What is the approximate area of the shaded region of the given figure?

Question

How do i find the area of the black shaded portion of the circle?

I noticed the 4 so i think that's the radius. The formula to find the area $$A=πr^2$$ so I thought of using that to find the area of the circle it was $$50.27$$ The next thing i thought of doing was subtracting the area of the triangle from the area of the circle? But looking at the problem I don't think I know how to do that? Have the steps I've taken been right so far? How would I solve this problem? Yes O is the center of the circle

Answer 1

The hypotenuse (longest side) of the triangle is 8 long. It is a right triangle, because all corners are on the circle and the longest side goes through the center of the circle. So the top angle of the triangle is 90°

Get both other sides of the triangle via sine and cosine:

$$a =8 \sin(30°)$$ $$b =8 \cos(30°)$$

The area of the triangle is now

$$A_{triangle}=\frac{ab}2$$ that's because it's a right triangle and a and b are the sides which are perpendicular to each other.

Answer 2

Use the Thales' theorem to infer that the top angle is right angle. Then calculating the area of the triangle should be straightforward.

Since the angle on the right is 30 degrees, the length of the side on the left is 4.

The side on the right is cos(30 degrees) = $\frac{\sqrt{3}}{2} 8 = 4 \sqrt{3} $. Therefore the are of the triangle is $\frac{1}{2} 4 \sqrt{3} \times 4 = 8 \sqrt{3}$.

Hence the are of the shaded region is $\frac{1}{2} \pi 4^2 - 8 \sqrt{3} = 8 (\pi - \sqrt{3}) \approx 11.28$.

Answer 3

Label the points of the diameter A, B and the third point of the triangle as E.

AO = EO = BO as they are radii. So by isosceles triangles. angle OEB = angle OBE = 30. So angle BOE = 120. So angle AOE = 60. But A0 = EO so angle OAE = angle AEO = 1/2(180 - 60) = 60. So triangle AOE is equilateral. so AE = 4.

And angle AEB = AEO + OEB = 60 + 30 = 90. So AEB is a right triangle. It has one side = 4, a hypotenuse = 8 so the third side is $\sqrt{64 - 16} = 4\sqrt 3$. The area of the triangle is $1/2 * 4 * 4\sqrt 3 = 8\sqrt 3$.

The area of the circle is $16*\pi$ so half the circle is $8*\pi$ so the shade region is $8\pi - 8\sqrt 3 = 8(\pi - \sqrt 3)$.

=======

A few things. Assuming O is the center of the circle so that bottom line is the diameter than the triangle is a right triangle (do you know why?) Assuming 30 is the degree of the angle the other angle is 60. and we have a 30-60-90 triangle. Assuming the radius if 4 (it's a really bad picture) then that diameter of the circle is 8 and the hypotenuse of the right triangle is eight.

That's enough.

1) the triangle is a right triangle. (I leave it to you to prove it. I'll give you a hint if you need it.)

2) the triangle is a 30-60-90 right triangle with hypotenuse of 8. I leave it to you to find the area of the triangle. (I'll give you hints if you need them.

3) the shaded area is 1/2 the area of the circle minus the area of the triangle.

====

Th: Any triangle inscribed in a circle where to points are opposite sites of a diameter is a right triangle. (And any right triangle inscribed in a circle will have its hypotenuse form the diameter.)

Proof. Let AB be the diameter of a circle. Let O be the center of the circle. Let E be the third point of a triangle. OA = OB = OE as they are radii so triangles AOE and EOB are isosceles. So angle AEO = angle BAE and angle BEO = angle ABE so angle AEB = AEO + BEO = BAE + ABE. So as AEB + BAE + ABE = 180 and as AEB = BAE + ABE => AEB = 90.

(And vice versa. If AEB is a right angle, we can construct a ray that splits AEB into two angles one equal to BAE and the other to ABE and the ray intersects AB at x. Then as base angles are equal Ax = Bx = Ex so A, B, E lie in a circle and as x is colinear to A and B, AB is the diameter.)

Answer 4

Okay, let's start from scratch.

FACT: A triangle with angles 30, 60, and 90, will have sides: hypotenuse = H. side opposite the 30 degree angle = 1/2 H. The side opposite the 60 degree angle will be length: $\sqrt 3/2$.

Another way of putting this is $\cos 30 = \sin 60 = \sqrt 3/2$ and $\sin 30 = cos 60 = 1/2$.

PROOF: Draw an equilateral triangle ABC with sides of length H. The angles will all be 60 degrees. Let M be the midpoint of BC. Then: AB = AC. Angle B = angle C. And BM = MC. so the triangles ABM and ACM are congruent. angles BMA and CMA are linear and equal so angle CMA is a right angle. and the angle MAC is 30 degrees.

So triangle CMA is a triangle with angles 30, 60, and 90 degrees. MC = 1/2 H. So by the pythagorean theorem AM = $\sqrt{H^2 - (1/2*H)^2} = \sqrt 3/2 * H$.

FACT: A triangle with the angles 30, 60, and 90 with hypotenuse H will have area = $\sqrt 3/8*H^2$.

PROOF: The area is $1/2*b*a$. By above $a = 1/2 H$ and $b = \sqrt 3 /2 H$. So Area = $1/2 * 1/2 H * \sqrt 3/2 *H = \sqrt 3/8 H^2$.

FACT: A triangle inscribed is a circe with one of its sides being the diameter is a right triangle. (This is Thale's theorem).

PROOF: Label the triangle ABE where AB is the diameter and E is the third point. Label the center of the circle O.

The sum of the angles ABE + EBA + AEB = ABE + EBA +AEO + OEB = 180.

But AO = EO = BO so ABE = AEO and EBA = OEB so AEB = AEO + OEB = 90. So the triangle is a right triangle.

SOOOO

In the diagram the triangle is a right triangle with angles 30, 60 and 90 and it has an hypotenuse of length 8.

THEREFORE its area is $\sqrt 3/8 * 8^2 = 8\sqrt 3$.

The circle has area of $\pi 4^2 = 16\pi$ so the half circle has area $8 \pi$.

The shaded area is the half circle minus the triangle

!!!ERGO!!! the area of the shaded region is $8\pi - 8\sqrt 3$.