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In the figure above, PQRS is a rectangle. The length of QS is 13 and Q
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28 Sep 2023, 01:12
28 Sep 2023, 01:12
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81% (02:04) correct
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In the figure above, PQRS is a rectangle. The length of QS is 13 and QR : PQ :: 5 : 12.
Quantity A |
Quantity B |
Area of the semi-circular region with center O and diameter QR |
\(3 \pi\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
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Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
Re: In the figure above, PQRS is a rectangle. The length of QS is 13 and Q
[#permalink]
08 Oct 2023, 15:08
08 Oct 2023, 15:08
Expert Reply
QR : PQ :: 5 : 12
=> Let QR = 5x and PQ = 12x, where x is a constant of proportionality.
We also know that QS = 13.
Thus, applying Pythagoras theorem in triangle SPQ, we have
PQ^2 + PS^2 = QS^2
=> PQ^2 + QR^2 = 13^2
=> (12x)^2 + (5x)^2 = 132
=> 144x^2 + 25x^2 = 169
=> x^2 = 1
=> x = 1 (x cannot be −1 since lengths must be positive)
=> QR = 5x = 5
=> Radius of the semi-circle =5/2
Thus, the area of the semi-circle\(=\frac{1}{2}× π ×(\frac{5}{2})^2
= 3.125π > 3π\)
=> Let QR = 5x and PQ = 12x, where x is a constant of proportionality.
We also know that QS = 13.
Thus, applying Pythagoras theorem in triangle SPQ, we have
PQ^2 + PS^2 = QS^2
=> PQ^2 + QR^2 = 13^2
=> (12x)^2 + (5x)^2 = 132
=> 144x^2 + 25x^2 = 169
=> x^2 = 1
=> x = 1 (x cannot be −1 since lengths must be positive)
=> QR = 5x = 5
=> Radius of the semi-circle =5/2
Thus, the area of the semi-circle\(=\frac{1}{2}× π ×(\frac{5}{2})^2
= 3.125π > 3π\)
gmatclubot
Re: In the figure above, PQRS is a rectangle. The length of QS is 13 and Q [#permalink]
08 Oct 2023, 15:08
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