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Triangles - Data Sufficiency #157, OG (12th Edition)

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Triangles - Data Sufficiency #157, OG (12th Edition)

by Chick » Mon Dec 07, 2009 11:07 pm
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 centimetres.

(2) The 2 legs of the triangle are of equal length.


The correct answer is D.

I understand that statement 2 is sufficient b/c it is tells us that the triangle is an isosceles, meaning that base ("x") and height ("y") are the same length and that we can find the perimeter. It appears more difficult to understand why statement 1 is sufficient. I understand it but it is a bit convoluted in OG's explanation because it states an additional formula to find x+y, namely:
(x+y)^2 = x^2 + y^2 + 2xy
(x+y)^2 = 100 + 2(50)
(x+y)^2 = 100 + 2(50)
(x+y)^2 = 100 + 2(50)
x+y = square root of 200 (this gives you the two missing sides x+y, and thus enables you to find the perimeter)

Is it necessary to use the formula above to find that statement 1 is sufficient? Can someone provide a different approach from the OG to why statement 2 is sufficient? Thanks.
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by Testluv » Tue Dec 08, 2009 1:06 am
No, it is not at all necessary to use that formula. In fact, you don't have to do ANY algebraic manipulation to solve this problem. In this question, we can employ the most powerful strategy in DS: the n number of unknowns requires n number of distinct linear equations rule.

Applied here:

Well, the question asks us to figure out the perimeter. In order to do that we need to know the lengths of the other two sides. Let's call the other two sides "a" and "b". So, let's just slightly rephrase the question to: "can we figure out a and b?"

How many equations are in the question stem? One. This one: a^2 + b^2 = 10^2. (Strictly speaking this is not a linear equation; however, because this is geometry we can't have negative values here!). And how many unknowns are in this equation? Two: a and b. That's two unknowns but one equation. Therefore, in order to be sufficient, a statement simply needs to provide one more equation. It is also important that this equation not introduce a third unknown (b/c then we would have three unknowns but only two equations).

(1) The area of the triangle is 25 square centimetres.

This provides us with another equation without introducing a third unknown; therefore, sufficient.

(The equation is: 0.5 * a*b = 25. We know that a and b are the two legs of a right triangle, and that, therefore, one of them is the height and the other the base. But because this is DS, we just need to recognize that we have this equation--no need to jot it down even).

(2) The 2 legs of the triangle are of equal length.

This, too, provides an additional equation: a = b. Therefore, sufficient.

Both statements independently sufficient: choose D.

Notice that under this approach, we did not need to manipulate any equations and we only needed to refer to the bare minimum of geometry concepts: the formula for the area of a triangle, the pythagoras formula, and that's it. That's why the n number of unknowns requires n distinct linear equations rule is the most powerful tactic in DS. What also helped was noticing that the real question was just whether we could figure out the lengths of the two legs of the triangle, and so rephrasing the question slightly. The OG writes explanations to prove that the accredited response is categorically correct; not necessarily to show the most efficient approach. That's why OG explanations are so technical, so textbook.
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by Ian Stewart » Sat Dec 12, 2009 3:16 am
I have to disagree with the post above; I find it very risky to try to apply some distorted version of the 'n linear equations/n unknowns' rule to similar questions at the higher level of the GMAT. Of course sometimes this will give the right answer, but as often as not it will land a test taker in a trap. In the question above, it does not ask for the values of our unknowns, but rather for their sum; we may need less information to find some combination of x and y than to find either x or y alone. Second, neither of our equations are linear, so the rule doesn't even apply. Third, while we cannot have negative values in geometry, that does not mean we can ignore the fact that our equations are non-linear, as the first question below illustrates. Were you to apply such a 'rule' to either of the following questions, for example, you would not get the right answer:

If the lengths of three sides of a right triangle are x, y and 10, where 10 is the length of the hypotenuse, what is the value of x?

1) x + y = 14
2) xy = 48

Despite the fact that we have two equations, two unknowns when we consider statement 1 alone, we still can't determine whether x is equal to 6 or to 8. The same is true when we use Statement 2 alone. Even combining the statements, we have three distinct equations, two unknowns, and still cannot find x. The answer is E.

If the lengths of the three sides of a triangle are x, y and 7, what is the perimeter of the triangle?

1) x = 10 - y
2) x - y = 2



This is a rather simple example, but illustrates that we don't always need to know the values of x or y alone to find the value of x+y; that is, we don't always need two distinct equations in our two unknowns if we are asked to find some combination of those unknowns. Using Statement 1, we have only one equation, two unknowns, but we can find x+y; the answer is A.

