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M10-05

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Re: M10-05  [#permalink]

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New post 27 Dec 2017, 13:49
1
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]

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New post 27 Dec 2017, 20:40
1
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]

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New post 27 Dec 2017, 23:42
Bunuel wrote:
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.


Thanks, Bunuel.
I have one more doubt.
If I am given just the value of diagonal of a rectangle and I am asked to determine the area of that rectangle, then what is the method of determining the area of that rectangle?
Thanks.
Math Expert
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Joined: 02 Sep 2009
Posts: 51121
Re: M10-05  [#permalink]

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New post 27 Dec 2017, 23:45
Deepshikha1907 wrote:
Bunuel wrote:
Deepshikha1907 wrote:

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.


Thanks, Bunuel.
I have one more doubt.
If I am given just the value of diagonal of a rectangle and I am asked to determine the area of that rectangle, then what is the method of determining the area of that rectangle?
Thanks.


You cannot get the area of a rectangle by just knowing the length of its diagonals.
_________________

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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

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Re: M10-05  [#permalink]

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New post 27 Dec 2017, 23:56
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.


Is the only way to know the area of a rectangle is to have its length and breadth known? So that one can use the formula area of rectangle= length*breadth?
Math Expert
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Joined: 02 Sep 2009
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Re: M10-05  [#permalink]

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New post 28 Dec 2017, 00:05
Deepshikha1907 wrote:
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.


Is the only way to know the area of a rectangle is to have its length and breadth known? So that one can use the formula area of rectangle= length*breadth?


There are also other cases. For example, the length of the diagonal and adjacent angles, which in turn will give you the length of the sides...
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
GMAT Math Book | Triangles | Polygons | Coordinate Geometry | Factorials | Circles | Number Theory | Remainders; 8. Overlapping Sets | PDF of Math Book; 10. Remainders | GMAT Prep Software Analysis | SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS) | Tricky questions from previous years.

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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M10-05  [#permalink]

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New post 05 Dec 2018, 02:22
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]

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New post 09 Dec 2018, 09:02
Since we know area of rectangle can be d1*d2/2 or d1^2/2. Isn't the information in the question enough?
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Re: M10-05 &nbs [#permalink] 09 Dec 2018, 09:02

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