January 20, 2019 January 20, 2019 07:00 AM PST 07:00 AM PST Get personalized insights on how to achieve your Target Quant Score. January 19, 2019 January 19, 2019 07:00 AM PST 09:00 AM PST Aiming to score 760+? Attend this FREE session to learn how to Define your GMAT Strategy, Create your Study Plan and Master the Core Skills to excel on the GMAT.
Author 
Message 
TAGS:

Hide Tags

Math Expert
Joined: 02 Sep 2009
Posts: 52294

In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 04:07
Question Stats:
66% (01:31) correct 34% (01:20) wrong based on 131 sessions
HideShow timer Statistics



Current Student
Joined: 25 Nov 2014
Posts: 98
Concentration: Entrepreneurship, Technology
GPA: 4

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 09:29
I approached this problem by proving that each of the smaller triangle is congruent to the other. 1: Tr(AHE) and Tr(CFG) > AE = CG (since mid points of opp sides of parallelogram, so they are equal) > AH = CF (same) > Angle A = Angle C (opp angles of parallelogram are equal) Thus Tr(AHE) and Tr(CFG) are congruent. Similarly we can prove that Tr(BFE) and Tr(DHG) are congruent. Now draw a line from E to G, and consider the triangles EHG and EFG. We can prove that these triangles are also congruent. (Common side and other 2 sides are equal bec of the congruence of other triangles proved above). In all, we get all the smaller triangles congruent to each other. Thus the asked ratio is obtained simply by counting the no of triangles that are shaded and that are not shaded. 5 portions are shaded and 3 are not. Thus 5:3. Ans : E
_________________
Kudos!!



Manager
Joined: 27 Dec 2013
Posts: 237

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 09:45
IMO the answer is D (8:5) Please let me whether my answer is correct or not. Bunuel wrote: Attachment: SimilarTriangles2.jpg 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 unshaded area? (A) 3:8 (B) 3:5 (C) 5:8 (D) 8:5 (E) 5:3 Kudos for a correct solution.
_________________
Kudos to you, for helping me with some KUDOS.



Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 09:48



Current Student
Joined: 25 Nov 2014
Posts: 98
Concentration: Entrepreneurship, Technology
GPA: 4

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 10:32
shriramvelamuri, Can you share your thoughts on how you got D?
_________________
Kudos!!



Board of Directors
Joined: 17 Jul 2014
Posts: 2599
Location: United States (IL)
Concentration: Finance, Economics
GPA: 3.92
WE: General Management (Transportation)

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 10:44
not calculating anything, we can eliminate A, B, and C right away. between D & E, I chose to go with D. we can aprox. divide everything into 8 small triangles. 5 shaded and 3 unshaded. but since the areas of all triangles are not the same, then it is not possible to be 5:3. therefore D makes more sense.



Intern
Joined: 02 Jan 2014
Posts: 38

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 17:00
IMO The answer is E. if the area of the shaded triangle is x , we can divide the figure in the same area triangles. So, it becomes 5x/3x which is 5/3.



Intern
Joined: 21 Nov 2013
Posts: 11
Location: Canada
Concentration: Strategy, Technology

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 18:09
A square is a type of parellelogram, so let's assume it's a 4by 4 square (total area of 16). You should see without calculation that means the area of the centre square (EFGH) is 8, and each triangle in the corner is 2 units of area.
Unshaded = 3*2 = 6 Shaded = 8 + 2 = 10
10/6 = 5/3
Answer: E



Manager
Joined: 27 Dec 2013
Posts: 237

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 19:23
Hi Sherlocked. My mistake. The answer is 5:3. My overworked brain has let me do another simple mistake. Kind Regards, SherLocked2018 wrote: shriramvelamuri, Can you share your thoughts on how you got D?
_________________
Kudos to you, for helping me with some KUDOS.



Retired Moderator
Status: On a mountain of skulls, in the castle of pain, I sit on a throne of blood.
Joined: 30 Jul 2013
Posts: 323

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 21:15
Bunuel wrote: Attachment: SimilarTriangles2.jpg 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 unshaded area? (A) 3:8 (B) 3:5 (C) 5:8 (D) 8:5 (E) 5:3 Kudos for a correct solution.Since all squares are parallelograms, let us assume that the mother figure is also a square for calculation ease. The figure formed by joining the midpoints of a square is always a square. The outer square's side (s)=Diagonal of inner square. Therefore the inner square's side (a) = \(\sqrt{2}\)a=s a=\(\frac{s}{\sqrt{2}}\) Area of the inner square = \(a^2\)=\(\frac{s^2}{2}\)...(1) Now all the triangles formed in the figure are congruent according the the SSS rule. Each triangle is also an isosceles triangle since 2 of the sides are s/2 and 1 remaining side(the side of the inner square) is \(\frac{s}{\sqrt{2}}\) Area of an Isosceles Triangle= \(\frac{y}{4} \sqrt{4x^2y^2}\) where x=equal side and y=unequal side. \(\frac{s}{4\sqrt{2}} \sqrt{4*\frac{s^2}{4}  \frac{s^2}{2}}\) Are of each triangle = \(\frac{s^2}{8}\)...(2) Shaded Area = (1) + (2) =\(\frac{s^2}{2}\) + \(\frac{s^2}{8}\) =\(\frac{5s^2}{8}\) Unshaded Area = 3* (2) =\(\frac{3s^2}{8}\) \(\frac{Shaded Area}{Unshaded Area} = \frac{5}{3}\) Answer: E



