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# M10:Q5 - DS

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M10:Q5 - DS [#permalink]

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22 Aug 2008, 14:33
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What is the ratio of the area of the TV screen with diagonal 18'' to that of the screen with diagonal 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

[Reveal] Spoiler: OA
D

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Re: M10:Q5 - DS [#permalink]

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23 Aug 2008, 02:43
balboa wrote:
What is the ratio of the area of the TV screen with diagonal 18'' to that of the screen with diagonal 15''?

a)The ratio of width to length is the same for both screens
b) The width of the 18''-screen is 20% greater than that of the 15''-screen

IMO C

Length of screen with 15'' diagonal = l
Breadth of screen with 15'' diagonal = b

Length of screen with 18'' diagonal = L
Breadth of screen with 18'' diagonal = B

we want (L*B)/(l*b)

1 is insuff. 1 gived us l/L = b/B

So area = L^2/l^2 ... insuff

2 is insuff because it gives information about width but no information abt the ratio.
2 gives L=1.2*l and B = 1.2*b

Both the statements taken together give
[(1.2 * l) * (1.2 *b) ]/ [l * b]
Ratio of area = 144/100
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Re: M10:Q5 - DS [#permalink]

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02 Sep 2008, 08:05
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LoyalWater wrote:
balboa wrote:
What is the ratio of the area of the TV screen with diagonal 18'' to that of the screen with diagonal 15''?

a)The ratio of width to length is the same for both screens
b) The width of the 18''-screen is 20% greater than that of the 15''-screen

IMO C

Length of screen with 15'' diagonal = l
Breadth of screen with 15'' diagonal = b

Length of screen with 18'' diagonal = L
Breadth of screen with 18'' diagonal = B

we want (L*B)/(l*b)

1 is insuff. 1 gived us l/L = b/B

So area = L^2/l^2 ... insuff

2 is insuff because it gives information about width but no information abt the ratio.
2 gives L=1.2*l and B = 1.2*b

Both the statements taken together give
[(1.2 * l) * (1.2 *b) ]/ [l * b]
Ratio of area = 144/100

Note that S1 is sufficient. $$\frac{l}{b} = \frac{L}{B}$$ equals $$\frac{l}{L} = \frac{b}{B}$$, which tells us that the rectangles are similar. Thus the ratios of the screens can be found.
S2 is also sufficient. It tells us that these rectangles are similar. $$\frac{b}{B}$$ = 1.2 and $$\frac{18}{15}$$ = 1.2 --> the rectangles are similar again.

The answer must be D.

Hope this helps.
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Re: M10:Q5 - DS [#permalink]

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07 Oct 2009, 15:22
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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Re: M10:Q5 - DS [#permalink]

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12 Oct 2009, 01:51
Here's a good piece of info about similar rectangles:
http://standards.nctm.org/document/eexa ... /index.htm

If we had to find the ratio of areas of the two rectangles, it would equal the squared ratio of the corresponding sides. If this ratio is equal to 1.2, the ratio of areas is equal to $$1.2^2 = 1.44$$.

Hope it helps.
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Re: M10:Q5 - DS [#permalink]

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14 Dec 2009, 08:15
(1) is sufficient.

Because the triangles are similar, the angles are equal for both rectangles.

L = 18 cos a
B = 18 sin a
l = 15 cos a
b = 15 sin a

(L*B)/(l*b) = [(18 cos a)(18 sin a)]/[(15 cos a)/15 sin a)] = 18^2/15^2 = 1.44
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Re: M10:Q5 - DS [#permalink]

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14 Dec 2009, 08:18
(2) is sufficient.

18/15 = 1.2
B/b = 1.2

Therefore, L/l = 1.2

(B*L)/(b*l) = (B/b)*(L/l) = 1.2^2 = 1.44
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Re: M10:Q5 - DS [#permalink]

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14 Dec 2009, 09:22
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So, let’s define our rectangles’ sides as L and W for length and width of the bigger and l and w for the smaller one. Also, let’s define the diagonal D (bigger) and d (smaller).
2 rectangles are similar when their sides’ ratios are the same (W/w=L/l) and when the ratio of width to length is the same for both rectangles (W/L=w/l) If so, then the ratios of their diagonals, perimeters and areas are also the same.
In our example, knowing that the rectangles are similar means that the ratio of the diagonals D/d is 18/15 = 6/5. The ratio of the sides W/w and L/l must be 6/5 as well. So, the ratio of the areas A/a can be calculated.
Statement 1 says that the ratio of width to length is the same for both screens, that is W/L = w/l.
That is sufficient to answer the question, but if you want to calculate exact value you have to take the ratio of either W/w or L/l or D/d or P/p (for perimeters) and square it. Here it is (6/5)^2 = 36/25

