area of shaded region of circle not covered by traingle

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wieb83

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Finding the area of a shaded region....

There is a circle centered at O, has a diameter at 10 cm and triangle ABC is equilateral. What is the area of the shaded regions?

So I am not sure how to post a picture of it but basically we have a circle that has a triangle on the right side of it and the circle is centered where the base of the triangle is. So the base of the triangle would be 10 cm along with the diameter of the circle. The triangle points beyond the right side of the circle and the shaded region would be the two small area the triangle doesn't cover on the right side of the circle.

I can find the area of the circle and the triangle and I know that the area of 1/2 the circle + the area of the triangle is missing the area of the shaded region. I just can't figure out how to solve this problem....Please help!
 
I was able to create a picture but with the triangle on the top of the circle instead of the side as described before. The shaded region is the unknown.


73038624.png
 
Draw the darn thing yourself: make the triangle stick out at top: EASIER to "see" !
Label the center M (hate to use O's!); diameter (or base of triangle) AB;
top of triangle C; intersection points D and E (D on AC, E on BC).

Join MD, ME and DE: get 4 equilateral triangles ADM, BEM, DEM and CDE: all with sides = r = 5

Now all you need to do is calculate area of one of the 3 segments (all same area).
Easy, since the arcs are all = 60/360 of circle's circumference: kapish?
One of the segments is covered by triangle ABC: two are not : OK??

EDIT: nice diagram; change the O to M :wink:
 
Look at my drawing. The new smaller triangle is also equilateral.
The area of one of the shaded circular segments is \(\displaystyle \frac{R}{2}\left( {\phi - \sin (\phi )} \right)\).
Where R is the radius.
http://mathworld.wolfram.com/CircularSegment.html
 

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Since the triangle's base is on a diameter of the circle, you know that the base is 10 units long.

From the center, draw the radius lines to the points on the non-base sides of the triangle, where these sides cross the circle at the top "ends" of the shaded areas. Label with "5". Note that these lines split the top half of the circle into three sectors. The left- and right-hand sectors contain triangles.

By nature of equilateral triangles, you know that the triangles' angles on the diameter line, where the original triangle's vertices met the circle, must measure sixty degrees.

Since the two new triangles have two sides each measuring 5 units (from the picture you've drawn), you know that these new triangles are isosceles, and thus must have base angles measuring sixty degrees. Then their third angles, at the center of the circle, must also be sixty degrees, so they are equilateral, too.

By nature of straight lines, you know that the diameter "angle" has a "measure" of 180 degrees. Since you have already accounted for the two new triangles, you can then find the measure of the sector between these two new triangles.

Find the areas of the two new triangles. Find the area of the sector between them. Add.

Subtract this sum from the area of the half-circle. The difference is the area of the shaded portion.

If you get stuck, please reply showing all of your steps and reasoning. Thank you! :D
 
Yikes Wieb: 3 replies!!
So you don't get all rattled, I suggest you use pka's: simplest :wink:
EXCEPT he kept center labelled O :shock:
 
Denis said:
… I suggest you use pka's: simplest …
Click to expand...


I'm not sure what the original poster will find simplest, but it's certainly simpler to be given the formula that PKA posted versus deriving it. :wink:

I like Elizebeth's strategy (which is really not very different from Denis'): find a bigger area and subtract a smaller area.

(1) Find the area of a sector (1/6th the area of the circle)

(2) Subtract the area of the triangle "within" the sector

(3) Double the result

I got [50 Pi - 75 sqrt(3)]/6 square centimeters. I did not verify this result.

What do you get?

 
I did get the answer {50pi -75sqrt(3)]/6 Square centimeters. The only problem is when I use the formula given by pka when I approximate the answer I am off my one decimal point. I get .45293 with pka's formula and with my answer it approximates out to 4.5293.
 
Nevermind....I figured it out! Thanks so much for all the wonderful help!
 
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