The area of the circle above, with center O, is 144π . If angle OZY m

Answer is A. Please refer to the attached.

You correctly subtracted the 240 degree area (which is not part of the triangle or the shaded region), however, you subtracted only half of the triangle. The triangle area to be subtracted is OYZ, and not OKZ, as you did.

Therefore, answer should be 48\(\pi\) - 36 \(\sqrt{3}\), answer A

Thanks for pointing out

How 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 = 12

What 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 OY

Because OY is a radius, label it 12

OY = 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 = ∠OZY

What 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.
_________________

The upper half of the Circle has 1/2 the Total Area = 72(pi)

We can Subtract out the sector area measured by the Central Angle XOY, which is 2 times the Inscribed Angle at OZY of 30 degrees

(60)/(360) * Area of Circle = Sector Area XOY

(1/6) * (144 (pi)) = 24(pi)

Next, we can connect a radius from Point O to Point Y on the Circumference of the Circle.

This will create an Isosceles Triangle YOZ with the 2 equal sides = radius = 12

Concept: the Perpendicular Height drawn from the Vertex between the 2 Equal Sides of an Isosceles Triangle to the Non Equal Side is also————> Angle Bisector at the Vertex YOZ and a Perpendicular Bisector of the NON equal
Side YZ


Since Angle XOY = 60 degrees, Angle YOZ = 120 degrees.

Thus, the Height = Angle Bisector = Perpendicular Bisector = Median = Line of Symmetry for the Isosceles Triangle will Subdivide Isosceles Triangle YOZ into TWO 30-60-90 Right Triangles.

Using the Fact that the radius = hypotenuse of Each right Triangle = 12

The 2 Legs will be:

6 and 6 * sqrt(3)

2 * (1/2) * 6 * 6 * sqrt(3) = 36 * sqrt(3) = Area of Triangle YOZ


Shaded Area = (Half the Circle Area) - (Sector Area YOX) - (Area of Isosceles Triangle YOZ)

72(pi) - 24(pi) - 36 * sqrt(3) =

48(pi) - 36 * sqrt(3)

-A-

EDIT: wow, generis put up a much better explanation. See above post from 3 years ago.

Posted from my mobile device

I did a slightly different method, which I believe is more straightforward. Based on the properties of circles, when one length of a triangle inscribed within circle is the diameter, and the vertice is along the circumference of circle, the angle formed is 90. (Pardon for the wrong terms used if any, I do not have the technical vocab for the geometric properties)

Thus XZ is the hypothenuse of triangle XYZ. Knowing that r =12, XZ = 24, and XY = 12, YZ = 12sqrt3. Area of right triangle = 72sqrt 3.

Moving onto the segment formed after discounting equilateral triangle XOY, area = 1/6 * 144pi (we know it is equilateral because OX and OY = radius = 12 = XY) - area of equilateral triangle XOY

To find area of triangle XOY = 1/2 * 12 * 6sqrt 3 (using 30-60-90 triangle)

So area of segment = 24pi - 36sqrt3
Area of right triangle = 1/2 * 12 * 12sqrt3 = 72sqrt 3
Area of shaded = 72pi - 72sqrt3 - (24pi - 36sqrt3) = 48pi - 36sqrt 3 (A)