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GRE Prep Club Team Member
Joined: 20 Feb 2017
Posts: 2506
Given Kudos: 1053
GPA: 3.39
Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q
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29 Mar 2021, 05:30
29 Mar 2021, 05:30
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Question Stats:
71% (01:58) correct
28% (03:08) wrong based on 7 sessions
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Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q = 135° what is the area of the quadrilateral?
A. \(\frac{25 + \sqrt{2}}{2}\)
B. \(\frac{1 + 25\sqrt{2}}{2}\)
C. \(\frac{25(1 + \sqrt{2})}{2}\)
D. \(25 + \sqrt{2}\)
E. \(25 + 25\sqrt{2}\)
A. \(\frac{25 + \sqrt{2}}{2}\)
B. \(\frac{1 + 25\sqrt{2}}{2}\)
C. \(\frac{25(1 + \sqrt{2})}{2}\)
D. \(25 + \sqrt{2}\)
E. \(25 + 25\sqrt{2}\)
Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q
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29 Mar 2021, 23:03
29 Mar 2021, 23:03
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GeminiHeat wrote:
Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q = 135° what is the area of the quadrilateral?
I can not share the figure with you since I am a new member in this forum and by its rule, I need to post at least 10 times before I can share. So, let me explain this verbally. Draw a figure. Join SP. Assume two triangles - triangle QRS and PQS.
In triangle QRS, angle R = 90, and QR = RS = 5. So \(area of triangle QRS = (1/2)*5*5\). Also, notice angle SQR = angle RSQ = 45, since it is a right isosceles triangle. Moreover, QS = 5\sqrt{2} by applying Pythagoras ratios in such a triangle.
In triangle PQS, angle PQS = 90, since 45° angle SQR is excluded from 135° angle PQR. Now PQ = 5, and QS = 5\sqrt{2}. So \(area of triangle PQS = (1/2)* 5 * 5\sqrt{2}\)
Add the areas of both triangle QRS and PQS.
C is the correct answer.
gmatclubot
Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q [#permalink]
29 Mar 2021, 23:03
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