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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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

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.

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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

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.

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.

Re: The hypotenuse of a right triangle is 10cm. What is the [#permalink]

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30 Mar 2012, 19:06

stmt 1 we know that bh=50 and x^2 + y^2 = 100. Also, we need to solve for x+y+100. If we were to make (x+y)^2 and then solve out from there we can manipulate it so we can create an equation to what we know so far. _________________

Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]

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03 Oct 2013, 09:09

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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I know that (2) is sufficient but I am having difficulty with (1).

The best approach to tackle statement questions in DS is as follows:

Step 1: Convert all the alphabetical statements in algebraic statements Step 2: Reduce the number of variable to minimum Step 3: Check how many variables are left. You may probably need that many statements to solve the questions but you might need lesser number of statements to answer the question.

Caution: Don't waste your time in solving the question. You have to analyse the data sufficiency and not solve the question.

Lets see how this approach works for this question:

Step 1: A right triangle is given with hypotenuses equal 10. A triangle has three side. Let the other two sides be a and b

We know in a right triangle a^2 + b^2 =100

and we have to determine perimeter = a + b + 10

Step 2: We introduced 2 variables a and b above. Can we reduce the constraint equation to one variable. Yes, we can.

p = a + \sqrt{100-a^2} + 10

Thus, we are left with one variable.

Step 3: Let us look at the statements and see if we can decipher this one variable

1) The area of the triangle is 25 square centimeters. Area of a right triangle is 1/2*a*b = 25 But we know the relationship between a and b as well. Therefore, we can calculate a and hence perimeter. Thus, this is a sufficient condition.\

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

This means a= b

Again, we can calculate a here and thus can find out the perimeter. Again sufficient.

Thus, the correct answer is D.

Note: We didn't solve for a or p nor did we enter any algebra here

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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

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.

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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

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.

Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]

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02 Aug 2014, 20:14

Bunuel wrote:

marcusaurelius wrote:

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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]

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06 Jan 2015, 16:41

x and y are the two unknown sides...

we can determine from the stem that x^2 + y^2 = 10^2

1.) we know triangle area = 1/2(base)(height) so 1/2(x)(y)=25. from this we can determine that (x)(y) = 50. x= 50/y. using equation above we have 2 equations 2 unknowns. sufficient.

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in [#permalink]

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08 Jul 2015, 13:20

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in cm, of the triangle?

(1) The area of the triangle is 25 cm^2 (2) The 2 legs of the triangle are of equal length

Doesn't statement 2 suggest that this right triangle is therefore a 45-45-90? Because if it has two equal legs, wouldn't that make it half a square? So if the hypotenuse is 10, then WHY can't we say that based on the 45-45-90 right triangle ratios that each leg is 10/root(2) and so the perimeter is 2*(10/root(2) + 10?

It seems that all explanations of this Q use the Pythagorean Theorem or algebraic equalities to solve.

Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]

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08 Jul 2015, 13:40

Expert's post

torontoclub15 wrote:

The hypotenuse of a right triangle is 10 cm. What is the perimeter, in cm, of the triangle?

(1) The area of the triangle is 25 cm^2 (2) The 2 legs of the triangle are of equal length

Doesn't statement 2 suggest that this right triangle is therefore a 45-45-90? Because if it has two equal legs, wouldn't that make it half a square? So if the hypotenuse is 10, then WHY can't we say that based on the 45-45-90 right triangle ratios that each leg is 10/root(2) and so the perimeter is 2*(10/root(2) + 10?

It seems that all explanations of this Q use the Pythagorean Theorem or algebraic equalities to solve.

Kudos for any help! Much appreciated.

Merging topics. Please refer to the discussion above and ask if anything remains unclear. _________________

Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink]

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08 Jul 2015, 14:24

Bunuel wrote:

marcusaurelius wrote:

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.

I know that (2) is sufficient but I am having difficulty with (1).

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 \(x+y\).

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...

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