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# In the diagram below, PQR is a triangle, right-angled at Q. A point S

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Joined: 02 Sep 2009
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In the diagram below, PQR is a triangle, right-angled at Q. A point S  [#permalink]

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23 Jul 2017, 23:22
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Difficulty:

75% (hard)

Question Stats:

55% (02:37) correct 45% (02:42) wrong based on 72 sessions

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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.

Attachment:

2017-07-24_1021.png [ 5.52 KiB | Viewed 1356 times ]

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Re: In the diagram below, PQR is a triangle, right-angled at Q. A point S  [#permalink]

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24 Jul 2017, 01:52
1
Bunuel wrote:
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.

Attachment:
2017-07-24_1021.png

(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.

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Re: In the diagram below, PQR is a triangle, right-angled at Q. A point S  [#permalink]

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06 Aug 2017, 00:40
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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In the diagram below, PQR is a triangle, right-angled at Q. A point S  [#permalink]

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05 Dec 2017, 12:07
1

Statement 1: PQ = 24, QR = 7, no idea of how the shape of SAQB => InSufficient
Statement 2: SQAB is square

Let the side 'x' be length of the side of the square.

Since, Triangle PAS, Triangle PQR are similiar.
By similarity property,
$$PQ/QR = (PQ - x)/x$$ => no info about PQ and QR to find value of x => InSufficient

Combining, 1+2, we can find the value of x as we know PQ and QR and hence area of square, which can be expressed as fraction of area of triangle PQR
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Re: In the diagram below, PQR is a triangle, right-angled at Q. A point S  [#permalink]

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07 Dec 2017, 02:05
sisirkant wrote:
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

Posted from my mobile device

Hi

May I know how you concluded on the basis of 'SAQB is a square' that triangle PQR is 45-90-45 triangle?
Re: In the diagram below, PQR is a triangle, right-angled at Q. A point S &nbs [#permalink] 07 Dec 2017, 02:05
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