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655-705 (Hard)|   Geometry|      
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Bunuel
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For the rectangular shown above, if BR = BQ = 5*2^(1/2) and 11 Nov 2021, 07:15
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55% (hard)

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68% (02:46) correct 32% (02:46) wrong based on 66 sessions

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For the rectangular shown above, if \(BR=BQ=5\sqrt{2}\) and \(AR=PB=CQ=\sqrt{119}\), what is the area of the triangle PQR?

A. 30
B. 45
C. 60
D. 75
E. 90

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For the rectangular shown above, if BR = BQ = 5*2^(1/2) and 11 Nov 2021, 20:11
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Bunuel

For the rectangular shown above, if \(BR=BQ=5\sqrt{2}\) and \(AR=PB=CQ=\sqrt{119}\), what is the area of the triangle PQR?

A. 30
B. 45
C. 60
D. 75
E. 90
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Take triangle BQR: It is an isosceles right angled triangle, so hypotenuse QR=\(5\sqrt{2}*\sqrt{2}=10\)

Take triangle BQP: The sides are \(\sqrt{119}\) and \(5\sqrt{2}\), so hypotenuse: \(PQ^2=(5\sqrt{2})^2+\sqrt{119}^2=50+119=169……….PQ=13\)

Take triangle PQR
PQ=PR=13, so it is an isosceles triangle.
Drop a perpendicular on QR at T from vertex P.
Now take triangle PQD, a right angled triangle : Side TP=\(\sqrt{13^2-5^2}=\sqrt{144}=12\)

Area of PQR = \(\frac{1}{2}*QR*TP=\frac{1}{2}*10*12=60\)



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For the rectangular shown above, if BR = BQ = 5*2^(1/2) and 11 Dec 2023, 00:56
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