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In the figure, ABCD and PQRS are rectangles inscribed in the
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19 Mar 2019, 09:42
19 Mar 2019, 09:42
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87% (02:19) correct
12% (01:27) wrong based on 8 sessions
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#GREpracticequestion In the figure, ABCD and PQRS are rectangles.jpg [ 27.01 KiB | Viewed 1836 times ]
In the figure, ABCD and PQRS are rectangles inscribed in the circle shown in the figure. If AB = 5, AD = 3, and QR = 4, then what is the value of l ?
(A) 3
(B) 4
(C) 5
(D) \(\sqrt{15}\)
(E) \(3 \sqrt{2}\)
Re: In the figure, ABCD and PQRS are rectangles inscribed in the
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13 Apr 2020, 10:30
13 Apr 2020, 10:30
Expert Reply
The main funda which we need to know in order to solve this question is:
"If a right triangle is inscribed in a circle, then its hypotenuse is a diameter of the circle"
There are two rectangles inscribed in the circle so the diagonals of each rectangle will be equal to the diameter of the circle.
Deduce the diagonal value using Pythagoras theorem for triangle ABD and use the same value for hypotenuse of triangle PQR.
=> PR = BD = \(\sqrt{{3^2} + {5^2}}\)= \(\sqrt{34}\)
Lets take triangle PQR, \({PR^2}\)= \({PQ^2}\) + \({QR^2}\)
=> 34 = \({L^2}\) + 16
=> \({L^2}\) = 34 - 16 = 18
=> L = \(\sqrt{18}\)= \(\sqrt{3*2*3}\) = \(3\sqrt{2}\)
"If a right triangle is inscribed in a circle, then its hypotenuse is a diameter of the circle"
There are two rectangles inscribed in the circle so the diagonals of each rectangle will be equal to the diameter of the circle.
Deduce the diagonal value using Pythagoras theorem for triangle ABD and use the same value for hypotenuse of triangle PQR.
=> PR = BD = \(\sqrt{{3^2} + {5^2}}\)= \(\sqrt{34}\)
Lets take triangle PQR, \({PR^2}\)= \({PQ^2}\) + \({QR^2}\)
=> 34 = \({L^2}\) + 16
=> \({L^2}\) = 34 - 16 = 18
=> L = \(\sqrt{18}\)= \(\sqrt{3*2*3}\) = \(3\sqrt{2}\)
gmatclubot
Re: In the figure, ABCD and PQRS are rectangles inscribed in the [#permalink]
13 Apr 2020, 10:30
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