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In the figure above, ABCD and BEFG are squares and points G and H are Updated on: 17 Jul 2020, 06:07
Originally posted by Frank28 on 17 Jul 2020, 03:46.
Last edited by Frank28 on 17 Jul 2020, 06:07, edited 1 time in total.
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In the figure above, ABCD and BEFG are squares and points G and H are on BC and EF, respectively. If the area of triangle DBH is 80 square meters, what is the sum of the areas of squares ABCD and BEFG?

(1) GF=8m
(2) GC=4m

Source: Topgmat

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In the figure above, ABCD and BEFG are squares and points G and H are 17 Jul 2020, 05:07
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We need the length of the sides of the square.

Area of DBH = 1/2* (DC) * (GB) + 1/2 * (GF) * (GB) = 80

Area of DBH = 1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80...... (1) [since GF = GB]



Statement 1. We know GF or GB. We can find DC using equation 1.

Sufficient

Statement 2- We know DC - GF. We have 2 equations and 2 variables. We can find DC and GF. [We are interested in +ve values]

Sufficient.

.
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In the figure above, ABCD and BEFG are squares and points G and H are 17 Jul 2020, 05:19
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nick1816
We need the length of the sides of the square.

Area of DBH = 1/2* (DC) * (GB) + 1/2 * (GF) * (GB) = 80

Area of DBH = 1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80...... (1) [since GF = GB]



Statement 1. We know GF or GB. We can find DC using equation 1.

Sufficient

Statement 2- We know DC - GF. We have 2 equations and 2 variables. We can find DC and GF. [We are interested in +ve values]

Sufficient.

BTW you don't suppose to share GMAT exam Questions.
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will you explain how you determined this?
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In the figure above, ABCD and BEFG are squares and points G and H are 17 Jul 2020, 05:26
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You can split the triangle DBH in 2 parts DGB and HGB

Area of DGB = 1/2 * height * base = 1/2 * DC *GB

Area of HGB = 1/2 * GF * GB

yashikaaggarwal
nick1816
We need the length of the sides of the square.

Area of DBH = 1/2* (DC) * (GB) + 1/2 * (GF) * (GB) = 80

Area of DBH = 1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80...... (1) [since GF = GB]



Statement 1. We know GF or GB. We can find DC using equation 1.

Sufficient

Statement 2- We know DC - GF. We have 2 equations and 2 variables. We can find DC and GF. [We are interested in +ve values]

Sufficient.

BTW you don't suppose to share GMAT exam Questions.
will you explain how you determined this?
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In the figure above, ABCD and BEFG are squares and points G and H are 17 Jul 2020, 05:37
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nick1816
You can split the triangle DBH in 2 parts DGB and HGB

Area of DGB = 1/2 * height * base = 1/2 * DC *GB

Area of HGB = 1/2 * GF * GB

yashikaaggarwal
nick1816
We need the length of the sides of the square.

Area of DBH = 1/2* (DC) * (GB) + 1/2 * (GF) * (GB) = 80

Area of DBH = 1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80...... (1) [since GF = GB]



Statement 1. We know GF or GB. We can find DC using equation 1.

Sufficient

Statement 2- We know DC - GF. We have 2 equations and 2 variables. We can find DC and GF. [We are interested in +ve values]

Sufficient.

BTW you don't suppose to share GMAT exam Questions.
will you explain how you determined this?
Show more
Did you used "the triangle drawn from same base and equal height have same area" Property here?

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In the figure above, ABCD and BEFG are squares and points G and H are 17 Jul 2020, 07:04
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In the figure above, ABCD and BEFG are squares and points G and H are 20 Jul 2020, 06:28
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This QUestion can also be done using Draw to decide approach :



Frank28
In the figure above, ABCD and BEFG are squares and points G and H are on BC and EF, respectively. If the area of triangle DBH is 80 square meters, what is the sum of the areas of squares ABCD and BEFG?

(1) GF=8m
(2) GC=4m

Source: Topgmat
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In the figure above, ABCD and BEFG are squares and points G and H are 21 Jul 2020, 22:33
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nick1816
We need the length of the sides of the square.

Area of DBH = 1/2* (DC) * (GB) + 1/2 * (GF) * (GB) = 80

Area of DBH = 1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80...... (1) [since GF = GB]



Statement 1. We know GF or GB. We can find DC using equation 1.

Sufficient

Statement 2- We know DC - GF. We have 2 equations and 2 variables. We can find DC and GF. [We are interested in +ve values]

Sufficient.

.
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Could you solve statement II. I’m unable to see how that is sufficient

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In the figure above, ABCD and BEFG are squares and points G and H are 22 Jul 2020, 01:54
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No...i haven't use that property.

The only altitude you can draw from D on BG is DC. Since G is an obtuse angle, altitude from D will lie outside of the triangle.

Also, GB is parallel to FE and FGB = 90*, GF is equal to the perpendicular drawn from H to GB.

yashikaaggarwal

Did you used "the triangle drawn from same base and equal height have same area" Property here?

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In the figure above, ABCD and BEFG are squares and points G and H are 22 Jul 2020, 02:04
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1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80 (Given in the question statement).......(1)

DC-GF = 4 (from statement 2)

DC = 4+GF........(2)

From (1) and (2)

\(\frac{1}{2} (4+GF) *GF + \frac{1}{2} * (GF)^2 = 80\)

\((GF)^2 +2*GF - 80 = 0\)

We can stop here. Since the product of the roots are negative, we will get 1 positive root and 1 negative root. (As i mentioned in my previous solution, we're interested only in positive root)

Anyways,

\((GF)^2 +10GF -8GF -80 = 0\)

(GF+10)(GF-8) = 0

GF = -10 or 8

Side can't be negative. So, we discard -10.

GF= 8 and DC = 4+8=12


SiddharthR
nick1816
We need the length of the sides of the square.

Area of DBH = 1/2* (DC) * (GB) + 1/2 * (GF) * (GB) = 80

Area of DBH = 1/2* (DC) * (GF) + 1/2 * (GF) * (GF) = 80...... (1) [since GF = GB]


Statement 1. We know GF or GB. We can find DC using equation 1.

Sufficient

Statement 2- We know DC - GF. We have 2 equations and 2 variables. We can find DC and GF. [We are interested in +ve values]

Sufficient.

.

Could you solve statement II. I’m unable to see how that is sufficient

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In the figure above, ABCD and BEFG are squares and points G and H are 23 Jul 2020, 01:28
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I saw this question on my online GMAT, although I saw this question after finishing the exam. I'm assuming whoever posted this question violated GMAC policy.

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Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
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