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.
Radius of the semi-circle shown above is 1 . What is the area of the s
[#permalink]
11 Jan 2025, 00:32
11 Jan 2025, 00:32
Expert Reply
00:00
Question Stats:
50% (01:23) correct
50% (02:55) wrong based on 2 sessions
Hide Show timer Statistics
Attachment:
GRE geometry semi circle practice question.png [ 19.18 KiB | Viewed 186 times ]
Radius of the semi-circle shown above is 1 . What is the area of the shaded region?
(A) $\(\frac{\pi}{6}-\frac{1}{4}\)$
(B) $\(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)$
(C) $\(\frac{\pi}{4}-\frac{\sqrt{3}}{4}\)$
(D) $\(\frac{\pi}{6}+\frac{1}{4}\)$
(E) $\(\frac{\pi}{4}+\frac{1}{4}\)$
- Part of the project: GRE Quant & Verbal ADVANCED Daily Challenge 2024 Edition - Gain 20 Kudos & Get FREE Access to GRE Prep Club TESTS
- Also replying to the unanswered questions
Radius of the semi-circle shown above is 1 . What is the area of the s
[#permalink]
19 Feb 2025, 09:22
19 Feb 2025, 09:22
Expert Reply
OFFICIAL EXPLANATION
As the sum of the angles on a straight line is 180 degrees, we get $\(\angle \mathrm{AOB}=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)\)$ $\(=180^{\circ}-120^{\circ}=60^{\circ}\)$
We know that the Area of sector having radius ' r ' \& angle $\theta$ degrees at the centre is $\(\frac{\theta}{360} \times \pi \mathrm{r}^2\)$ so, the Area of the sector OACB having radius 1 and angle 60 degrees at the centre is $\(=\frac{60}{360} \times \pi \times(1)^2=\frac{\pi}{6}\)$
Next the area of the equilateral triangle OAB is $\(\frac{\sqrt{3}}{4} \times { Side }^2=\frac{\sqrt{3}}{4} \times(1)^2=\frac{\sqrt{3}}{4}\)$
So, the area of the shaded region is = Area of sector $\(\mathrm{OACB}\)-$ Area of equilateral triangle $\(\mathrm{OAB}=\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)$
Hence the answer is (B).
As the sum of the angles on a straight line is 180 degrees, we get $\(\angle \mathrm{AOB}=180^{\circ}-\left(90^{\circ}+30^{\circ}\right)\)$ $\(=180^{\circ}-120^{\circ}=60^{\circ}\)$
We know that the Area of sector having radius ' r ' \& angle $\theta$ degrees at the centre is $\(\frac{\theta}{360} \times \pi \mathrm{r}^2\)$ so, the Area of the sector OACB having radius 1 and angle 60 degrees at the centre is $\(=\frac{60}{360} \times \pi \times(1)^2=\frac{\pi}{6}\)$
Next the area of the equilateral triangle OAB is $\(\frac{\sqrt{3}}{4} \times { Side }^2=\frac{\sqrt{3}}{4} \times(1)^2=\frac{\sqrt{3}}{4}\)$
So, the area of the shaded region is = Area of sector $\(\mathrm{OACB}\)-$ Area of equilateral triangle $\(\mathrm{OAB}=\frac{\pi}{6}-\frac{\sqrt{3}}{4}\)$
Hence the answer is (B).
gmatclubot
Radius of the semi-circle shown above is 1 . What is the area of the s [#permalink]
19 Feb 2025, 09:22
Moderators:
