Two sides of a triangle have lengths x and y and meet at a r

Question

Two sides of a triangle have lengths x and y and meet at a right angle. If the perimeter of the triangle is 4x, what is the ratio of x to y ?

a) 2 : 3
b) 3 : 4
c) 4 : 3
d) 3 : 2
e) 2 : 1

pqhai · Accepted Answer

E is correct. Here is my solution.
Hope it helps.

EMPOWERgmatRichC · Answer

Hi All,

When right triangles show up on Test Day, they're rarely random - they're often one of the right triangles that are frequently tested: 3/4/5, 5/12/13, 30/60/90, 45/45/90.

In this prompt, the fact that the perimeter is equal to 4 times one of the shorter sides is an interesting (and rather specific) piece of information....MAYBE it matches one of the special right triangles I listed....?

reetskaur correctly realized that this description is a match for a 3/4/5 right triangle, so we can AVOID all of the crazy geometry/algebra and say that...

X = 3
Y = 4
Hypoteneuse = 5
Perimeter = 12

X:Y = 3:4

The GMAT is based heavily on established Quant and Verbal patterns - keep an eye open for pattern-based shortcuts throughout the Test - they're everywhere....

Final Answer:  B

GMAT assassins aren't born, they're made,
Rich

mau5 · Answer

There is a small typo there in your solution.

Actually B is the right answer.

reetskaur · Answer

Let's assume the simple set of sides for a right triangle as x=3, y=4 and hypotenuse= 5. Now the perimeter adds to 3+4+5= 12, which is equal to 3x. Therefore x:y= 3/4

pqhai · Answer

Mau5
Ahhhh....my bad. Thank you, you're correct. 8x = 6y --> x/y = 3/4.

Posted from my mobile device

Temurkhon · Answer

Two equations

x^2+y^2=c^2

x+y+c=4x

we need  x/y  ratio, so should express  c  through  x and y

c=4x-x-y=3x-y

substitute  c  in first equation

x^2+y^2=(3x-y)^2 => 8x=6y => x/y=6/8=3/4

B

ZLukeZ · Answer

How can you assume x to be 3 and not 4. 
Could it not be x 4 y3 z 5? I mean its basically the same triangle just a different distributionß

Going with the algebraic approach x^2+y^2=(3x-y)^2 => 8x=6y => x/y=6/8=3/4
I don't get how x^2+y^2=(3x-y)^2 becomes => 8x=6y. Could someone add some more steps?

Lucky2783 · Answer

Let  y=K*x 
Then the hypotnuse will be  4x- x - Kx = (3-K)x 
So, 
 1+ K^2 = (3-K)^2 
 K= \frac{4}{3}  
so  y= \frac{4}{3} * x 
Answer : B

EMPOWERgmatRichC · Answer

Hi ZLukeZ,

In the prompt, we're told that X and Y are the two "legs" of the right triangle and that the PERIMETER = 4X

IF.....
the legs are X=3 and Y=4, then the perimeter is 4(3) = 12.
A 3/4/5 triangle has a perimeter of 3+4+5 = 12, so this "matches" what we were told.

IF....
the legs are X=4 and Y=3, then the perimeter is 4(4) = 16
This does NOT match the perimeter of a 3/4/5 triangle, so X CANNOT be 4.

GMAT assassins aren't born, they're made,
Rich

Tmoni26 · Answer

Hi ZuleZ,

I also had to do a double take on that part of the question

So x^2 + y^2 = (3x-y)^2
Let us focus on the (3x-y)^2 part of the equation
FOIL it and you get (3x-y)(3x-y) ------> 9x^2-6xy+y^2, lets put this back into the main equation

x^2+y^2 = 9x^2 - 6xy +y^2 ---------> y^2-y^2 + 6xy = 9^2- x^2 --------------> 6xy = 8x^2 ---divide by x -----> 6y=8x

Hope this provides some clarification

Shruti0805 · Answer

The question can very easily be solved using the answer options, in my opinion. It would take lesser time than the algebra would.

MikeScarn · Answer

Why are you assuming it's a 3/4/5 triangle right off the bat? Could be a 45, 45, 90? I understand it works if we make it a 3,4,5 and make x=3 and y=4, but how do we know that's the only possible solution?

CC:  Bunuel

mandy0rhtdm · Answer

I think the empower provide method should ideally be avoided since it is not always applicable and at some times if you apply it wrongly it will lead you to wrong answers 
Its best to follow the basics here!

