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09 Jun 2021, 01:30
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C
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Difficulty:
25%
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Question Stats:
84% (02:29) correct
16%
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wrong
based on 63
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In rectangle ABDE above, line segment FC is parallel to side AB. If DC = 3, CG = 8, and GB = 10, what is the area of ABDE ?
A. 72
B. 96
C. 108
D. 120
E. 144
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asishron29181
Joined: 04 May 2020
Last visit: 04 Nov 2022
Posts: 185
Given Kudos: 68
Location: India
Schools: Great Lakes (I)
GMAT 1: 610 Q47 V27
WE:Programming (Computer Software)
09 Jun 2021, 04:57
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FC is parallel to side AB. If DC = 3, CG = 8, and GB = 10
FC parallel to AB and parallel to ED since it is a rectangle
DC =3 , CG = 8 , GB = 10
BC = 6 (Pythagoras theorem)
In triangle BED
Using Apollonius's theorem
GC/ED = BC/BD
ED = 12
Now area of ABED = b * h
12 * 9 = 108
IMO C
FC parallel to AB and parallel to ED since it is a rectangle
DC =3 , CG = 8 , GB = 10
BC = 6 (Pythagoras theorem)
In triangle BED
Using Apollonius's theorem
GC/ED = BC/BD
ED = 12
Now area of ABED = b * h
12 * 9 = 108
IMO C
09 Jun 2021, 05:43
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Solution:
Since we are asked the area of the rectangle, we need its length and breadth.
So, if we can somehow find out the value of BD and ED, we can get the answer.
For BD,
We have \(GC=8\) and \(BG=10\). We can use the Pythagoras Theorem on \(\triangle BGC\) to get the value of BC.
\(⇒BG^2=GC^2+BC^2\)
\(⇒100=64+BC^2\)
\(⇒BC^2=36\)
\(⇒BC=6\)
Therefore,\( BD=3+BC=3+6=9. \)
For ED,
We can prove that \(\triangle BCG\) ~ \(\triangle BED\)
as \(\angle C = \angle D\) (\(90^o\) each)
\(\angle B = \angle B\) (common)
\(\angle G = \angle E\) (alternate because GC || ED)
Hence \(\triangle BCG\) ~ \(\triangle BED\)
Now we can say:
\(\frac{GC}{ED}=\frac{BC}{BD}\)
\(⇒\frac{8}{ED}=\frac{6}{9}\)
\(⇒ ED=\frac{9*8}{6}\)
\(⇒ ED=12\)
Area of the reactangle ABED = \(BD*ED=9*12=108\)
HEnce the right answer is Option C.
Since we are asked the area of the rectangle, we need its length and breadth.
So, if we can somehow find out the value of BD and ED, we can get the answer.
For BD,
We have \(GC=8\) and \(BG=10\). We can use the Pythagoras Theorem on \(\triangle BGC\) to get the value of BC.
\(⇒BG^2=GC^2+BC^2\)
\(⇒100=64+BC^2\)
\(⇒BC^2=36\)
\(⇒BC=6\)
Therefore,\( BD=3+BC=3+6=9. \)
For ED,
We can prove that \(\triangle BCG\) ~ \(\triangle BED\)
as \(\angle C = \angle D\) (\(90^o\) each)
\(\angle B = \angle B\) (common)
\(\angle G = \angle E\) (alternate because GC || ED)
Hence \(\triangle BCG\) ~ \(\triangle BED\)
Now we can say:
\(\frac{GC}{ED}=\frac{BC}{BD}\)
\(⇒\frac{8}{ED}=\frac{6}{9}\)
\(⇒ ED=\frac{9*8}{6}\)
\(⇒ ED=12\)
Area of the reactangle ABED = \(BD*ED=9*12=108\)
HEnce the right answer is Option C.
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