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M10-05

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Bunuel
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M10-05 [#permalink] New post   16 Sep 2014, 00:41
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What is the ratio of the area of a rectangular TV screen with a diagonal of 18 inches to that of a rectangular screen with a diagonal of 15 inches?


(1) Both screens have the same width-to-length ratio.

(2) The width of the 18-inch screen is 20% greater than the width of the 15-inch screen.
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Bunuel
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Re M10-05 [#permalink] New post   16 Sep 2014, 00:41
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Official Solution:


What is the ratio of the area of a rectangular TV screen with a diagonal of 18 inches to that of a rectangular screen with a diagonal of 15 inches?

In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their corresponding sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals act as the hypotenuses in right triangles formed by width and length).

(1) Both screens have the same width-to-length ratio.

This implies that the rectangles are similar. Consequently, the right triangles formed by the diagonals are similar as well, which means that the ratio of the areas of the triangles equals the ratio of the areas of the rectangles \(= (\frac{18}{15})^2 = 1.44\). Sufficient.

(2) The width of the 18-inch screen is 20% greater than the width of the 15-inch screen.

Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse). Thus, the right triangles formed by the width and length are similar (in right triangles, if two corresponding sides have the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore, the rectangles are similar as well, which means that the ratio of the areas of the triangles equals the ratio of the areas of the rectangles \(= (\frac{18}{15})^2=1.44\). Sufficient.


Answer: D
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Re: M10-05 [#permalink] New post   09 Oct 2014, 15:07
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Bunuel wrote:
Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D



I did not follow the highlighted part. Ratio of widths is same as ratio of hypotenuse.
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Re: M10-05 [#permalink] New post   10 Oct 2014, 02:41
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earnit wrote:
Bunuel wrote:
Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D



I did not follow the highlighted part. Ratio of widths is same as ratio of hypotenuse.


Stem says that the ratio of the diagonals is 18/15 = 1.2 and (2) says that the ration of widths is also 1.2. So, the ratio of widths = 1.2 = ratio of diagonals (hypotenuse).
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Re: M10-05 [#permalink] New post   16 Jan 2015, 03:03
I think this question is good and helpful.
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Re: M10-05 [#permalink] New post   27 Dec 2017, 14:49
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Bunuel wrote:
What is the ratio of the area of a rectangular TV screen with a diagonal of 18'' to that of a rectangular 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.



Hi Bunuel,
I have encountered a genuine doubt while solving this question. Are not the diagonals of a rectangle equal in length? If it's so, then in any case the area of each rectangle can be determined using the formula 1/2*(diagonal)^2 and the ratio of their areas can be determined. This would make the 2 statements not required to answer the question itself.
Please make this clear to me. I am really confused.
Thanks.
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Re: M10-05 [#permalink] New post   27 Dec 2017, 21:40
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Deepshikha1907 wrote:
Bunuel wrote:
What is the ratio of the area of a rectangular TV screen with a diagonal of 18'' to that of a rectangular 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.



Hi Bunuel,
I have encountered a genuine doubt while solving this question. Are not the diagonals of a rectangle equal in length? If it's so, then in any case the area of each rectangle can be determined using the formula 1/2*(diagonal)^2 and the ratio of their areas can be determined. This would make the 2 statements not required to answer the question itself.
Please make this clear to me. I am really confused.
Thanks.


You can find the area with diagonal^2/2 ONLY for squares. Yes, the diagonals of a rectangle are equal, but rectangles with equal diagonals have different areas.
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Re: M10-05 [#permalink] New post   05 Dec 2018, 03:22
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For those who need more clarity on concepts tested in this sum -

Property - In two similar triangles, the ratio of their areas is the square of the ratio of their sides.

Take example of \(3-4-5\) triangle. Area will be \(6\).

Now if its a \(6-8-10\) triangle. Area will be \(24.\)

Ratio of sides =\(6/3 = 8/4 = 10/5 = 2\)
Ratio of area = \(24/6 = 4\)

Hence (Ratio of sides)\(^2\) = Ratio of the area

For statement 2 -
Special note on right angled triangles. These triangles follow 4 cases on similarity -
Case 1 - Side - Angle - Side case
Case 2 - At least 2 angles are proportional
Case 3 - All sides are proportional
Case 4 - Hypotenuse - Leg condition : If the lengths of the hypotenuse and a leg of a right triangle are proportional to the corresponding parts of another right triangle, then the triangles are similar.

Bonus note - In two similar triangles, their perimeters and corresponding sides, medians and altitudes will all be in the same ratio.
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Re: M10-05 [#permalink] New post   02 Oct 2020, 22:53
Bunuel wrote:
Official Solution:


In two similar triangles (or rectangles), the ratio of their areas is the square of the ratio of their sides.