Director
Joined: 07 Aug 2011
Posts: 533
Concentration: International Business, Technology

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 21:16
Bunuel wrote: Attachment: The attachment SimilarTriangles2.jpg is no longer available 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 unshaded area? (A) 3:8 (B) 3:5 (C) 5:8 (D) 8:5 (E) 5:3 Kudos for a correct solution.Answer E .
Attachments
similar_triangles.jpg [ 55.81 KiB  Viewed 8611 times ]



Director
Joined: 07 Aug 2011
Posts: 533
Concentration: International Business, Technology

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
16 Apr 2015, 21:29
SherLocked2018 wrote: I approached this problem by proving that each of the smaller triangle is congruent to the other. 1: Tr(AHE) and Tr(CFG) > AE = CG (since mid points of opp sides of parallelogram, so they are equal) > AH = CF (same) > Angle A = Angle C (opp angles of parallelogram are equal) Thus Tr(AHE) and Tr(CFG) are congruent.
Similarly we can prove that Tr(BFE) and Tr(DHG) are congruent. Now draw a line from E to G, and consider the triangles EHG and EFG. We can prove that these triangles are also congruent. (Common side and other 2 sides are equal bec of the congruence of other triangles proved above).
In all, we get all the smaller triangles congruent to each other. Thus the asked ratio is obtained simply by counting the no of triangles that are shaded and that are not shaded.
5 portions are shaded and 3 are not. Thus 5:3.
Ans : E note all the smaller triangles need not to be congruent, it depends on the angles subtended by sides of the gram. however, all the smaller triangles have to have the same area.
Attachments
similar_triangles.jpg [ 27.64 KiB  Viewed 8597 times ]



Intern
Joined: 29 Mar 2015
Posts: 11

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
17 Apr 2015, 02:23
Refer the triangle GCF. The area is 1/2 * Base * Height. Which is 1/2 * GC * x (let x be the height).
Note that this is the same height for the triangle HDG. Also since DG=GC (mid point), the two triangles have equal area. Lets call it A.
Now consider the triangle HGF (join F & H). The area of that triangle can similarly be written as
1/2*x*2GC. (as CD=FH & CD=2CG).
Which is equal to 2A.
Similarly for the other half.
Adding, we have the shaded portion as 5A & the unshaded portion is 3A, hence the answer is 5:3. That is E.



Manager
Joined: 15 May 2014
Posts: 62

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
18 Apr 2015, 04:00
Varignon's Theorem: The midpoints of the sides of an arbitrary quadrangle form a parallelogram. If the quadrangle is convex or reentrant, i.e. not a crossing quadrangle, then the area of the parallelogram is half the area of the quadrangle.
Let \(x\) be the area of the parallelogram \(ABCD\) Then Area of parallelogram \(EFGH\,=\,\frac{x}{2}\) Area of each triangle \(=\,\frac{1}{4}\,(\frac{x}{2})\)
Area of shaded region:Area of unshaded region \(=\) [Area of parallelogram \(EFGH\) + Area of \(\triangle\)EFB]\(:\)[Area of \(\triangle\)s \(AEH\), \(HGD\), \(FGC\)] \(=\) [\(\frac{x}{2}\,+\,\frac{x}{8}\)]\(:\)[\(3\,*\,\frac{x}{8}\)] \(=\) \(\frac{5x}{8}\)\(:\)\(\frac{3x}{8}\) \(=\) \(\,5:3\)
Answer E



Math Expert
Joined: 02 Sep 2009
Posts: 52294

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
20 Apr 2015, 04:27
Bunuel wrote: 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 unshaded area? (A) 3:8 (B) 3:5 (C) 5:8 (D) 8:5 (E) 5:3 Kudos for a correct solution.Attachment: The attachment SimilarTriangles2.jpg is no longer available VERITAS PREP OFFICIAL SOLUTIONThere 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) Attachment:
SimilarTriangles31 (1).jpg [ 53.84 KiB  Viewed 8377 times ]
Now look at the original figure. HE is formed by joining the midpoints 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)
_________________
New to the Math Forum? Please read this: Ultimate GMAT Quantitative Megathread  All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics



NonHuman User
Joined: 09 Sep 2013
Posts: 9450

Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid
[#permalink]
Show Tags
30 Sep 2018, 17:27
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: In the given figure, ABCD is a parallelogram and E, F, G and H are mid &nbs
[#permalink]
30 Sep 2018, 17:27






