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08 Jun 2016, 10:06
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
73% (02:16) correct
27%
(02:32)
wrong
based on 56
sessions
History
Date
Time
Result
Not Attempted Yet
In a rectangular coordinate system, what is the square root of the area of a trapezoid whose vertices have the coordinates (2, -2), (2, 3), (20, 2), (20, -2)?
A. 7.5
B. 9
C. 10.22
D. 12.25
E. 14
A. 7.5
B. 9
C. 10.22
D. 12.25
E. 14
06 May 2020, 05:59
Kudos
Bookmarks
Bunuel
Solution
- • Based on the co-ordinates given in the question, let’s draw the trapezoid.
• Now, we can see that AD is parallel to BC and height of the trapezoid is equal to AB.
• Thus, area of the trapezoid = \(\frac{1}{2}\)*(sum of parallel sides)*height \(= \frac{1}{2}*(AD +BC)*AB = \frac{1}{2}*(5 + 4) *18 = 9^2\)
• Therefore, the square root of the area = 9
General Discussion
08 Jun 2016, 10:10
Kudos
Bookmarks
B.
we can form rectangle and then minus the area of triangle
18*5 -9 =81 sq rt of which is 9
we can form rectangle and then minus the area of triangle
18*5 -9 =81 sq rt of which is 9