Statement 2 says the width of the 18''-screen is 20% greater than that of the 15''-screen, that is W=1,2w. From this we get W/w = 1,2. So, we know the ratio and can say that rectangles are similar.
That is sufficient to answer the question. To check yourself square 1,2 and you’ll get 1,44 which is the same as 36/25.
Each statement is sufficient.
The answer is D.
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Re: M10:Q5 - DS [#permalink]

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17 Oct 2010, 14:48
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Orange08 wrote:
I am completely lost on this. Any better explanation please?

First note that the diagnol of a rectangle is given by $$d=sqrt{l^2+w^2}$$. And we need ratio of areas or lw. We know the diagnols.

(1) $$w_1/l_1 = w_2/l_2 = k$$
$$w_1/w_2=l_1/l_2$$
$$d_1=sqrt{l_1^2(1+k^2)}$$
$$d_2=sqrt{l_2^2(1+k^2)}$$
$$d_1/d_2 = l_1/l_2=w_1/w_2$$
Hence we know the ratio of areas as well :
$$l_1w_1/l_2w_2 = d_1^2/d_2^2=36/25$$
Sufficient !

(2) $$w_1=1.2w_2$$
$$l_1^2=d_1^2-w_1^2$$
$$l_2^2=d_2^2-w_2^2=d_1^2(5/6)^2-w_1^2(5/6)^2=(5/6)^2 * (d_1^-w_1^2) = (5/6 * l_1)^2$$
$$l_2 = (5/6) l_1$$
Hence we know both ratio of lengths and ratio of widths, so we can find out ratio of areas by mutliplying.
Sufficient !

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Re: M10:Q5 - DS [#permalink]

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19 Oct 2010, 14:09
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Orange08 wrote:
I am completely lost on this. Any better explanation please?

What is the ratio of the area of the TV screen with diagonal 18'' to that of the screen with diagonal 15''?

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

Given: ratio of diagonals = 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 --> Rectangles are similar --> right triangles made by diagonals are similar as well --> ratio of areas of triangles = ratio of areas of rectangles = (18/15)^2=1.44. Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen --> Ratio of widths = 1.2 = ratio of diagonals (hypotenuse) --> right triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio (in our case W/w=D/d=1.2), then these right triangles are similar --> rectangles are similar too --> ratio of areas of triangles = ratio of areas of rectangles = (18/15)^2=1.44. Sufficient.

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Re: M10:Q5 - DS [#permalink]

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21 Dec 2010, 06:30
A.

S1: let the sides be a and xa , and b and xb
we need to find --> a*xa / b*xb = a^2/b^2
using pythagoran theorem: a^2 + x^2a^2 = 18^2
and b^2 + x^2b^2 = 15^2

dividing both : we can cancel 1 + x^2 as the common term and hence are left with a^2/b^2 = 18^2/15^2

S2: does not lead us to anywhere
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Re: M10:Q5 - DS [#permalink]

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21 Dec 2010, 06:34
sleekmover wrote:
A.

S1: let the sides be a and xa , and b and xb
we need to find --> a*xa / b*xb = a^2/b^2
using pythagoran theorem: a^2 + x^2a^2 = 18^2
and b^2 + x^2b^2 = 15^2

dividing both : we can cancel 1 + x^2 as the common term and hence are left with a^2/b^2 = 18^2/15^2

S2: does not lead us to anywhere

OA for this question is D, not A. Refer to the above solutions to see why (2) is sufficient too.
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Re: M10:Q5 - DS [#permalink]

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25 Dec 2011, 23:04
Bunuel wrote:
Orange08 wrote:
I am completely lost on this. Any better explanation please?

What is the ratio of the area of the TV screen with diagonal 18'' to that of the screen with diagonal 15''?

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

Given: ratio of diagonals = 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 --> Rectangles are similar --> right triangles made by diagonals are similar as well --> ratio of areas of triangles = ratio of areas of rectangles = (18/15)^2=1.44. Sufficient.

