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TV Screen

by crackgmat007 » Tue Sep 29, 2009 5:44 pm
What is the ratio of the area of a TV screen with a diagonal of 18'' to that of a screen with a diagonal of 15''?

1. The ratio of width to length is the same for both screens.
2. The width of the 18''-screen is 20% greater than that of the 15''-screen.

OA - D
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by xcusemeplz2009 » Thu Oct 01, 2009 1:34 am
IMO D

staement 1)
let width of TV with diag 18" =x; then L=nx
similarly for 15" W=y;L=ny(as raio of W/L is same foer both)
x^2+(nx)^2=18^2=>x^2(1+n^2).............(1)
y^2+(ny)^2=15^2=>y^2(1+n^2)...............(2)
dividing 1 AND 2 (x/y)^2=18^2/15^2

A1(area of 18")/A2 (area of 15")=2*square of ratio of corresponding sides=2*(x/y)^2=2*(18/15)^2
suff.

Statement 2) x=6/5y
x/y=6/5
A1/A2=2*(6/5)^2
SUFF.
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by uptowngirl92 » Wed Oct 07, 2009 3:22 pm
Completly missed this one :(

The property of similar triangles is that the ratio of the areas=Ratio of the squares of the corresponding sides..
But here we have a rectangle..Hmm..How would one calculate this if need be?Will it be multiplied by 2 since triangle is half that of the rectangle?(We are seeing the diagnol as a side of the rectangle)

I am sure I am messing this up than need be but the explanation is not clear at all:(

Similiar triangles concept anyone?!!
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by ChrisHinkle » Wed Oct 07, 2009 3:36 pm
We have two rectangles and we know the diagonal of each. Also the diagonals form the hypotenuses of right triangles whose legs are the sides of the rectangle.

Statement 1: If the ratio of base to height (width to length) of the two right triangles is the same, the triangles are similar. Like 3-4-5 is similar to 6-8-10. The given values on the diagonals give us the ratio between the triangles: 15 to 18 (5 to 6). So the ratio of the areas will be 5^2 to 6^2. (Try out a couple of values to confirm - as long as you keep a constant ratio of 5 to 6 between corresponding sides of the two triangles, the ratios of the areas will be constant.)

Statement 2: Notice that the given 20% increase matches the 15 to 18 relationship of the diagonals. Again I can prove that the right triangles are similar since two sets of sides have the same ratio (and therefore the third does as well). From there the problem is the same.

Answer D
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by uptowngirl92 » Wed Oct 07, 2009 6:36 pm
Hey Chris,

Suppose the same question has triangles instead of rectangles,then the answer would have been EXACTLY the same right?The ratio of the areas will be ratio of the areas will be 5^2 to 6^2,applying similiar triangles.

I am not clear on this fact:Since its a rectangles should'nt it be TWICE the ratio??

I hope my question is clear!
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by uptowngirl92 » Fri Oct 09, 2009 12:11 am
:?:
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by Katrusya » Fri Oct 09, 2009 12:24 pm
ChrisHinkle, I have a question for you:
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by mmslf75 » Tue Dec 15, 2009 11:09 pm
ChrisHinkle wrote:We have two rectangles and we know the diagonal of each. Also the diagonals form the hypotenuses of right triangles whose legs are the sides of the rectangle.

Statement 1: If the ratio of base to height (width to length) of the two right triangles is the same, the triangles are similar. Like 3-4-5 is similar to 6-8-10. The given values on the diagonals give us the ratio between the triangles: 15 to 18 (5 to 6). So the ratio of the areas will be 5^2 to 6^2. (Try out a couple of values to confirm - as long as you keep a constant ratio of 5 to 6 between corresponding sides of the two triangles, the ratios of the areas will be constant.)

Statement 2: Notice that the given 20% increase matches the 15 to 18 relationship of the diagonals. Again I can prove that the right triangles are similar since two sets of sides have the same ratio (and therefore the third does as well). From there the problem is the same.

Answer D
I do not understand this ..can u please explain a bit more !!
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by Stuart@KaplanGMAT » Wed Dec 16, 2009 1:01 am
uptowngirl92 wrote:Hey Chris,

Suppose the same question has triangles instead of rectangles,then the answer would have been EXACTLY the same right?The ratio of the areas will be ratio of the areas will be 5^2 to 6^2,applying similiar triangles.

I am not clear on this fact:Since its a rectangles should'nt it be TWICE the ratio??

I hope my question is clear!
Let's use numbers to test your hypothesis!

To keep things simple, let's say our small triangle is 3/4/5 and our big triangle is 6/8/10, so we have a 1:2 ratio in lengths.

The area of the small triangle is (1/2)(3)(4) = 6.
The area of the big triangle is (1/2)(6)(8) = 24.

So, the ratio of areas is 1:4.

Now let's compare the rectangles:

Area of the small rectangle is 3*4 = 12.
Area of the big rectangle is 6*8 = 48.

Again, the ratio of the areas is 1:4.

