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Square ABCD has area 50. The points P, Q, R, and S are the midpoints o
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07 Apr 2022, 02:45
07 Apr 2022, 02:45
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Square ABCD has area 50. The points P, Q, R, and S are the midpoints of \(\overline{AB}, \overline{BC}, \overline{DC},\) and \(\overline{AD}\) respectively. What is the area of square PQRS?
A. 12.5
B. 25
C. \(20 \sqrt{2}\)
D. \(25 \sqrt{2}\)
E. 37.5
Part of the project: GRE Quantitative Reasoning Daily Challenge - (2022) EDITION
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GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
GRE Math Essentials for a TOP Quant Score - Q170!!! (2021)
GRE Geometry Formulas for a Q170
GRE - Math Book
Square ABCD has area 50. The points P, Q, R, and S are the midpoints o
[#permalink]
07 Apr 2022, 19:40
07 Apr 2022, 19:40
1
Let's take the side of the square \(= x\)
\(x^2 = 50\)
\(x = 5\sqrt{2} = AB = BC = CD = AD\)
Now in triangle \(APS, AP\) & \(AS =\frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}\)
So according to the Pythagoras theorem, \(PS = \frac{5}{\sqrt{2}} * \sqrt{2} = 5\)
Area of \(PQRS = 25\)
Answer B
\(x^2 = 50\)
\(x = 5\sqrt{2} = AB = BC = CD = AD\)
Now in triangle \(APS, AP\) & \(AS =\frac{5\sqrt{2}}{2} = \frac{5}{\sqrt{2}}\)
So according to the Pythagoras theorem, \(PS = \frac{5}{\sqrt{2}} * \sqrt{2} = 5\)
Area of \(PQRS = 25\)
Answer B
gmatclubot
Square ABCD has area 50. The points P, Q, R, and S are the midpoints o [#permalink]
07 Apr 2022, 19:40
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