(2) The width of the 18''-screen is 20% greater than that of the 15''-screen --> Ratio of widths = 1.2 = ratio of diagonals (hypotenuse) --> right triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio (in our case W/w=D/d=1.2), then these right triangles are similar --> rectangles are similar too --> ratio of areas of triangles = ratio of areas of rectangles = (18/15)^2=1.44. Sufficient.

perfect explanation in my view,
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Re: M10:Q5 - DS [#permalink]

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26 Jun 2012, 19:29
What is the property that applies to similar rectangles? How can we prove two rectangles similar?
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Re: M10:Q5 - DS [#permalink]

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05 Jul 2012, 02:25
(2) The width of the 18''-screen is 20% greater than that of the 15''-screen --> Ratio of widths = 1.2 = ratio of diagonals (hypotenuse) --> right triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio (in our case W/w=D/d=1.2), then these right triangles are similar --> rectangles are similar too --> ratio of areas of triangles = ratio of areas of rectangles = (18/15)^2=1.44. Sufficient.

For Bunnel,

In reference to the above explanation that you had posted, I have a question,

When you say - " right triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio (in our case W/w=D/d=1.2), then these right triangles are similar"

Which theorem of similar triangles is that? Neither does AAA, SSS or SAS apply in this case? For statement 2 you are using the logic that two sides ( diagonals and widths are in the same ratio) but the angle included between these two sides is not necessarily equal so SAS theorem does not apply?

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Re: M10:Q5 - DS [#permalink]

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05 Jul 2012, 02:55
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teal wrote:
(2) The width of the 18''-screen is 20% greater than that of the 15''-screen --> Ratio of widths = 1.2 = ratio of diagonals (hypotenuse) --> right triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio (in our case W/w=D/d=1.2), then these right triangles are similar --> rectangles are similar too --> ratio of areas of triangles = ratio of areas of rectangles = (18/15)^2=1.44. Sufficient.

For Bunnel,

In reference to the above explanation that you had posted, I have a question,

When you say - " right triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio (in our case W/w=D/d=1.2), then these right triangles are similar"

Which theorem of similar triangles is that? Neither does AAA, SSS or SAS apply in this case? For statement 2 you are using the logic that two sides ( diagonals and widths are in the same ratio) but the angle included between these two sides is not necessarily equal so SAS theorem does not apply?

Solution says: RIGHT triangles made by width and length are similar, as in right triangles if 2 corresponding sides are in the same ratio, then these right triangles are similar.

You can apply SAS here since the angle between the sides is 90 degrees for both triangles.
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Re: M10:Q5 - DS [#permalink]

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05 Jul 2012, 04:18
well the sides you are considering are width and diagonal NOT the length and the width - the included angle for Width and diagonal cannot be 90?

I am not comfortable with statement 2 similarity proof.
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Re: M10:Q5 - DS [#permalink]

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05 Jul 2012, 04:33
teal wrote:
well the sides you are considering are width and diagonal NOT the length and the width - the included angle for Width and diagonal cannot be 90?

I am not comfortable with statement 2 similarity proof.

OK. Say the three sides of one right triangle are: $$a$$, $$b$$ and $$c$$ --> $$a^2+b^2=c^2$$ --> $$b^2=c^2-a^2$$.

Now, if the three sides of another triangle are $$ka$$, $$x$$, and $$kc$$ --> $$(ka)^2+x^2=(kc)^2$$ --> $$x^2=k^2(c^2-a^2)=k^2*b^2$$ --> $$x=kb$$.

So, as you can see $$b$$ and $$x$$ are also in the same ratio. So, these two triangles must be similar.

Hope it's clear now.
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Re: M10:Q5 - DS [#permalink]

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05 Jul 2012, 04:44
Thanks, makes sense. So is it fine to generalize that if two sides of two right right triangles are in the same proportion, third side will be also in the same proportion. This will be true always for a right triangles.

Is this true for any other types of triangles?
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Re: M10:Q5 - DS [#permalink]

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08 Jul 2012, 03:17
teal wrote:
Thanks, makes sense. So is it fine to generalize that if two sides of two right right triangles are in the same proportion, third side will be also in the same proportion. This will be true always for a right triangles.

Is this true for any other types of triangles?

Hi Bunuel,
are we good to generalize the above as stated by Teal.

Thanks!
Re: M10:Q5 - DS   [#permalink] 08 Jul 2012, 03:17

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