Questions 123, 150 and 168 in the DS section of OG12, to give just three official examples among many that I have seen, seem specifically designed to trap test takers who simply count equations and unknowns. Questions 150 and 168 in particular ask about combinations of unknowns; we don't need to find the value of each unknown to answer these questions.

Back to the original question:
Chick wrote: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 centimetres.

(2) The 2 legs of the triangle are of equal length.
I do think the algebraic solution in the OG is worth looking at here. The (x+y)^2 = x^2 + 2xy + y^2 relationship appears very frequently on the GMAT, and if you ever see any two of the expressions x+y, xy and x^2 + y^2 in a question, you'll be able to use this relationship (and similarly, if you see two of x-y, xy and x^2 + y^2 you'll be able to use the fact that (x-y)^2 = x^2 - 2xy + y^2).

There are certainly other ways to look at the problem. As an alternative to the algebraic solution in the OG, you might notice that, if x and y are the lengths of the legs of the triangle, then x^2 + y^2 = 100 and xy = 50, so x^2 + y^2 = 2xy, and x^2 - 2xy + y^2 = 0. Factoring, (x-y)^2 = 0, so x = y, and we have the information given in Statement 2.

I can see a couple of purely geometric (non-algebraic) ways to solve the problem, but I think an algebraic approach may be simpler here. Anyway, one geometric solution: draw a circle around our right triangle. Since we have a right triangle, the hypotenuse must be a diameter. So the radius of our circle is 5. Now draw a height from the hypotenuse to the opposite vertex. Since the area of the triangle is 25, the height is of length 5. So this height must also be a radius of the circle, and therefore emanates from the midpoint of the hypotenuse. From here, we must have an isosceles triangle, and the result follows. Much easier to explain with a diagram!
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by Testluv » Sat Dec 12, 2009 7:18 am
Yes, for sure, Ian. As it happens, in this particular problem it is not problematic, but I definitely shouldn't have suggested that just b/c there isn't a risk of negative values that it is okay to treat a quadratic equation as linear--too eager to simplify the problem and huge oversight on my part--thank you for pointing it out.
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by Chick » Mon Dec 14, 2009 4:51 pm
Ian,

Thank you very much for your explanation! It is great to know the algebraic and non-algebraic approaches to solve this problem.
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by sumanr84 » Sun Jul 11, 2010 4:16 am
Ian Stewart wrote:
If the lengths of three sides of a right triangle are x, y and 10, where 10 is the length of the hypotenuse, what is the value of x?

1) x + y = 14
2) xy = 48
This is a real tricky question and might land up in consuming huge time in exam..Worth knowing..thanks a lot Ian..
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by GMATGuruNY » Mon Jul 12, 2010 8:15 pm
Chick wrote: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 centimetres.

(2) The 2 legs of the triangle are of equal length.


The correct answer is D.

I understand that statement 2 is sufficient b/c it is tells us that the triangle is an isosceles, meaning that base ("x") and height ("y") are the same length and that we can find the perimeter. It appears more difficult to understand why statement 1 is sufficient. I understand it but it is a bit convoluted in OG's explanation because it states an additional formula to find x+y, namely:
(x+y)^2 = x^2 + y^2 + 2xy
(x+y)^2 = 100 + 2(50)
(x+y)^2 = 100 + 2(50)
(x+y)^2 = 100 + 2(50)
x+y = square root of 200 (this gives you the two missing sides x+y, and thus enables you to find the perimeter)

Is it necessary to use the formula above to find that statement 1 is sufficient? Can someone provide a different approach from the OG to why statement 2 is sufficient? Thanks.
Sometimes you can streamline the process of evaluating statement 1 by looking ahead to statement 2.

Statement 2: The 2 legs are equal.

What happens if we make the 2 legs equal in statement 1?

Statement 1: The area is 25. If the 2 legs are equal, then (1/2)(x)(x) = 25, x^2=50, and x = sqrt(50). Would this value give us a hypotenuse of 10? Yes, because sqrt(50)^2 + sqrt(50)^2 = 10^2.

Because we're restricted to positive values, no other combination of values could satisfy the two conditions we've been given: xy = 50 and x^2 + y^2 = 100. Since we don't need to know the value of x or y individually but only their sum, statement 1 is sufficient.

Using the info in statement 2 allows us to see the only combination of values that will satisfy statement 1.

But a big caveat: This approach can be dangerous. The two statements have to be logically consistent, so the information in statement 2 will never contradict the information in statement 1. Just because statement 2 presents a possible scenario that will work in statement 1 doesn't mean you will have found the only scenario that could work in statement 1. So tread lightly.
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