The hypotenuse of a right triangle is 10 cm. What is the

For (1):
Let the legs be \(x\) and \(y\).

Given: \(h=10\) and \(\frac{xy}{2}=25\) (area of the right triangle \(area=\frac{leg_1*leg_2}{2}\)) --> \(h^2=100=x^2+y^2\) (Pythagoras) and \(xy=50\).

Q: \(P=x+y+10=?\) So we have to calculate the value \(x+y\).

Square \(x+y\) --> \((x+y)^2=x^2+2xy+y^2=(x^2+y^2)+2xy=100+2*50=200\) --> \(x+y=\sqrt{200}\).

Thus \(P=x+y+10=\sqrt{200}+10\).

Sufficient.
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First of all, we are asked to find the value of \(x+y\) not \(x\) or \(y\) and that's what we did: \(x+y=\sqrt{200}\). Now, \((x+y)^2=200\) has two solutions: \(-\sqrt{200}\) and \(\sqrt{200}\), but the first one is not valid since \(x\) and \(y\) must be positive. So there is only one acceptable numerical value of \(x+y\) possible, regardles of the individual values of \(x\) and \(y\). Which makes this statement sufficient.

Next, even if we were asked to find the value of \(x\) or \(y\) then yes, \(xy=50\) and \(x^2+y^2=100\) gives two values for \(x\) and \(y\) BUT in this case the answer would still be sufficient since again one of the values would be negative, thus not a valid solution for a length.

As for the solving DS questions: when dealing with DS problems try to avoid calculations as much as possible. Remember DS problems do not ask you to solve, but rather to determine if you are ABLE to solve and in many cases you can determine that a statement is sufficient without working out all of the math. So if you are able to see from \(xy=50\) and \(x^2+y^2=100\) that it's possible to solve for \(x+y\), then you don't need to actually do the math.

Hope it's clear.
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Bunuel, if we factored out the equation in (A) and ended up with two different values for X, does that make (A) insufficient? I think it would. Would the GMAT be cruel enough to pull this trick on us? Because if we have to solve it all the way to determine the value of x+y then it's really a PS problem rather than a DS problem.

I am asking this because I was wondering whether we really needed to solve it for A. If we end up with an non linear equation in a similar DS statement, do we really have to solve it? Aren't we supposed to just determine if we have sufficient data to solve the problem.

x^2+y^2= 100 (Pyth Theorem).....1)
2xy=100.........................2)
Adding 1) and 2)
(x+y)^2= 200
or x+y=10 sqroot 2 (Hence statement 1 is sufficient)
or Perimeter = x+y+10 = 10(1+sq.rt2)
Statement 2. 2x^2=100; x=5 sqrt2
Perimeter = 10 (1+sqrt2)..(Hence, statement 2 is sufficient)
As both statements are sufficient. Hence, D

Hi Bunuel,

Once you outlined the steps above, it's rather easy to solve. My question lies with the strategy -- how did you make the leap to "square" (x+y). What is the problem tipped you off that you had to solve a quadratic or at least, re-arrange it? Highlighted the area in question above.

Thanks!

It should come with practice...

We know the values of xy and x^2+y^2, while need to get the value of x+y. Now, if you add twice xy to x^2+y^2 you get the square of x+y, hence squaring x+y is quite natural thing to do.

Similar questions to practice:
if-p-is-the-perimeter-of-rectangle-q-what-is-the-value-of-p-135832.html
if-the-diagonal-of-rectangle-z-is-d-and-the-perimeter-of-104205.html
what-is-the-area-of-rectangular-region-r-166186.html
what-is-the-perimeter-of-rectangle-r-96381.html

Hope this helps.
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How: 1) is sufficient..
But x and y are coming out to be imaginary, so such sides of a right triangle don't exist.....so not point of calculating perimeter from (1)

thought we can calculate ...x+y....because imaginary parts cancelled out in addition...but if x and y are not real how can we think of perimeter...

\(x+y=\sqrt{200}\) so it's an irrational number, not imaginary number.
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Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Let a and b be two legs of the right triangle and c be its hypotenuse.
Then we have \(a^2 + b^2 = c^2\) and \(c = 10\).
Thus there are 3 variables and 1 equation. By using VA method, D is the answer most likely.

Condition 1) : \(\frac{1}{2}ab = 25\) or \(ab = 50\)
\((a+b)^2 = a^2 + b^2 + 2ab = c^2 + 2*50 = 100 + 100 = 200\).
Thus \(a + b = \sqrt{200} = 10\sqrt{2}\).
Hence \(a + b + c = 10 +10\sqrt{2}\).
It is sufficient.

