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What percentage of the square's area is shaded?

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What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 02:52
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A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

51% (02:43) correct 49% (01:33) wrong based on 41 sessions

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What percentage of the square's area is grey-shaded, with AB = 4 and DE = 1? (see picture)

A: 30%

B: 37,5%

C: 40%

D: 45%

E: 50%



*found it on brilliant.org and I guess it's a tough gmat-style questions
[Reveal] Spoiler: OA

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GeoProblemTriangle.jpg
GeoProblemTriangle.jpg [ 47.75 KiB | Viewed 649 times ]

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What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 03:05
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Option C

The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.

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Re: What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 06:31
Sasindran wrote:
Option C

The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.

Posted from my mobile device


Dear Sasindran , would u please elaborate more why these triangles are proportionate?
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Re: What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 06:57
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septwibowo wrote:
Sasindran wrote:
Option C

The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.

Posted from my mobile device


Dear Sasindran , would u please elaborate more why these triangles are proportionate?


Sure.

We all know that AB and CD are parallel lines. When two lines intersect between parallel lines the triangles so formed by them are always proportional to each other. Also the distance of intersection from either of the lines are also proportional to the length of the parallel lines.

For our case lets consider the intersecting point as O. Here we have AB=4 and DE=1. Hence by the above theorem we have perpendicular Distance between O and line AB is proportional to the perpendicular distance between O and the line DE.

i.e., AB/DE = per dist between O and AB/per dist between O and DE

4/1=height of triangle ABO/height of triagle DEO
also we know that height of triangle AEO+height of triagle DEO=4 (divided at O in the ratio 4:1)

Hence height of triangle AEO=4/5=3.2
height of triangle DEO=1/5=0.8

Hence area of triangle AEO=1/2*4*3.2=6.4

And you know the remaining septwibowo

Hope i made clear!

Thanks
_________________

Help with kudos if u found the post useful. Thanks

Kudos [?]: 38 [1], given: 89

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Re: What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 07:11
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septwibowo

Although may find 1000s of resources I am providing one:

https://www.varsitytutors.com/hotmath/h ... rtionality
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Re: What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 08:34
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septwibowo

Why the two triangles are indeed similar is shown graphically below.

Angles a, b, and c are identical, making the two triangles similar. Hence, the base to height ratio is similar and we can set up a proportion to find the height:

\(\dfrac {4-h}{4}=\dfrac {h}{1}\)

h = 0.8

So, height of the shaded triangle is 3.2 and since we have the base (4) and the height (3.2) we can calculate the area.

\(3.2 = \dfrac{16}{5}\)

Area of grey-shaded triangle:

\(\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}\)

Percentage of the square's area:

\(\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%\)
Attachments

GeoProblemTriangleSol1.jpg
GeoProblemTriangleSol1.jpg [ 55.06 KiB | Viewed 472 times ]

GeoProblemTriangleSol2.jpg
GeoProblemTriangleSol2.jpg [ 62.96 KiB | Viewed 465 times ]

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Re: What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 22:09
Sasindran wrote:
septwibowo wrote:
Sasindran wrote:
Option C

The two triangles are proportionate. Hence the height of the shaded triangle is 4/(4+1)*4 =3.2. Hence area of the shaded region =1/2*4*3.2=6.4. The ratio is 6.4/(4*4)*100=40%.

Posted from my mobile device


Dear Sasindran , would u please elaborate more why these triangles are proportionate?


Sure.

We all know that AB and CD are parallel lines. When two lines intersect between parallel lines the triangles so formed by them are always proportional to each other. Also the distance of intersection from either of the lines are also proportional to the length of the parallel lines.

For our case lets consider the intersecting point as O. Here we have AB=4 and DE=1. Hence by the above theorem we have perpendicular Distance between O and line AB is proportional to the perpendicular distance between O and the line DE.

i.e., AB/DE = per dist between O and AB/per dist between O and DE

4/1=height of triangle ABO/height of triagle DEO
also we know that height of triangle AEO+height of triagle DEO=4 (divided at O in the ratio 4:1)

Hence height of triangle AEO=4/5=3.2
height of triangle DEO=1/5=0.8

Hence area of triangle AEO=1/2*4*3.2=6.4

And you know the remaining septwibowo

Hope i made clear!

Thanks


Thank you very much Sasindran for your generosity! Kudo for you!
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Kudos [?]: 50 [0], given: 226

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Joined: 27 Dec 2016
Posts: 198

Kudos [?]: 50 [0], given: 226

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Re: What percentage of the square's area is shaded? [#permalink]

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New post 29 Nov 2017, 22:10
drexxie wrote:
septwibowo

Why the two triangles are indeed similar is shown graphically below.

Angles a, b, and c are identical, making the two triangles similar. Hence, the base to height ratio is similar and we can set up a proportion to find the height:

\(\dfrac {4-h}{4}=\dfrac {h}{1}\)

h = 0.8

So, height of the shaded triangle is 3.2 and since we have the base (4) and the height (3.2) we can calculate the area.

\(3.2 = \dfrac{16}{5}\)

Area of grey-shaded triangle:

\(\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}\)

Percentage of the square's area:

\(\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%\)


Thank you drexxie ! I forgot this theorem, my bad!
_________________

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Re: What percentage of the square's area is shaded?   [#permalink] 29 Nov 2017, 22:10
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