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10 Aug 2017, 01:16
Kudos
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
93% (01:04) correct
7%
(01:35)
wrong
based on 56
sessions
History
Date
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Not Attempted Yet
Attachment:
8.10.png [ 4.39 KiB | Viewed 3252 times ]
What is the area of a trapezoid ABCD as above figure?
A. 84
B. 96
C. 104
D. 120
E. 150
Sirakri
GMAT 2: 700 Q48 V38
Posts: 104
10 Aug 2017, 01:33
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MathRevolution
Consider perpendicular from a point E to B.
EC = DC-DE = 16-10=6
Using Pythagorean's Theorem:
BE^2 = BC^2-EC^2 = 100-36 = 64
BE = 8
Area of trapezoid= BE*(AB+CD)*1/2 = 8*13 = 104.
C.
10 Aug 2017, 03:41
Kudos
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A right angled triangle formed if the rectangle is completed(which is otherwise missing from the picture)
The right angled triangle has hypotenuse - 10 and height 6(16-10).
Therefore the base can be got by \(\sqrt{10^2 - 6^2} = 8\)
The area of the triangle is \(\frac{1}{2}\)*Base*Height = \(\frac{1}{2}*8*6 = 24\)
The rectangle has area of \(16*8 = 128\)
Area of trapezoid = Area(Rectangle) - Area(Triangle) \(= 128 - 24 = 104\)(Option C)
The right angled triangle has hypotenuse - 10 and height 6(16-10).
Therefore the base can be got by \(\sqrt{10^2 - 6^2} = 8\)
The area of the triangle is \(\frac{1}{2}\)*Base*Height = \(\frac{1}{2}*8*6 = 24\)
The rectangle has area of \(16*8 = 128\)
Area of trapezoid = Area(Rectangle) - Area(Triangle) \(= 128 - 24 = 104\)(Option C)