Condition 2) : \(a = b\)
We have \(a^2 + b^2 = 2a^2 = c^2 = 100\).
Then \(a^2 = 50\) or \(a = 5\sqrt{2}\).
Thus \(a + b + c = 2 \cdot 5\sqrt{2} + 10 = 10 + 10\sqrt{2}\).
It is also sufficient.

D is the answer as expected.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Aside: 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 the video below.

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:

RELATED VIDEO FROM OUR COURSE

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We are given that the hypotenuse of a right triangle is 10 cm and we need to determine the perimeter of the triangle. We can let the other two sides (i.e., the two legs) of the right triangle be a and b. Since it’s a right triangle, by the Pythagorean theorem, we have a^2 + b^2 = 10^2, or a^2 + b^2 = 100. If we can find the values of a and b, then we can determine the perimeter of the triangle, since it will be a + b + 10.

Statement One Alone:

The area of the triangle is 25 square centimeters.

We are given a right triangle and we’ve let the two non-hypotenuse sides be a and b. Recall that the two non-hypotenuse sides of a right triangle are actually the base and height of the triangle, so the area of this triangle is A = ab/2. From the aforementioned equation a^2 + b^2 = 100, we can solve b as b = √(100 - a^2). Since we are given that A = 25, and substituting b = √(100 - a^2) in A = ab/2, we can say:

25 = [a√(100 - a^2)]/2

We see that we can solve for a (though we don’t have to actually solve for it). And once we’ve solved for a, we can determine the value of b, since b = √(100 - a^2). Therefore, we can determine the perimeter of the triangle, since we can determine the values of both a and b. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The 2 legs of the triangle are of equal length.

Since a^2 + b^2 = 100 and we are given that a = b, we can say:

a^2 + a^2 = 100

We see that we can solve for a (though we don’t have to actually solve for it). And once we’ve solved for a, we can determine the value of b, since b = a. Therefore, we can determine the perimeter of the triangle, since we can determine the values of both a and b. Statement two alone is sufficient to answer the question.

Answer: D
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Check out our detailed video solution to this problem here:
https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html-soluti ... ciency_383
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For 1 there is a much easier approach:

Think the hypotenuse as the diameter of a circle. In this case the area will be hy*height/2=25 --> height = 5.
height = 5 --> half of hy --> half of diameter = radius --> therefore the height fall in the middle point, and so leg1 = leg2.
1 is SUFF.

Hope it helps.

Matt

Hi All,

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.

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 of the triangle.

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

X^2 + Y^2 = 10^2

**Note: 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.

Fact 2 also 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

Final Answer:

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 certain 'math work.'

GMAT assassins aren't born, they're made,
Rich
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Bunuel
chetan2u
Plz help me on this doubt:
if triangle is a right triangle and hyp is 10 , should not the legs be 6 and 8 as per Pythagorean triplets?

You are assuming with no ground for it that the lengths of the sides are integers. Knowing that hypotenuse equals to 10 DOES NOT mean that the sides of the right triangle necessarily must be in the ratio of Pythagorean triple - 6:8:10. Or in other words: if \(a^2+b^2=10^2\) DOES NOT mean that \(a=6\) and \(b=8\), certainly this is one of the possibilities but definitely not the only one. In fact \(a^2+b^2=10^2\) has infinitely many solutions for \(a\) and \(b\) and only one of them is \(a=6\) and \(b=8\).

For example: \(a=1\) and \(b=\sqrt{99}\) or \(a=2\) and \(b=\sqrt{96}\) ...
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ty so much,
so if sides are 6,8,10 then it is definitely right trngle but
if it is right traingles and one side is 10 we cnt assume viceversa

Note that the reverse of a Pythagorean theorem is also true: if the lengths of the sides of a triangle are a, b, and c, and a^2 + b^2 = c^2, then we have a right triangle.

Also, if the lengths of the sides of a triangle are a, b, and c, where the largest side is c, then:

For a right triangle: \(a^2 +b^2= c^2\).
For an acute (a triangle that has all angles less than 90°) triangle: \(a^2 +b^2>c^2\).
For an obtuse (a triangle that has an angle greater than 90°) triangle: \(a^2 +b^2<c^2\).
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Hello Bunuel
Can we consider hypotenuse=10; and height=10 in statement 2? I asked this question because there is no strict condition that the base and height MUST be the equal legs.
Thanks__
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