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

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Intern
Joined: 03 Oct 2016
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29 Nov 2017, 03:52
1
12
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Difficulty:

95% (hard)

Question Stats:

38% (02:47) correct 62% (02:51) wrong based on 148 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

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GeoProblemTriangle.jpg [ 47.75 KiB | Viewed 6757 times ]

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29 Nov 2017, 09:34
5
1
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.

$$3.2 = \dfrac{16}{5}$$

$$\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}$$

Percentage of the square's area:

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

GeoProblemTriangleSol1.jpg [ 55.06 KiB | Viewed 6531 times ]

GeoProblemTriangleSol2.jpg [ 62.96 KiB | Viewed 6546 times ]

##### General Discussion
Senior Manager
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29 Nov 2017, 04:05
4
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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29 Nov 2017, 07:31
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%.

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Dear Sasindran , would u please elaborate more why these triangles are proportionate?
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29 Nov 2017, 07:57
3
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%.

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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

Thanks
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29 Nov 2017, 08:11
1
septwibowo

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

https://www.varsitytutors.com/hotmath/h ... rtionality
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29 Nov 2017, 23:09
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

Thanks

Thank you very much Sasindran for your generosity! Kudo for you!
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29 Nov 2017, 23:10
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.

$$3.2 = \dfrac{16}{5}$$

$$\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}$$

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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04 Jan 2018, 01:17
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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25 Mar 2018, 07:45
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:

$$\dfrac {4-h}{4}=\dfrac {h}{1}$$

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.

$$3.2 = \dfrac{16}{5}$$

$$\dfrac {\dfrac{16}{5}\times 4}{2} = \dfrac{32}{5}$$

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?   [#permalink] 25 Mar 2018, 07:45
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