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.
_________________

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.

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!

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...

Hi Bunuel, I also have the same doubt, Knowing 2 of the 3 (Diagonal, perimeter, or area), we can find the third one por Triangles, Squares and rectangles? Is that correct. Could you help us?

Thanks a lot.

Regards.

Luis Navarro
Looking for 700

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.
_________________

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
_________________

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

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?

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