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11 Nov 2021, 07:15
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C
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Difficulty:
55%
(hard)
Question Stats:
70% (02:46) correct
30%
(02:44)
wrong
based on 64
sessions
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Not Attempted Yet
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
Attachment:
1.jpg [ 7.01 KiB | Viewed 2012 times ]
11 Nov 2021, 20:11
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Bunuel
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 = \(\frac{1}{2}*QR*TP=\frac{1}{2}*10*12=60\)
C
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EF002CA2-BFB7-4355-9128-652D465ADE24.jpeg [ 9.48 KiB | Viewed 1858 times ]
11 Dec 2023, 00:56
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