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07 Mar 2016, 16:55
Kudos
Bookmarks
C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
45%
(medium)
Question Stats:
71% (02:45) correct
29%
(03:01)
wrong
based on 167
sessions
History
Date
Time
Result
Not Attempted Yet
Attachment:
trapezoid ABCD.jpg [ 4.86 KiB | Viewed 5567 times ]
A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7
* A solution will be posted in two days.
09 Mar 2016, 03:07
Kudos
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MathRevolution
HI..
Although it should be given that O is the center, I take it as center..
Join AO.. we have an isosceles triangle with equal sides as radius or 4 and the third side as 6
so throw an altitude on AB from O as shown..
the ALTITUDE= \(\sqrt{4^2-3^2}=[m]\sqrt{7}\)[/m]..
now we know two side as 6 and 8 and ALTITUDE as \(\sqrt{7}\)..
Area= \(\frac{(6+8)}{2}*\sqrt{7}\)=\(7\sqrt{7}\)
C
Attachments
New Microsoft Office Word Document (3).docx [10.63 KiB]
Downloaded 144 times
General Discussion
11 Mar 2016, 05:51
Kudos
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If AB=6, DC=8, what is the area of the trapezoid ABCD?
A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7
->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.
A. √7
B. 7
C. 7√7
D. 5√7
E. 6√7
->Since AB=6, HC=1 and OH=3. Then, BH=√(4^2-3^2 )=√7. So, the area of the trapezoid ABCD is (1/2)(6+8)√7=7√7.
Thus, C is the answer.
