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23 Jul 2017, 23:22
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C
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Difficulty:
95%
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Question Stats:
45% (02:30) correct
55%
(02:29)
wrong
based on 154
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In the diagram below, PQR is a triangle, right-angled at Q. A point S is chosen on the hypotenuse and two perpendiculars, one each on to PQ and QR, are drawn from S such that they meet PQ and QR at A and B, respectively. What is the area of SAQB as a fraction of the area of triangle PQR?
(1) PQ = 24 and QR = 7.
(2) SAQB is a square.
2017-07-24_1021.png [ 5.52 KiB | Viewed 5700 times ]
(1) PQ = 24 and QR = 7.
(2) SAQB is a square.
Attachment:
2017-07-24_1021.png [ 5.52 KiB | Viewed 5700 times ]
24 Jul 2017, 01:52
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Bunuel
(1) We could calculate the area of PQR. However, we can't calculate the area of SAQB. Insufficient.
(2) Since SAQB is a square, we could calculate the length of SA=SB based on the length of QP and QR.
We have \(S_{PQR}=\frac{1}{2} * PQ * QR\)
We also have \(S_{PQR} = S_{PQS} + S_{SQR}\\
= \frac{1}{2}*PQ * SA + \frac{1}{2} * QR * SB \\
= \frac{1}{2} * SA * (PQ + QR)\)
Hence \(\frac{1}{2} * PQ * QR = \frac{1}{2} * SA * (PQ + QR) \implies SA = \frac{PQ * QR}{PQ + QR}\)
However, we can't know the length of PQ and QR. Hence insufficient.
Combine (1) and (2):
From (1) we could calculate the area of PQR
From (2) we could calculate SA=SB so we could calculate the area of SAQB.
Hence, we could calculate "the area of SAQB as a fraction of the area of triangle PQR". Sufficient.
The answer is C.
06 Aug 2017, 00:40
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Why the answer is not B? If saqb is a square than pqr is 45 90 45 angled. In that case area of saqb is half the area of pqr. Am I missing anything here? Please explain. Thanks
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