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Radius of the semi-circle shown above is 1 . What is the area of the s
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11 Jan 2025, 00:32
11 Jan 2025, 00:32
Expert Reply
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50% (00:59) correct
50% (04:24) wrong based on 4 sessions
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GRE geometry semi circle practice question.png [ 19.18 KiB | Viewed 591 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) $\(\frac{\pi}{4}+\frac{1}{4}\)$
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Radius of the semi-circle shown above is 1 . What is the area of the s
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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
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