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| FROM Veritas Prep Blog: Looking for Similar Triangles on the GMAT |
 Our Geometry book discusses the various rules we use to recognize similar triangles such as SSS, AA, SAS and RHS so we are assuming that we needn’t take those up here. We are also assuming that you are comfortable with the figures that beg you to think about similar triangles such as  Try to figure out the similar triangles and the reason they are similar in each one of these cases. (Angles that look 90 are 90). Most of the figures have right angles/parallel lines. This topic was also discussed by David Newland in a rather engaging post last week. You must check it out for its content as well as its context! What we would like to discuss today are situations where most people do not think about similar triangles but if they do, it would make the question very easy for them. But before we do that, we would like to discuss a concept related to similar triangles which is very useful but not discussed often. We already know that sides of similar triangles are in the same ratio. Say two triangles have sides a, b, c and A, B, C respectively. Then, a/A = b/B = c/C = k Note that the altitudes of the two triangles will also be in the same ratio, ‘k’, since all lengths have the ratio ‘k’. Then what is the relation between the areas of the two triangles? Since the ratio of the bases is k and the ratio of the altitudes is also k, the ratio of the areas will be k*k = k^2. So if there are two similar triangles such that their sides are in the ratio 1:2, their areas will be in the ratio 1:4. Now we are all ready to tackle the question we have in mind. Question: In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area?  (A) 3:8 (B) 3:5 (C) 5:8 (D) 8:5 (E) 5:3 Solution: There are many ways to do this question but we will look at the method using similar triangles (obviously!). Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below)  Now look at the original figure. HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4. Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P. The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P. Hence, area of shaded region : Area of unshaded region = 5:3 Answer (E) Try to think of other ways in which you can solve this question. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
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18 Feb 2015, 05:03
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Amit989
Joined: 10 Nov 2014
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Schools: ISB'22 (I) Said '22 Cambridge '22 IIMA PGPX'22 (WL)
GMAT 1: 730 Q47 V44
WE:Information Technology (Computer Software)
Schools: ISB'22 (I) Said '22 Cambridge '22 IIMA PGPX'22 (WL)
GMAT 1: 730 Q47 V44
Posts: 26
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26 Sep 2017, 22:32
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[quote="Bunuel"][rss2posts title="Veritas Prep Blog" title_url="https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/03/looking-for-similar-triangles-on-the-gmat/" sub_title="Looking for Similar Triangles on the GMAT"]
Our Geometry book discusses the various rules we use to recognize similar triangles such as SSS, AA, SAS and RHS so we are assuming that we needn’t take those up here.
We are also assuming that you are comfortable with the figures that beg you to think about similar triangles such as
Try to figure out the similar triangles and the reason they are similar in each one of these cases. (Angles that look 90 are 90). Most of the figures have right angles/parallel lines.
This topic was also discussed by David Newland in a rather engaging post last week. You must check it out for its content as well as its context!
What we would like to discuss today are situations where most people do not think about similar triangles but if they do, it would make the question very easy for them. But before we do that, we would like to discuss a concept related to similar triangles which is very useful but not discussed often.
We already know that sides of similar triangles are in the same ratio. Say two triangles have sides a, b, c and A, B, C respectively. Then, a/A = b/B = c/C = k
Note that the altitudes of the two triangles will also be in the same ratio, ‘k’, since all lengths have the ratio ‘k’.
Then what is the relation between the areas of the two triangles? Since the ratio of the bases is k and the ratio of the altitudes is also k, the ratio of the areas will be k*k = k^2.
So if there are two similar triangles such that their sides are in the ratio 1:2, their areas will be in the ratio 1:4.
Now we are all ready to tackle the question we have in mind.
Question: In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area?
(A) 3:8
(B) 3:5
(C) 5:8
(D) 8:5
(E) 5:3
Solution: There are many ways to do this question but we will look at the method using similar triangles (obviously!).
Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below)
Now look at the original figure.
HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4.
Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P.
The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P.
Hence, area of shaded region : Area of unshaded region = 5:3
Answer (E)
Try to think of other ways in which you can solve this question.
--------------------
Hi Bunuel,
Thanks for the explanation. Though I did it in a similar approach the only thing that am stuck at while reviewing is : how did we conclude that we have 4 small Triangles and all 4 of them are congruent?
""Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF"".
AHE and DGH can have different areas. Cannot they? We are not given that the sides of the \\gm are equal. Could you please explain how you found them to be similar. I understand that two sides are similar but what is the 3rd property?
Best-
Amit
Our Geometry book discusses the various rules we use to recognize similar triangles such as SSS, AA, SAS and RHS so we are assuming that we needn’t take those up here.
We are also assuming that you are comfortable with the figures that beg you to think about similar triangles such as
Try to figure out the similar triangles and the reason they are similar in each one of these cases. (Angles that look 90 are 90). Most of the figures have right angles/parallel lines.
This topic was also discussed by David Newland in a rather engaging post last week. You must check it out for its content as well as its context!
What we would like to discuss today are situations where most people do not think about similar triangles but if they do, it would make the question very easy for them. But before we do that, we would like to discuss a concept related to similar triangles which is very useful but not discussed often.
We already know that sides of similar triangles are in the same ratio. Say two triangles have sides a, b, c and A, B, C respectively. Then, a/A = b/B = c/C = k
Note that the altitudes of the two triangles will also be in the same ratio, ‘k’, since all lengths have the ratio ‘k’.
Then what is the relation between the areas of the two triangles? Since the ratio of the bases is k and the ratio of the altitudes is also k, the ratio of the areas will be k*k = k^2.
So if there are two similar triangles such that their sides are in the ratio 1:2, their areas will be in the ratio 1:4.
Now we are all ready to tackle the question we have in mind.
Question: In the given figure, ABCD is a parallelogram and E, F, G and H are midpoints of its respective sides. What is the ratio of the shaded area to that of the un-shaded area?
(A) 3:8
(B) 3:5
(C) 5:8
(D) 8:5
(E) 5:3
Solution: There are many ways to do this question but we will look at the method using similar triangles (obviously!).
Assume the area of the parallelogram is 8P. In a parallelogram, the lengths of opposite sides are the same. The two triangles formed by the diagonal and two sides are similar by SSS and the ratio of their sides is 1. So they will have equal areas of 4P each (look at the figures in second row below)
Now look at the original figure.
HE is formed by joining the mid-points of AD and AB. So AH/AD = AE/AB = 1/2 and included angle A is common. Hence by SAS rule, triangle AHE is similar to triangle ADB. If the ratio of sides is 1/2, ratio of areas will be 1/4.
Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF which are not shaded so the area of these three triangles together will be 3P.
The total area of parallelogram is 8P and the unshaded region is 3P. So the shaded region must be 5P.
Hence, area of shaded region : Area of unshaded region = 5:3
Answer (E)
Try to think of other ways in which you can solve this question.
--------------------
Hi Bunuel,
Thanks for the explanation. Though I did it in a similar approach the only thing that am stuck at while reviewing is : how did we conclude that we have 4 small Triangles and all 4 of them are congruent?
""Since area of triangle ADB is 4P, area of AHE is P. We have 3 such triangles, AHE, DHG and CGF"".
AHE and DGH can have different areas. Cannot they? We are not given that the sides of the \\gm are equal. Could you please explain how you found them to be similar. I understand that two sides are similar but what is the 3rd property?
Best-
Amit
