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, points P, Q, R, S, and T lie on a line. Q is the
[#permalink]
13 Dec 2023, 01:15
13 Dec 2023, 01:15
Expert Reply
00:00
Question Stats:
90% (02:15) correct
9% (04:18) wrong based on 11 sessions
Hide Show timer Statistics
Attachment:
GRE circle (17).jpg [ 13.29 KiB | Viewed 1464 times ]
In the figure above, points P, Q, R, S, and T lie on a line. Q is the center of the smaller circle and R is the center of the larger circle. P is the point of contact of the two circles, S is a point on the smaller circle, and T is a point on the larger circle. If RS = 1 and ST = 4, what is the area of the region inside the larger circle but outside the smaller circle?
Show: ::
50.24
Re: In the figure above, points P, Q, R, S, and T lie on a line. Q is the
[#permalink]
13 Jan 2024, 06:25
13 Jan 2024, 06:25
Expert Reply
Post A Detailed Correct Solution For these FRESH Quant & Verbal Questions And Get A Kudos
Re: In the figure above, points P, Q, R, S, and T lie on a line. Q is the
[#permalink]
09 Jun 2024, 12:29
09 Jun 2024, 12:29
Expert Reply
Required area = (Area of bigger circle with center R)– (Area of smaller circle with center Q
\(= \pi \times PR^2 - \pi \times PQ^2\)
We know that:
RS = 1and ST = 4
\(>RT=1+4=5\)
\(>PR=RT=5\)
Also, we have:
\(PS = PT– ST\)
\(= 2 * PR–ST\)
\(= 10– 4=6\)
\(>PQ+QS=6\)
\(>2 * PQ=6\)
\(>PQ=3\)
Thus, we have:
PR = 5 and PQ=3
Thus, required area
\(= \pi \times PR^2 - \pi \times PQ^2\)
\(= \pi (5^2-3^2)\)
\(=16 \pi\)
\(= \approx 50.24\)
\(= \pi \times PR^2 - \pi \times PQ^2\)
We know that:
RS = 1and ST = 4
\(>RT=1+4=5\)
\(>PR=RT=5\)
Also, we have:
\(PS = PT– ST\)
\(= 2 * PR–ST\)
\(= 10– 4=6\)
\(>PQ+QS=6\)
\(>2 * PQ=6\)
\(>PQ=3\)
Thus, we have:
PR = 5 and PQ=3
Thus, required area
\(= \pi \times PR^2 - \pi \times PQ^2\)
\(= \pi (5^2-3^2)\)
\(=16 \pi\)
\(= \approx 50.24\)
