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In the figure above, what is the area of the shaded region?
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05 Sep 2017, 22:53
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In the figure above, what is the area of the shaded region?
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Updated on: 05 Sep 2017, 23:26
We have to find area of shaded region. area of shaded region = Area of big triangle  area of small triangle. ie area of shaded region = Area( Triangle ABC)  Area( Triangle ADE) < Refer attached figure As both triangles are right triangle and both have an angle common ( Angle A) =>Big triangle is similar to small triangle => triangle ABC ~ triangle ADE Area formula for 2 similar triangles => \(\frac{Ar(ABC)}{Ar(ADE)} = \frac{AB^2}{AD^2} = \frac{BC^2}{DE^2} = \frac{AC^2}{AE^2}\) Ar(ABC) = 1/2 *1*1 = 1/2 BC=1 DE=2/3 =>\(\frac{Ar(ABC)}{Ar(ADE)} = \frac{1^2}{(2/3)^2}\) => Ar(ADE)= Ar(ABC)* 4/9 = 1/2 * 4/9 = 2/9 Area of shaded region = Ar(ABC)  Ar(ADE) = 1/2 2/9 =5/18 Answer: B
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Originally posted by Nikkb on 05 Sep 2017, 23:12.
Last edited by Nikkb on 05 Sep 2017, 23:26, edited 1 time in total.



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Re: In the figure above, what is the area of the shaded region?
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05 Sep 2017, 23:22
Bunuel wrote: In the figure above, what is the area of the shaded region? (A) 5/36 (B) 5/18 (C) 1/3 (D) 5/12 (E) 5/6 Attachment: 20170906_1052.png Divide the shaded region into one triangle on the top with (12/3) = 1/3 height and x as base area of triangle will be 1/3x area of rectangle with 2/3 length and x width will be 2/3x area of shaded region = 5/6x 5/6x + (1x)*1/3 = 1/2 3x/6 = 1/6 x=1/3 5/6*1/3 = 5/18 B
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In the figure above, what is the area of the shaded region?
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05 Sep 2017, 23:41
Attachment:
20170906_1052.png [ 6.09 KiB  Viewed 3925 times ]
In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1 The triangle ABC is similar to ADE(AA similarity) It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\) (sides opposite equal angles are equal) Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\) Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\) Therefore, are of shaded region is \(\frac{1}{2}  \frac{2}{9} = \frac{(94)}{18} = \frac{5}{18}\) (Option B)
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Re: In the figure above, what is the area of the shaded region?
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06 Sep 2017, 02:52
Bunuel wrote: In the figure above, what is the area of the shaded region? (A) 5/36 (B) 5/18 (C) 1/3 (D) 5/12 (E) 5/6 Attachment: 20170906_1052.png Since the unshaded triangle and the whole triangle are similar . Ratio of their area = (2/3)^2 = 4/9 Ratio of area of shaded region to full triangle = 14/9 = 5/9 Area of complete triangle = 1/2*1*1 = 1/2 Area of shaded region = 5/9 *1/2 = 5/18 Answer B
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In the figure above, what is the area of the shaded region?
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06 Sep 2017, 13:28
pushpitkc wrote: Attachment: 20170906_1052.png In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1 The triangle ABC is similar to ADE(AA similarity) It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\) (sides opposite equal angles are equal) Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\) Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\) Therefore, are of shaded region is \(\frac{1}{2}  \frac{2}{9} = \frac{(94)}{18} = \frac{5}{18}\) (Option B)(E) 5/6 Attachment: 20170906_1052.png pushpitkc , (and the rest of the posters, all of whom used triangles' similarity or assumed right angles at some point), I don't think we know we have two right angles. We are not told that the two vertical lines are parallel. If the two vertical lines were parallel, then the transversal which forms the base of both triangles, and which would cut the parallel lines, would create two congruent right angles. We know from the diagram's mark that one angle is a right angle. That angle and the unmarked angle at the smaller triangle's base would be corresponding angles created by parallel lines cut by a transversal, and would be congruent. In that case, the small and large triangles would share a right angle and a vertex angle, and by AA property of triangles, the triangles would be similar. We are not told that the shorter vertical line is perpendicular to the horizontal line. Same result as above, i.e., if short vertical line were perpendicular to base, triangles would be similar. Nor does the figure display a right angle mark where the shorter vertical line meets the base to create a small triangle with one side of length \(\frac{2}{3}\). It seems all we know is that the two triangles share one angle, at the vertex on the left. The short vertical line might not be vertical. I'm not sure how we can conclude that the triangles are similar. Am I missing something?



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Re: In the figure above, what is the area of the shaded region?
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06 Sep 2017, 23:10
genxer123 wrote: pushpitkc wrote: Attachment: 20170906_1052.png In the figure, ABC is a right angled isosceles triangle, as AB = BC = 1 The triangle ABC is similar to ADE(AA similarity) It has been given that DE = \(\frac{2}{3}\). Similarly, AD = \(\frac{2}{3}\) (sides opposite equal angles are equal) Area of triangle ABC = \(\frac{1}{2}*1*1 = \frac{1}{2}\) Area of triangle ADE = \(\frac{1}{2}*(\frac{2}{3})^2 = \frac{2}{9}\) Therefore, are of shaded region is \(\frac{1}{2}  \frac{2}{9} = \frac{(94)}{18} = \frac{5}{18}\) (Option B)(E) 5/6 Attachment: 20170906_1052.png pushpitkc , (and the rest of the posters, all of whom used triangles' similarity or assumed right angles at some point), I don't think we know we have two right angles. We are not told that the two vertical lines are parallel. If the two vertical lines were parallel, then the transversal which forms the base of both triangles, and which would cut the parallel lines, would create two congruent right angles. We know from the diagram's mark that one angle is a right angle. That angle and the unmarked angle at the smaller triangle's base would be corresponding angles created by parallel lines cut by a transversal, and would be congruent. In that case, the small and large triangles would share a right angle and a vertex angle, and by AA property of triangles, the triangles would be similar. We are not told that the shorter vertical line is perpendicular to the horizontal line. Same result as above, i.e., if short vertical line were perpendicular to base, triangles would be similar. Nor does the figure display a right angle mark where the shorter vertical line meets the base to create a small triangle with one side of length \(\frac{2}{3}\). It seems all we know is that the two triangles share one angle, at the vertex on the left. The short vertical line might not be vertical. I'm not sure how we can conclude that the triangles are similar. Am I missing something? genxer123 , You are 100% correct. There is no mention of shorter line being parallel to the larger line in the shaded region. We just assumed it to be parallel and solved the question which is not at all acceptable on GMAT. Also there were option available, so we just solved it. Had it been a data sufficiency question, then the situation could have created a real problem. So the question should say that smaller line is parallel to longer line...
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Re: In the figure above, what is the area of the shaded region?
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11 Sep 2017, 10:02
Bunuel wrote: In the figure above, what is the area of the shaded region? (A) 5/36 (B) 5/18 (C) 1/3 (D) 5/12 (E) 5/6 Attachment: 20170906_1052.png We see that we have two similar triangles. The ratio of the side length of the larger triangle to the smaller triangle is 1/(2/3). Since the base and height of the larger triangle is 1, the base and height of the smaller triangle is 2/3. Thus, the area of the larger triangle is 1 x 1 x 1/2 = ½ and the area of the smaller triangle is 2/3 x 2/3 x 1/2 = 4/18. So, the area of the shaded region is 1/2  4/18 = 9/18  4/18 = 5/18. Answer: B
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Re: In the figure above, what is the area of the shaded region?
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19 Sep 2018, 09:22
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