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11 Dec 2017, 23:04
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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:
25%
(medium)
Question Stats:
76% (02:02) correct
24%
(01:55)
wrong
based on 118
sessions
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Not Attempted Yet
In the figure above, if PQ is parallel to OR, what is the area of quadrilateral PQRO? Note: typo in the diagram, the coordinates of R are (5, 0).
(A) 9
(B) 14
(C) 18
(D) 36
(E) cannot be determined
Attachment:
2017-12-12_1001.png [ 7.97 KiB | Viewed 3548 times ]
rahul16singh28
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11 Dec 2017, 23:14
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Bunuel
Area should be 4*4 + 0.5*4*1 = 18
13 Dec 2017, 20:45
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Bunuel
We have a trapezoid with height OP and bases OR and PQ
Length of PQ
∆ OPQ is right isosceles with equal sides OP and PQ
--The angle at P is 90°
PQ || OR. The two segments are cut by traversal OP. Same-side interior angles sum to 180.
∠ POR = 90, hence the angle at P is also 90
-- ∠ PQO = 45. Parallel lines are cut by the traversal OQ, and ∠ QOR = 45. As an alternate interior angle, ∠ PQO also = 45°
-- ∠ POQ is 45: the right angle at vertex O has one angle marked as 45°; 90 - 45 = 45 = ∠ POQ
--OP = PQ: sides opposite equal angles of a triangle are equal
--OP = 4 = PQ, which is one base of the trapezoid
Height, other base, and area
The height of the trapezoid is OP = 4 (Height is perpendicular to base; OP is perpendicular to OR and PQ.)
The other base is OR = 5
Area of trapezoid: \(\frac{(base_1 + base_2)}{2} * h\)
\((\frac{4 + 5}{2}) * 4 = (4.5 * 4)=\\
18\)
Answer C