Here is an alternative method of doing it. 
X and y are the right angled sides so the third side is the sqrt of their squares.
Since the perimeter is 4x

That is. X + y + z = 4x
X^2 + y^2 = z^2

Now for this sort of equation to solve either x y and z all have to be rational numbers or a rational multiple of same irrational number.

In either case their ratio will be rational over another rational.
And since the third side is the square root of their squares sum the ratio of x to y in pure number form needs to yield a rational number .
That is if a/b is the ratio of x and y then a^2 + b^2. = a rational number squared.

Only option b satisfies this.

Posted from my mobile device

EMPOWERgmatRichC · Answer

Hi MikeScarn,

You ask some really good questions. To answer them, you have to remember a few things:

1) This is the GMAT, not a general "math test." The GMAT is built around patterns and is predictable (NOTHING in a question is there by chance - the wording is always really specific, the concepts tested are really specific and even the answer choices are specific), so you can use the patterns used by the question-writers to your advantage.
2) The answer choices can often provide a clue as to how to approach the question. Here, they help to define what is NOT possible. If we were dealing with a 45/45/90 triangle, then the ratio of the legs would be 1:1. That's clearly not an option here, so the triangle cannot be a 45/45/90.

GMAT assassins aren't born, they're made,
Rich

ekrizek16 · Answer

Thank you for these great explanations - I finally understand how to get to 6y = 8x. Then, as I understand it, I am looking for the ratio of x:y, or x/y. In that case, I simply divide both sides by y to get 6 = 8x/y and then divide both sides by 8 to get 6/8 = x/y. Obviously somehow I have reversed the two. Could someone help me with where I went wrong?

Thanks a ton!

Thanks a ton!

NinetyFour · Answer

Since it is a right trangle:
 x^2 + y^2 = c^2 
 c = sqrt(x^2 + y^2)

Perimeter = 4x so  x + y + sqrt(x^2 + y^2) = 4x

Simplify this down and we get  \frac{x}{y} = \frac{3}{4}

philipssonicare · Answer

Please don't skip heaps of steps when posting. It can be much too hard to follow.

x+y+√(x^2+y^2)=4x 
 y+√(x^2+y^2)=3x 
 √(x^2+y^2)=3x-y 
 x^2+y^2=(3x-y)^2 
 x^2+y^2=9x^2-6xy+y^2 
 0=8x^2-6xy 
 6xy=8x^2 
 6y=8x 
 3y=4x 
 3y/4=x 
 3/4=x/y

Fdambro294 · Answer

Call the Hypotenuse = Z

Perimeter = X + Y + Z = 4X

Pythagoras ---> (X)^2 + (Y)^2 = (Z)^2

re-arranging the Perimeter:
Y + Z = 3X
Z = 3X - Y

----Substituting Z = (3X - Y) in for Z into Pythagoras-----

(X)^2 + (Y)^2 = (3X - Y)^2

(X)^2 + (Y)^2 = 9X^2 - 6XY + (Y)^2

----subtract (Y)^2 from both sides----

(X)^2 = 9(X)^2 - 6XY

6XY = 8(X)^2

----since we know that X = (+)Positive Length, we can Subtract X from Both Sides----

6Y = 8X

6/8 = X/Y

Ratio of: X to Y ------> 6 to 8 -----> 3 to 4

-B-

Fdambro294 · Answer

Using the Answer Choices and Ratio Methodology:

X + Y + Z = Perimeter of 4X

Ratio of:

X..... : Y ....... : Z .........TOTAL
_________________________
X..................................4X

Which means Y + Z = 3X of the Ratio Units

(A) X :Y = 2:3

in Ratio Units:

X = 2 --------> then Total 4X = 8
Y = 3

which means Hypotenuse Z = 8 - (2 + 3) = 2

the Hypotenuse Z can NOT be Less Than < Leg Y

Eliminate A

(C) X : Y = 4 : 3

X = 4 ---------> Total Ratio Units = 16
Y = 3

Hypotenuse Z = 16 - (4 + 3) = 11

the Sides would have to be in the Ratio of: 3m - 4m - 11m

there is no Unknown Ratio Multiplier m that would satisfy Pythagoras (b/c of the 3-4-5 Triplet)

you can keep checking Answer Choices....

(B) X : Y = 3 : 4

X = 3 Ratio Units ----------> Total Perimeter Ratio Units = 4X = 12

Y = 4

Z = 12 - (4 + 3) = 12 - 7 = 5

This means the Triangle would be in the Ratio of:

3m - 4m - 5m ----------> we know this is a Valid Right Triangle

-B-

3m : 4m : 11m

Two sides of a triangle have lengths x and y and meet at a r

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Two sides of a triangle have lengths x and y and meet at a r

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