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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.
_________________

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.

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...

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...

\(x+y=\sqrt{200}\) so it's an irrational number, not imaginary number.
_________________

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

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11 Jul 2015, 15:07

1

This post received KUDOS

We can solve it using topic about similar triangles 1)Area= (10*Height)/2 = 25 --> Height=5, so if this 3 triangles are similar, then x/5=5/10-x, x^2-10x+25=0 --> x=5 The hypotenuses of small traingles => 1st: x^2+25, 2nd:125-20x+x^2..... So we can calculate Perimiter after pluging X=5 in the stated equations (no need to calculate) SUFFICIENT

2) 2 Legs are equal => pythagoras theorem: x^2+x^2=100, x=5*sqrt 2, so again it's enougj to calculate the perimeter (10+10*sqrt 2)

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

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18 Jul 2015, 15:56

russ9 wrote:

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 \(x+y\).

Please correct me if i'm wrong but it seems though if we know 2 of the 3 (Diagonal, perimeter, or area), we can find the third one. Is that correct?

Does this work for rectangles/squares and other figures as well?

Also, can you suggest similar problems -- one where the perimeter is given and we are required to find something else?

Thanks!

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?

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

Let me try to answer this question.

The answer to your question (in general) is as follows:

1. For ANY triangle, perimeter = P = a+b+c ---> so given 2 of the 3 sides and P, you can definitely calculate the 3rd side. For triangles that are not right angled or equilateral or isosceles, you would 1 angle or some sort of a relation between the sides and the area or a combination of these to sufficiently answer DS questions for these triangles.

i) For right triangles you also have the hypotenuse ^2 = base^2 + perpendicular ^2 as another equation. Thus for a right angled triangle, you have 2 equations.

ii) Similarly for isosceles and equilateral triangles (by drawing an a right angle, as shown in picture attached), you divide these 2 triangles into 2 same or congruent right angled triangles. So, for these 2 triangles as well you have 2 equations as mentioned above.

FYI, also remember that for right angled triangles and isosceles/equilateral triangles, you can calculate the 3rd side or the perimeter if the area is given as shown in the question discussed in this thread (by applying the relation, \((a+b)^2 = a^2+b^2+2ab\) and noticing that 0.5*ab represents the area of the triangle).

2. For rectangles and squares, the diagonals of these 2 shapes, divide these rectangles and squares into 2 congruent or same right angled triangles. The points mentioned above in 1) thus also apply these shapes as well.

Hope this answers your question. Let me know if there are any other questions.

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Rect, Square and Triangle.jpg [ 36.37 KiB | Viewed 10396 times ]

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

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18 Jul 2015, 16:44

Engr2012 wrote:

luisnavarro wrote:

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

Let me try to answer this question.

The answer to your question (in general) is as follows:

1. For ANY triangle, perimeter = P = a+b+c ---> so given 2 of the 3 sides and P, you can definitely calculate the 3rd side. For triangles that are not right angled or equilateral or isosceles, you would 1 angle or some sort of a relation between the sides and the area or a combination of these to sufficiently answer DS questions for these triangles.

i) For right triangles you also have the hypotenuse ^2 = base^2 + perpendicular ^2 as another equation. Thus for a right angled triangle, you have 2 equations.

ii) Similarly for isosceles and equilateral triangles (by drawing an a right angle, as shown in picture attached), you divide these 2 triangles into 2 same or congruent right angled triangles. So, for these 2 triangles as well you have 2 equations as mentioned above.

FYI, also remember that for right angled triangles and isosceles/equilateral triangles, you can calculate the 3rd side or the perimeter if the area is given as shown in the question discussed in this thread (by applying the relation, \((a+b)^2 = a^2+b^2+2ab\) and noticing that 0.5*ab represents the area of the triangle).

2. For rectangles and squares, the diagonals of these 2 shapes, divide these rectangles and squares into 2 congruent or same right angled triangles. The points mentioned above in 1) thus also apply these shapes as well.

Hope this answers your question. Let me know if there are any other questions.

That mean that this kind of question (Given area and diagonal of the examples above mentioned) in DS are "automatic", for instance; No matter system equation is quadratic always is going to be one answer and would be sufficient (I do not need to check the specific roots of the quadratic). Is it correct?

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