So, just because the rectangle is composed of two triangles, that doesn't affect the ratio of the small one to the big one. The only time our ratios increase is when we add another dimension, such as going from the ratio of the sides (one-dimensional) to the ratio of the areas (two-dimensional).
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by mmslf75 » Sat Dec 19, 2009 8:02 am
Stuart Kovinsky wrote:
uptowngirl92 wrote:Hey Chris,

Suppose the same question has triangles instead of rectangles,then the answer would have been EXACTLY the same right?The ratio of the areas will be ratio of the areas will be 5^2 to 6^2,applying similiar triangles.

I am not clear on this fact:Since its a rectangles should'nt it be TWICE the ratio??

I hope my question is clear!
Let's use numbers to test your hypothesis!

To keep things simple, let's say our small triangle is 3/4/5 and our big triangle is 6/8/10, so we have a 1:2 ratio in lengths.

The area of the small triangle is (1/2)(3)(4) = 6.
The area of the big triangle is (1/2)(6)(8) = 24.

So, the ratio of areas is 1:4.

Now let's compare the rectangles:

Area of the small rectangle is 3*4 = 12.
Area of the big rectangle is 6*8 = 48.

Again, the ratio of the areas is 1:4.

So, just because the rectangle is composed of two triangles, that doesn't affect the ratio of the small one to the big one. The only time our ratios increase is when we add another dimension, such as going from the ratio of the sides (one-dimensional) to the ratio of the areas (two-dimensional).

Area of TRIANGLE 1/Area of TRIANGLE 2 = Side^ 2 / Side ^2 right ??

EDITED THE POST...sorry for that typo
Why we require 1 or 2 either of it ?
Last edited by mmslf75 on Sun Dec 20, 2009 8:22 am, edited 2 times in total.
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by Stuart@KaplanGMAT » Sat Dec 19, 2009 12:11 pm
mmslf75 wrote:
Area of Rect = Side^ 2 / Side ^ right ??

Why we require 1 or 2 either of it ?
Area of a rectangle is length * width, not sure where you got that formula.

Another formula for the area of a square is:

diagonal^2 / 2

but that only works because each side of the square is equal. Just knowing the diagonal of a rectangle does not give the area.

For example, if we know that the diagonal is 15, our sides could be :

9, 12, 15 (blow up of a 3/4/5 triangle)

15/root2, 15/root2, 15 (right isosceles triangle, making our TV a square)

5, root200, 15 (since 5^2 + (root200)^2 = 15^2)

and an infinite number of other combinations.

So, just knowing the diagonals of the TVs doesn't tell us about their other specific proportions.
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by mmslf75 » Sun Dec 20, 2009 8:22 am
Stuart Kovinsky wrote:
mmslf75 wrote:
Area of Rect = Side^ 2 / Side ^ right ??

Why we require 1 or 2 either of it ?
Area of a rectangle is length * width, not sure where you got that formula.

Another formula for the area of a square is:

diagonal^2 / 2

but that only works because each side of the square is equal. Just knowing the diagonal of a rectangle does not give the area.

For example, if we know that the diagonal is 15, our sides could be :

9, 12, 15 (blow up of a 3/4/5 triangle)

15/root2, 15/root2, 15 (right isosceles triangle, making our TV a square)

5, root200, 15 (since 5^2 + (root200)^2 = 15^2)

and an infinite number of other combinations.

So, just knowing the diagonals of the TVs doesn't tell us about their other specific proportions.
I have edited that post
forgot to mention TRIANGLE 2 in the denominator....

St 1 says that Ratio of WIDTH to LENGTH is same for both
Now if rect 1 has 9 12 15 type, side-side-diagonal.

Inorder to have same ratio and diagonal 18, I need to have L,W, 18...
Now L and W cannot take values of the type 3,4 for same ratio right as we have 18 there...

SO how is 1 alone correct ???
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by ssuarezo » Mon Dec 21, 2009 12:36 pm
mmslf75 wrote:
ChrisHinkle wrote:We have two rectangles and we know the diagonal of each. Also the diagonals form the hypotenuses of right triangles whose legs are the sides of the rectangle.
Answer D
I do not understand this ..can u please explain a bit more !!
Chris just applied the concept of similar triangles, the known ones are 3,4,5 or 6,8,10 (multiplied by 2).
Thanks for your explanation Chris.
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by ssuarezo » Mon Dec 21, 2009 12:50 pm
mmslf75 wrote:
I have edited that post
forgot to mention TRIANGLE 2 in the denominator....

St 1 says that Ratio of WIDTH to LENGTH is same for both
Now if rect 1 has 9 12 15 type, side-side-diagonal.

Inorder to have same ratio and diagonal 18, I need to have L,W, 18...
Now L and W cannot take values of the type 3,4 for same ratio right as we have 18 there...

SO how is 1 alone correct ???
mmslf75:
st1 says in summary that triangles are similar. We tried 3,4,5 and 9,12,15. the ratios between l-w in each individual rectangle are the same. Yes, they can take values of type 3,4,5 because they are similar, same ratio between l-w.
The question asks you for ratios, not real areas, so knowing that they're similar, then, they share a ratio, then u can get the ratio of areas between rectangles (divided in triangles) as Stuart demonstrated it.
Silvia
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