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What percentage of the square's area is shaded?
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29 Nov 2017, 02:52
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37% (02:45) correct 63% (02:49) wrong based on 146 sessions
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What percentage of the square's area is greyshaded, 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 gmatstyle questions
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Re: What percentage of the square's area is shaded?
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29 Nov 2017, 08:34
septwibowoWhy 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: \(\dfrac {4h}{4}=\dfrac {h}{1}\) 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 greyshaded triangle: \(\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}\) Percentage of the square's area: \(\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%\)
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What percentage of the square's area is shaded?
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29 Nov 2017, 03:05
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
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Re: What percentage of the square's area is shaded?
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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?
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29 Nov 2017, 06:57
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 septwibowoHope i made clear! Thanks
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Re: What percentage of the square's area is shaded?
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29 Nov 2017, 07:11
septwibowoAlthough 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?
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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 septwibowoHope i made clear! Thanks Thank you very much Sasindran for your generosity! Kudo for you!
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Re: What percentage of the square's area is shaded?
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29 Nov 2017, 22:10
drexxie wrote: septwibowoWhy 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: \(\dfrac {4h}{4}=\dfrac {h}{1}\) 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 greyshaded 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?
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04 Jan 2018, 00:17
well, this question is hardly a gmat question. The source is also unreliable. However, this question is really a tough question.



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Re: What percentage of the square's area is shaded?
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25 Mar 2018, 06:45
drexxie wrote: 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:
\(\dfrac {4h}{4}=\dfrac {h}{1}\)
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 greyshaded triangle:
\(\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}\)
Percentage of the square's area:
\(\dfrac {\dfrac {32}{5}}{16} = \dfrac{2}{5} \rightarrow \dfrac{2}{5}\times 100 = 40\%\)
This is real compact and accurate answer. Thanks mate. Kudo for you!




Re: What percentage of the square's area is shaded? &nbs
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25 Mar 2018, 06:45






