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enter image description here

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

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    $\begingroup$ The steps look right so far to me. What's stopping you from calculating the area of the triangle? (It's supposed to be a right triangle, I bet). $\endgroup$
    – Eli Rose
    Jan 5 '16 at 21:16
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    $\begingroup$ Is this a bad Illustration or is $O$ really not the center of the circle? $\endgroup$
    – adjan
    Jan 5 '16 at 21:17
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    $\begingroup$ @Eli Rose if $O$ is the center, then Thales' theorem applies, thus it's a right angle $\endgroup$
    – adjan
    Jan 5 '16 at 21:19
  • $\begingroup$ O should be the center of the circle $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 21:20
  • $\begingroup$ @Eli Rose How would I calculate the area of the triangle? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 21:20
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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.

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  • $\begingroup$ Why did you multiply 8sin(30)? should'nt it just be sin(30)? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 22:25
  • $\begingroup$ @MATHASKER recall that $\sin(\alpha)=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos(\alpha)=\frac{\text{adjacent}}{\text{hypotenuse}}$ you want to find the adjacent and the opposite side of the triangle $\endgroup$
    – null
    Jan 5 '16 at 22:49
  • $\begingroup$ So than I would have to find the opposite right? so $$ Sin(30) = opp/8$$ $$Sin(30)*8=opp$$ Opp = 4. Now how would I find the height? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 22:53
  • $\begingroup$ @MATHASKER I explained that in my answer and comment already: Get both other sides of the triangle The other side is the "height", because it is a right triangle. $\endgroup$
    – null
    Jan 5 '16 at 22:57
  • $\begingroup$ So than the opposite would be the height right? so $$(6.92)(4)/2$$ Would that give me the answer? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 23:02
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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 $\frac{1}{2} \pi 4^2 - 8 \sqrt{3} = 8 (\pi - \sqrt{3}) \approx 11.28$.

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  • $\begingroup$ I don't know what Thale's theorem is? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 21:20
  • $\begingroup$ en.wikipedia.org/wiki/Thales'_theorem $\endgroup$
    – SSF
    Jan 5 '16 at 21:23
  • $\begingroup$ I never learned it in school ever. But consider the three radii from the center to to the three points. They are equal so the form isoceles triangles. So they have equal base angles. So the far angle of the triangle is equal to the sum of the other two angles. So the total of the angles, 180, is twice the far angle. So the far angle is 90. $\endgroup$
    – fleablood
    Jan 5 '16 at 21:35
  • $\begingroup$ How did you get $$(√3/2)8$$ for cosine of 30? shouldn't it be $$cos(30)=adj/8$$ adj=6.92 right? and then what would you do after that? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 22:38
  • $\begingroup$ A 30 60 90 triangle is half of an equilateral triangle. Cut an equilateral triangle in half, and you get a right triangle. The hypotenuse is one of the original sides. Hypotenuse = 1. The bottom leg is 1/2 the original side = 1/2.. So sin30 = 1/2. The third side is the altitude which we need the pythagorean theorem to find. cos 30 = \sqrt 1 - sinn^2 30 = \sqrt 1 - (1/2)^2 = sqrt 3/4 = (sqrt 3)/2. $\endgroup$
    – fleablood
    Jan 6 '16 at 0:25
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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 $8\pi - 8\sqrt 3 = 8(\pi - \sqrt 3)$.

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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.

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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.)

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  • $\begingroup$ How do i know that its a right angle? Hint? and also how would i find the area of the triangle? its $$1/2bh$$ so the b would be 8 but what about h? I looked up the formula to calculate area of a right triangle it was $$ab/2$$ where a was the straight line in the right triangle and b was the base. Any Hint would be enough $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 21:52
  • $\begingroup$ I gave the proof that the triangle is a right angle. A 30 60 90 triangle is half of an equilateral triangle. So the hypotenuse is 1, the base is 1/2 and the height is $\sqrt 3/2$. Here the hypotenuse is 8 so the sides are 8 and $\sqrt 3 * 8/2$. $\endgroup$
    – fleablood
    Jan 5 '16 at 22:02
  • $\begingroup$ Lable the three points A, B and E. AB = 8 and ABE = 30 so AE = 4 and EB = 4$\sqrt 3$. BAE = 60 AE = 4 so h = 4*$\sqrt 3/2$ so b = 8 and h= 4*$\sqrt 3/2$. !OR! turn the triangle on its side. b = AE = 4 and h = EB = 4$\sqrt 3$. $\endgroup$
    – fleablood
    Jan 5 '16 at 22:09
  • $\begingroup$ How did you get 1 for hypotenuse? I thought it was eight and what if point E was not straight above point O? How would you find the height of the triangle?How did you get √3/2 for height? $\endgroup$
    – MATH ASKER
    Jan 5 '16 at 22:12
  • $\begingroup$ 1 is for 30 60 90 triangles in general. This one is 8 times larger so the sides are 8 times longer. E is not above point O. E is above some unkown 5th point. But it doesn't matter as AE is 4 and EAB is 60 the height above that unknown 5th point is therefore $\sqrt 3/2$*AE. OR you can turn the triangle side wise and use b = AE and h = EB. (It is a right triangle, after all.) $\endgroup$
    – fleablood
    Jan 5 '16 at 22:16
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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 $8\pi - 8\sqrt 3$.

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