 50

Combined with the original equation, we now have 2 variables and 2 unique equations, so we CAN solve this and calculate the Length and Width.
Fact 1 is SUFFICIENT.

Fact 2: The 2 legs are of equal length.

This tells us that L = W.

Combined with the original equation, we again have 2 variables and 2 unique equations, so we CAN solve this and calculate Length and Width.
Fact 2 is SUFFICIENT.

Final Answer: D

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

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Last edited by Rich.C@EMPOWERgmat.com on Sun Jun 05, 2016 2:30 pm; edited 1 time in total

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Post Mon Apr 14, 2014 9:21 am
Hi Rich,

That part about our ability to solve "2 variables and 2 unique equations" [and get only one unique solution] really only applies to linear equations. In this question, we have two quadratic equations, and some systems of quadratic equations can yield up to 4 unique solutions.

I mention this because I wouldn't want students to apply this concept to other systems that aren't linear.

Cheers,
Brent

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Post Mon Apr 14, 2014 11:47 am
Hi Brent,

I absolutely agree, which is why I noted that the idea is normally more complex (such as in a graphing question). Since we have a fixed hypotenuse of 10 in a right triangle and we're not doing anything but talking about the lengths of the remaining 2 sides, the options are severely limited. We can't have a "negative" side length, so the options/possibilities that you warn about are not possible in this scenario.

I too would not want a Test Taker to misapply any concepts that he/she learns on this site; if a tactic does apply though, and I haven't seen it discussed by any of the other experts here, then I'll present it.

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

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Post Tue Apr 15, 2014 5:37 am
Fair enough.

Cheers,
Brent

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Pazoki Junior | Next Rank: 30 Posts Default Avatar
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Post Sun Jun 05, 2016 12:28 pm
Rich.C@EMPOWERgmat.com wrote:
Hi shibsriz,

This Geometry question is actually based on an algebra concept: System math.

We're given the hypotenuse of a right triangle (10cm) and we're asked for the PERIMETER of the triangle. This means that we'll need to know the length and width of the triangle.

To start, we can use the Pythagorean Theorem: L^2 + W^2 = 10^2. This is one equation with 2 variables. Normally, this would be a much more complex idea, but since we're dealing with Geometry rules, there's no such thing as a "negative side length", so we know that both L and W must be positive. We'll need another formula to combine with this one if we want to answer this question.

Fact 1: The area of the triangle is 25 cm^2.

This tells us LW = 25.

Combined with the original equation, we now have 2 variables and 2 unique equations, so we CAN solve this and calculate the Length and Width.
Fact 1 is SUFFICIENT.

Fact 2: The 2 legs are of equal length.

This tells us that L = W.

Combined with the original equation, we again have 2 variables and 2 unique equations, so we CAN solve this and calculate Length and Width.
Fact 2 is SUFFICIENT.

Final Answer: D

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

We have 1/2BH=Area so is your quote "Fact 1: The area of the triangle is 25 cm^2.

This tells us LW = 25." right? I have just started so I can't understand it.

Thanks

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Post Sun Jun 05, 2016 2:32 pm
Hi Pazoki,

There was a slight error in my original post. Since Area = 25, the product LW actually equals 50. So to answer your question - YES:

(1/2)(Base)(Height) = 25
(1/2)(L)(W) = 25

so (L)(W) = 50

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

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Post Tue Jun 07, 2016 11:24 pm
Pazoki wrote:
We have 1/2BH=Area so is your quote "Fact 1: The area of the triangle is 25 cm^2.

This tells us LW = 25." right? I have just started so I can't understand it.

Thanks
BTW nice catch Pazoki - an eye for detail like this is a great predictor of success on the actual test! Smile

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nasahtahir Newbie | Next Rank: 10 Posts
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Post Sat Nov 05, 2016 8:30 pm
Just to get it out of the way, every time in a Geometry question where A STATEMENT LOCKS in the diagram that statement is considered to be sufficient?

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Post Sat Nov 05, 2016 8:51 pm
Hi nasahtahir,

You have to be careful about thinking in these types of 'generalities' - in DS questions, you have to make sure that you have PROOF of whatever you think is the correct answer. A statement/Fact is considered sufficient if that additional information results in JUST ONE answer (either just one numerical answer to a Value question or a consistent Yes - or a consistent No - answer to a YES/NO question).

In this prompt, each of the two individual Facts, when combined with what we already knew in the prompt, was enough to prove that there was just one answer to the given question. Thus, each Fact was sufficient.

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

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Post Sun Nov 06, 2016 6:48 am
nasahtahir wrote:
Just to get it out of the way, every time in a Geometry question where A STATEMENT LOCKS in the diagram that statement is considered to be sufficient?
If a statement "locks" the diagram in a way that creates only one possible answer to the target question, then that statement is sufficient. Statement 1 locks our triangle into an isosceles right triangle with a hypotenuse of length 10. Since there is only one such right triangle in the universe, there can only be 1 answer to our target question. So, that statement is sufficient.

Cheers,
Brent

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nasahtahir Newbie | Next Rank: 10 Posts
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Post Sun Nov 06, 2016 6:07 pm
Brent@GMATPrepNow wrote:
nasahtahir wrote:
Just to get it out of the way, every time in a Geometry question where A STATEMENT LOCKS in the diagram that statement is considered to be sufficient?
If a statement "locks" the diagram in a way that creates only one possible answer to the target question, then that statement is sufficient. Statement 1 locks our triangle into an isosceles right triangle with a hypotenuse of length 10. Since there is only one such right triangle in the universe, there can only be 1 answer to our target question. So, that statement is sufficient.

Cheers,
Brent
I understand but again correct me if I'm wrong isn't our target question "What is the perimeter" ?
Sure with the process we found out x=y that means it is an isosceles triage so what? We would still require the 1 more side if not two. Please explain :s

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Post Mon Nov 07, 2016 4:46 am
nasahtahir wrote:
Brent@GMATPrepNow wrote:
nasahtahir wrote:
Just to get it out of the way, every time in a Geometry question where A STATEMENT LOCKS in the diagram that statement is considered to be sufficient?
If a statement "locks" the diagram in a way that creates only one possible answer to the target question, then that statement is sufficient. Statement 1 locks our triangle into an isosceles right triangle with a hypotenuse of length 10. Since there is only one such right triangle in the universe, there can only be 1 answer to our target question. So, that statement is sufficient.

Cheers,
Brent
I understand but again correct me if I'm wrong isn't our target question "What is the perimeter" ?
Sure with the process we found out x=y that means it is an isosceles triage so what? We would still require the 1 more side if not two. Please explain :s
We know that x = y, AND it is given that the hypotenuse has length 10.

Cheers,
Brent

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