Prove that an expression for the area of the shaded region is A = (r^2/2)(pi - √3)

EddyBenzen122

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1631390775940.pngI'm only having difficulties finding the area of the equilateral traingle of ABC, I've got the rest solved. I know that the area for each sector of the three circles is (pi * r^2/6). As a result I get (pi * r^2/6) * 3 for all of the sectors combined. Next, I need to find what the area of the equilateral traingle of ABC is, which is what I'm having trouble with. So how could I figure that out?
 
I need to find what the area of the equilateral traingle (sic) of ABC is
Bisect triangle ABC and the resulting pair of right triangles have base r/2 and hypotenuse r. Calculate the height of the right triangles, then use the area formula for a triangle.

Or, you could use Heron's Formula.

Or, you could search for the formula that gives the area of an equilateral triangle in terms of its side length.

What class are you taking?

?
 
The sides of the triangle are r. All angles are 60 degrees.
Draw a line from point C to line AB which is perpendicular to C
This line will bisect angle C and create two right triangles. You can easily find the area of each of these two triangles.

Out of curiosity why do you want to know this area?
 
Bisect triangle ABC and the resulting pair of right triangles have base r/2 and hypotenuse r. Calculate the height of the right triangles, then use the area formula for a triangle.

Or, you could use Heron's Formula.

Or, you could search for the formula that gives the area of an equilateral triangle in terms of its side length.

What class are you taking?

?
I'm taking Math IB
 
The sides of the triangle are r. All angles are 60 degrees.
Draw a line from point C to line AB which is perpendicular to C
This line will bisect angle C and create two right triangles. You can easily find the area of each of these two triangles.

Out of curiosity why do you want to know this area?
I need to prove whatever it says on the image above.
 
I need to prove whatever it says
It says to verify that the shaded area is (r^2/2)(Pi - sqrt(3)).

I had time today for checking your work so far. Have you confused the names 'sector' and 'segment'?

You are correct that sector ABC has area r^2*Pi/6, but I don't understand why you've multiplied that area by 3 (you haven't explained what you're thinking very well).

A circular sector looks like a slice of pie. Here is sector ABC (shaded in blue).

sectrX.jpg
You do not want to count that blue area three times, unless your plan is to add three of those sector areas followed by subtracting the area of triangle ABC twice. Is that your plan?


A circular segment looks like this:

sectrX.jpg

In other words, the shaded area in this exercise is the combination of the area of triangle ABC (red) plus the areas of three identical segments (pink).

sec2seg.JPG

The area formula for a circular segment is:

Asegment = r^2/2 * (θ - sin(θ))

Did you try any of the previous suggestions for calculating the area of triangle ABC?

One suggestion is to bisect triangle ABC into two identical right triangles.

equatri.JPG
The height is h.

The base of each right triangle is R/2

Have you learned the Pythagorean Theorem? We use that to express h in terms of R.

The area of any triangle is: 1/2 * base * height

Please explain your plan and show us what you can try for getting the area of triangle ABC. Thank you!

?
 
It says to verify that the shaded area is (r/2)(Pi - sqrt(3)).

I had time today for checking your work so far. Have you confused the names 'sector' and 'segment'?

You are correct that sector ABC has area r^2*Pi/6, but I don't understand why you've multiplied that area by 3 (you haven't explained what you're thinking very well).

A circular sector looks like a slice of pie. Here is sector ABC (shaded in blue).

View attachment 28884
You do not want to count that blue area three times, unless your plan is to add three of those sector areas followed by subtracting the area of triangle ABC twice. Is that your plan?
Yes, that was my plan.
A circular segment looks like this:

View attachment 28885

In other words, the shaded area in this exercise is the combination of the area of triangle ABC (red) plus the areas of three identical segments (pink).

View attachment 28886

The area formula for a circular segment is:

Asegment = r^2/2 * (θ - sin(θ))

Did you try any of the previous suggestions for calculating the area of triangle ABC?

One suggestion is to bisect triangle ABC into two identical right triangles.

View attachment 28887
The height is h.

The base of each right triangle is R/2

Have you learned the Pythagorean Theorem? We use that to express h in terms of R.

The area of any triangle is: 1/2 * base * height

Please explain your plan and show us what you can try for getting the area of triangle ABC. Thank you!

?
I got everything figured out.
Also, how do I mark a response as solved in this forum?
 
Glad you worked it out. You also now have a formula for the area of any equilateral triangle in terms of the side length.

how do I mark a response as solved in this forum?
The unsolved/solved option needs to be selected when creating a thread, if it's still available. (The forum lost some functionality, after a software update.)

I just did a test, and I didn't find the option (but I'm currently on the mobile version).

?‍♂️
 
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