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

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The hypotenuse of a right triangle is 10 cm. What is the [#permalink] New post 09 May 2010, 10:43
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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 centimeters.
(2) The 2 legs of the triangle are of equal length.


[Reveal] Spoiler:
I know that (2) is sufficient but I am having difficulty with (1).
[Reveal] Spoiler: OA
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Re: DS-Geometry [#permalink] New post 09 May 2010, 11:33
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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.

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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Re: DS-Geometry [#permalink] New post 09 May 2010, 13:43
Very nice, thank you.
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Re: DS-Geometry [#permalink] New post 28 Feb 2012, 07:57
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.

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.
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Re: DS-Geometry [#permalink] New post 28 Feb 2012, 08:23
Expert's post
Lstadt 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.

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.


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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Re: The hypotenuse of a right triangle is 10cm. What is the [#permalink] New post 30 Mar 2012, 18:06
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.
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink] New post 03 Oct 2013, 08:09
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink] New post 18 May 2014, 10:01
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
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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink] New post 18 May 2014, 23:18
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marcusaurelius 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 centimeters.
(2) The 2 legs of the triangle are of equal length.


[Reveal] Spoiler:
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
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Re: DS-Geometry [#permalink] New post 24 May 2014, 08:13
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.

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.


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!
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Re: DS-Geometry [#permalink] New post 24 May 2014, 09:08
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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.

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.


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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NEW TO MATH FORUM? PLEASE READ THIS: ALL YOU NEED FOR QUANT!!!

PLEASE READ AND FOLLOW: 11 Rules for Posting!!!

RESOURCES: [GMAT MATH BOOK]; 1. Triangles; 2. Polygons; 3. Coordinate Geometry; 4. Factorials; 5. Circles; 6. Number Theory; 7. Remainders; 8. Overlapping Sets; 9. PDF of Math Book; 10. Remainders; 11. GMAT Prep Software Analysis NEW!!!; 12. SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) NEW!!!; 12. Tricky questions from previous years. NEW!!!;

COLLECTION OF QUESTIONS:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS ; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: The hypotenuse of a right triangle is 10 cm. What is the [#permalink] New post 02 Aug 2014, 19:14
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.

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.


Hi Bunuel,

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!
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The hypotenuse of a right triangle is 10 cm. What is the [#permalink] New post 16 Sep 2014, 04:54
marcusaurelius 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 centimeters.
(2) The 2 legs of the triangle are of equal length.


[Reveal] Spoiler:
I know that (2) is sufficient but I am having difficulty with (1).



i applied like this..

1:1:\sqrt{2} for option 1 .. since hypo is 10 then

other 2 are 5\sqrt{2}

(5\sqrt{2} *5\sqrt{2}) /2 = 25.. suff.

1:\sqrt{3}:2 wont suffice here ...

option b . same logic

1:1:\sqrt{2}

Hence D.
The hypotenuse of a right triangle is 10 cm. What is the   [#permalink] 16 Sep 2014, 04:54
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