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Similar Figures Problem - Geometry

Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Similar Figures Problem - Geometry

by omair_bba » Sun Sep 09, 2012 9:11 am
Hi, Below Q is not from any OG.


Q: In the figure, BD is parallel to EG, AD = 6, DG = 4, and triangle AEF has an area of 75. What is the area of triangle ABC ?

A) 27
B) 36
C) 45
D) 54
E) 63

Image

Answer = A .

Please explain answer too !

Thankx...
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by Javoni » Sun Sep 09, 2012 10:47 am
The triangles of ABC and AEF are similar because angle B and angle E are equal (BD paraller to EG and BC to EF). Also, angle A belongs to both triangles, no doubts, yep?! From here BC/EF = 6/10 = 3/5

There is one miraculous ratio for similar triangles that states following:

BC/EF = Square Root of (Area of ABC)/Square Root of (Area of AEF), This ratio is applicable to all similar triangles, I guess:

So we get: 3/5 = Sqr Root (X)/Sqr Root (75) >>> We will get 9*75 = 25*X, Hence X or the Area of triangle of ABC equals: 9*75/25 = 27
Answer: Area of triangle of ABC would be 27 - A

Please, correct me if I am wrong
omair_bba wrote:Hi, Below Q is not from any OG.


Q: In the figure, BD is parallel to EG, AD = 6, DG = 4, and triangle AEF has an area of 75. What is the area of triangle ABC ?

A) 27
B) 36
C) 45
D) 54
E) 63

Image

Answer = A .

Please explain answer too !

Thankx...
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by huuquan » Mon Sep 10, 2012 7:49 am
Javoni wrote:The triangles of ABC and AEF are similar because angle B and angle E are equal (BD paraller to EG and BC to EF). Also, angle A belongs to both triangles, no doubts, yep?! From here BC/EF = 6/10 = 3/5

There is one miraculous ratio for similar triangles that states following:

BC/EF = Square Root of (Area of ABC)/Square Root of (Area of AEF), This ratio is applicable to all similar triangles, I guess:

So we get: 3/5 = Sqr Root (X)/Sqr Root (75) >>> We will get 9*75 = 25*X, Hence X or the Area of triangle of ABC equals: 9*75/25 = 27
Answer: Area of triangle of ABC would be 27 - A

Please, correct me if I am wrong
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by ygdrasil24 » Wed Sep 12, 2012 9:53 am
Step1: Consider TRi ABD, AEG : AD/DG = AB/BE = 3/5
Step2 : Consider Tri ABC & AEF : (AB/AE)^2=Area ABC/Area AEF ;
Solve 75*9/25 = 27.

Hence A
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by vk_vinayak » Wed Sep 12, 2012 10:29 am
omair_bba wrote:Hi, Below Q is not from any OG.


Q: In the figure, BD is parallel to EG, AD = 6, DG = 4, and triangle AEF has an area of 75. What is the area of triangle ABC ?

A) 27
B) 36
C) 45
D) 54
E) 63

Image

Answer = A .

Please explain answer too !

Thankx...
RULE: For similar triangles, the ratio of areas = the ratio of squares of the corresponding sides.

From the figure we can see that AB=6 and BE=4. Hence AE = 10. Let the area of ABC be x.

According to the rule: (AE)^2 / (AB)^2 = 75 /x => 100/36 = 75/x

Solving for x, we get x=27. Hence Ans is A.
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by Jim@StratusPrep » Sat Sep 15, 2012 9:45 am
Yup-

Good one to remember:

For similar triangles, the ratio of areas = the ratio of squares of the corresponding sides.
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by omair_bba » Sun Sep 16, 2012 8:39 am
Thanks @Javoni & @vk_vinayak for ratio rule of similar triangles....
i.e RULE: For similar triangles, the ratio of areas = the ratio of squares of the corresponding sides.

I also solved below Q with the same rule...helped a lot !

Q: The figure shows a wedge-shaped holding tank that is partially filled with water. If the tank is 1/16 full, what is the depth of the water at the deepest part ?

A) 3
B) 2
C) 1.5
D) 1
E) 0.75

Image

Answer = A .

Please confirm that this rule is also applicable here....
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