Given: ratio of diagonals \(= \frac{18}{15}=1.2\) (note that diagonals are the hypotenuses in right triangles made by width and length).

(1) The ratio of width to length is the same for both screens. This implies that the rectangles are similar, hence right triangles made by diagonals are similar as well, which means that the ratio of areas of triangles = ratio of areas of rectangles \(= ( \frac{18}{15} )^2 = 1.44\). Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen. Given: the ratio of widths = 1.2 = ratio of diagonals (hypotenuse), so right triangles made by width and length are similar, (in right triangles if 2 corresponding sides are in the same ratio, in our case \(\frac{W}{w}=\frac{D}{d}=1.2\), then these right triangles are similar). Therefore the rectangles are similar too, which means that the ratio of areas of triangles = the ratio of areas of rectangles \(= ( \frac{18}{15} )^2=1.44\). Sufficient.


Answer: D


HI GMATGuruNY, AndrewN , GMATCoachBen , Bunuel

I'm not clear about Statement (2)The width of the 18''-screen is 20% greater than that of the 15''-screen. We are not aware of the length?

So I marked as A
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Re: M10-05 [#permalink] New post   03 Oct 2020, 04:08
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Rule for any two right triangles ABC and DEF:
If \(\frac{AB}{DE} = \frac{BC}{EF}\), then \(\frac{AB}{DE} =\) \(\frac{BC}{EF} = \frac{AC}{DF}\).
As a result, the two triangles are SIMILAR.
In other words:
If two sets of corresponding sides are in the same ratio, then the third set of corresponding sides must also be in that ratio, with the result that the two triangles are similar.

Rule for two similar triangles:
If each side of the larger triangle is \(x\) times the corresponding side in the smaller triangle, then the area of the larger triangle is \(x^2\) times the area of the smaller triangle.

NandishSS wrote:

(2)The width of the 18''-screen is 20% greater than that of the 15''-screen. We are not aware of the length?

So I marked as A


Prompt:
Ratio of the diagonals \(= \frac{greater-dagonal}{smaller-diagonal} = \frac{18}{15} = \frac{6}{5}\)
Statement 2:
Ratios of the widths \(= \frac{greater-width}{smaller-width} = \frac{1.2}{1} = \frac{6}{5}\)

The two diagonals and the two widths are in the same ratio.
As a result, the two lengths must also be in this ratio:
\(\frac{greater-length}{smaller-length} = \frac{6}{5}\)
Since all of the sides are in the same ratio, the two triangles are SIMILAR.

Since the length of each side in the larger side is \(\frac{6}{5}\) the length of the corresponding side in the smaller triangle, the area of the larger triangle is \((\frac{6}{5})^2\) the area of the smaller triangle.
By extension, the area of the larger screen is \((\frac{6}{5})^2\) the area of the smaller screen.
SUFFICIENT.
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Re: M10-05 [#permalink] New post   03 Oct 2020, 14:26
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NandishSS wrote:
HI GMATGuruNY, AndrewN , GMATCoachBen , Bunuel

I'm not clear about Statement (2)The width of the 18''-screen is 20% greater than that of the 15''-screen. We are not aware of the length?

So I marked as A

Hello, NandishSS. I see that GMATGuruNY has already supplied a fine response to your query. I took a less rigorous approach but arrived at the same conclusion within a minute. I started with the idea that the two rectangular screens might be similar figures, if only I had enough information to prove it one way or the other. Now, the given diagonals fit a proportion:

\(\frac{18}{15}=\frac{6}{5}\)

The decimal equivalent is 1.2. If we knew that either of the other sides of the larger rectangle were 1.2 times greater than its corresponding side on the smaller rectangle, we would be able to deduce that the third side also had to fit the same proportion, using the Pythagorean theorem. Consider:

Smaller right triangle:

\(a^2+b^2=c^2\)

Larger right triangle:

\((1.2a)^2+(1.2b)^2=(1.2c)^2\)

Notice that statement (2) gives us just what we want. If we let w represent the width of the 15" screen, then 1.2w would represent the width of the 18" screen. That is as far as I went with it, to be honest. I knew I had the information I needed, so I picked (D).

I hope that helps. Thank you for thinking to tag me.

- Andrew
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Re: M10-05 [#permalink] New post   15 Jun 2021, 18:09
I think this is a high-quality question and I agree with the explanation.
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Re: M10-05 [#permalink] New post   31 May 2023, 15:04
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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Re: M10-05 [#permalink]
31 May 2023, 15:04
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