Bunuel wrote:
The area of the circle above, with center O, is 144π. If angle OZY measures 30 degrees, what is the area of the shaded region?
A. \(48π−36\sqrt{3}\)
B. \(48π−18\sqrt{3}\)
C. \(54π−18\sqrt{3}\)
D. \(72π−18\sqrt{3}\)
E. \(72π−27\sqrt{3}\)
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nishantt7How can we do it in case we do not want to apply trigonometry?
nishantt7, that's a good question. I didn't use trigonometry. The process and the math aren't too hard; that said, explaining clearly makes both process and math
look hard.
I'm going to preempt the peanut gallery because I've been bothered by this issue for a year and remained silent: yes, this explanation is lengthy and thorough. Yes, I could have written it in about five cryptic sentences. Yes, I expect a few of you will do exactly that. Last time I checked, however, the point of collective exchange of information in pursuit of learning was neither to show off nor to intimidate nor to condescend. A five-line answer might assist slightly those who already know the method. Given the peril of impenetrability in compressed details, however, such compression risks confusing those who don't know the method. Were I not already a teacher, and were I nervous about others' expertise and my relative lack thereof, I would be hesitant to post, perhaps even to ask questions. I have the impression that the degree of camaraderie among members on this forum has declined noticeably. That's a shame. Onward.
nishantt7, the
area of the shaded portion equals the area of
sector ZOY minus the area of
triangle YOZ. That's not immediately apparent. Start sketching, and it will be.
Sketch the figure, then take these steps
1. Calculate radius from area, where \(A=\pi r^2\), \(144 \pi = r^2\), \(r = 12\)
2. Write in what you know.
∠OZY = 30, r = 12, so OX and
OZ = 12What now? The answer choices all contain \(\sqrt{3}\). GMAT loves to test special triangles. \(\sqrt{3}\) should alert you immediately that, in all likelihood, you are looking for or will need a 30-60-90 triangle. You already have a 30° angle; that should tip you off to draw a triangle of some kind.
3. Draw the triangle: Draw a line from center O to Y to create
OYBecause
OY is a radius, label it
12OY =
OZ = 12. Indicate that equality with hatch marks
|| You have an isosceles triangle with one angle of 30°. Angles opposite equal sides of isosceles triangle are equal. For ∠OYZ, write in
30, and indicate that
∠OYZ = ∠OZYWhat is the measure of third angle ∠YOZ? 30 + 30 + ∠YOZ = 180, so ∠YOZ = 120. I noted it.
4. With two 30° angles, and a large angle of 120°, bisect the 120° angle to get two 30-60-90 triangles.
Draw a perpendicular bisector from center O to YZ to create
AY. Now
∠YOA = ∠ZOA = 60 5. Find the area of triangle YOZ:
The two 30-60-90 triangles have side ratios \(x : x \sqrt{3} : 2x\)
OZ, the hypotenuse opposite the 90° angle, corresponds with 2x. Use that to figure out other two sides' length.
OZ = 12. So
OA = (1)x = 6, and
AZ = x \(\sqrt{3}\) = 6\(\sqrt{3}\)
CAREFUL here. Either find the area of one 30-60-90 triangle and double, or note that YZ = (AZ) * 2 =\(12 \sqrt{3}\) (which is the base of triangle YOZ)
I chose former. Area of triangle AOZ = \((b * h)/2\), or \((6 * 6\sqrt{3}) / 2\) = \(18 \sqrt{3}\). Double it and
area of triangle
YOZ = \(36\sqrt{3}\)
6. Find area of sector ZOY
Its central angle is 120°.
Sector area = \(\frac{120}{360} * 144\pi\) =
\(48 \pi\)7. Shaded area therefore equals \(48 \pi\) - \(36 \sqrt{3}\)
Hope it helps.
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