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Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q [#permalink]
All, please check my work but here is how I solved...

Draw line parallel to RS from Q to opposite side, calling the new point T. New angle T is 90 because R is 90 and therefore S is 90. Supplementary angle above T is also 90.

This splits parallelogram into rectangle and a triangle above it.

Since original parallelogram has R=90, S=90 and Q=135, P must be 45.

If P is 45 and T's supplementary angle is 90, angle above Q is also 45. This creates a 45-45-90 triangle, TPQ. Legs are TP and QT. Hypotenuse is PQ.

Since RS=5, then QT=5 (since TS perpendicular to RS). Since QT opposite P, which is 45, TP is also 5 since its opposite a 45 degree angle as well.

Applying Pythagorean Theorem, 5^2 + 5^2 = PQ^2. PQ is 5 sq root 2. Therefore QR is same length since PQ=QR.

Therefore, to get area of rectangle we multiply RS (5) * QR (5 sq root 2) for 25 sq root 2.

Add that to area of triangle (5*5)/2

(25 sqroot 2 +25)/2

This factors to [25(1 + sqroot2)]/2

Option C.

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Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q [#permalink]
Expert Reply
Bunuel wrote:
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. \(25 + 25\sqrt{2}\)




To draw this quadrilateral, first, we are given angles R and Q. If we focus on point R, R is connected by two lengths of 5 and they create an angle of 90 degrees. Similarly, Q is connected by two lengths of 5 and they create an angle of 135 degrees. Then we connect PS and we have our graph.

Notice that triangle QRS is a right triangle, if we connect QS it splits angle PQR into a 90 and 45-degree angle, so triangle PQS is also a right triangle.

\(QS = 5\sqrt{2}\), the areas are \(\frac{5^2}{2}\) and \(\frac{25\sqrt{2}}{2}\).

Ans: C
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Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q [#permalink]
Expert Reply
Bunuel wrote:
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. \(25 + 25\sqrt{2}\)




The explanations already in the thread are a-okay. Just want to point out that once you draw the figure, you can Ballpark the actual area. The bottom part is almost a full 5x5 square. That's 25. Then we have to account for the top part, which looks like maybe about a quarter the size of the square. Okay, 25+6 = 31.

A. \(\frac{25 + 1.4}{2} = \frac{26}{2} = 13\)
Note even close.

B. \(\frac{1 + 25(1.4)}{2}=\frac{1 + 35}{2}=\frac{36}{2}=18\)
Not even close.

C. \(\frac{25(1 + 1.4)}{2}=\frac{25(2.4)}{2} = \frac{60}{2} = 30\)
Keep it.

D. \(25 + 1.4 = 26\)
Close but clearly too small.

E. \(25 + 25(1.4) = 25+35 = 60\)
Not even close.

Answer choice C.
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Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q [#permalink]
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Re: Quadrilateral PQRS has PQ = QR = RS = 5, If angle R = 90° and angle Q [#permalink]
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