GRE Prep Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
In the figure above, PQRS is a rectangle. The length of QS is 13 and Q
[#permalink]
28 Sep 2023, 01:12
28 Sep 2023, 01:12
1
Expert Reply
00:00
Question Stats:
81% (02:04) correct
18% (01:26) wrong based on 11 sessions
Hide Show timer Statistics
Attachment:
GRE polygon (4).jpg [ 9.89 KiB | Viewed 1149 times ]
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.
Part of the project: The Butler-GRE Daily New Quant and Verbal Questions to Practice (2023) - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Free Materials for the GRE General Exam - Where to get it!!
GRE Geometry Formulas
GRE - Math Book
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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
Moderators:
