SOLUTION: An equilateral triangle, each side of which measures 12, is inscribed in a circle. What is the area of the other part of the circle?

Algebra ->  Surface-area -> SOLUTION: An equilateral triangle, each side of which measures 12, is inscribed in a circle. What is the area of the other part of the circle?       Log On


   



Question 1009441: An equilateral triangle, each side of which measures 12, is
inscribed in a circle. What is the area of the other part
of the circle?

Found 2 solutions by Edwin McCravy, AnlytcPhil:
Answer by Edwin McCravy(20055) About Me  (Show Source):
You can put this solution on YOUR website!
We will need: 

1. the height of the triangle so that we can find its area, and 
2. the radius of the circle so that we can find its area.

Then we can subtract to find the other part of the circle. 



First we find the height of the triangle. 
We draw the altitude h.



By the Pythagorean theorem,

6%5E2%2Bh%5E2=12%5E2
36%2Bh%5E2=144
h%5E2=108
h=sqrt%28108%29
h=sqrt%2836%2A3%29
h=6sqrt%283%29

The area of the equilateral triangle

A=expr%281%2F2%29base%2Aheight

A=expr%281%2F2%29%2812%29%286sqrt%283%29%29

A=36sqrt%283%29

Next we draw the radii of the circle to the vertices
of the triangle.  Notice that this divides the height of
the triangle into the radius r and the apothem a which is
the height of the lower triangle:



We know that r%2Ba+=+h, so a=h-r
We know that h=6sqrt%283%29
So a=6sqrt%283%29-r

By the Pythagorean theorem, 

6%5E2%2Ba%5E2=r%5E2
36%2B%286sqrt%283%29-r%29%5E2=r%5E2
36%2B%2836%2A3-12sqrt%283%29%2Ar%2Br%5E2%29=r%5E2
36%2B108-12sqrt%283%29%2Ar%2Br%5E2=r%5E2
144-12sqrt%283%29%2Ar%2Br%5E2=r%5E2

Subtract r%5E2 from both sides:

144-12sqrt%283%29%2Ar=0

Solve for r:

144=12sqrt%283%29%2Ar

144%2F%2812sqrt%283%29%29=r

12%2Fsqrt%283%29=r

The area of the circle is given by

A=pi%2Ar%5E2

A=pi%2A%2812%2Fsqrt%283%29%29%5E2

A=pi%2A%28144%2F3%29

A=pi%2A48

A=48pi

So the area of the other part of the circle besides the
equilateral triangle is found by subtracting the area of 
the equilateral triangle from the area of the circle:

Answer: 48pi-36sqrt%283%29, or about 88.4

If we like, we can factor out 12

Answer:  12%284pi-3sqrt%283%29%29 but that isn't necessary.

Edwin



Answer by AnlytcPhil(1806) About Me  (Show Source):
You can put this solution on YOUR website!
We will need: 

1. the height of the triangle so that we can find its area, and 
2. the radius of the circle so that we can find its area.

Then we can subtract to find the other part of the circle. 



First we find the height of the triangle. 
We draw the altitude h.



By the Pythagorean theorem,

6%5E2%2Bh%5E2=12%5E2
36%2Bh%5E2=144
h%5E2=108
h=sqrt%28108%29
h=sqrt%2836%2A3%29
h=6sqrt%283%29

The area of the equilateral triangle

A=expr%281%2F2%29base%2Aheight

A=expr%281%2F2%29%2812%29%286sqrt%283%29%29

A=36sqrt%283%29

Next we draw the radii of the circle to the vertices
of the triangle.  Notice that this divides the height of
the triangle into the radius r and the apothem a which is
the height of the lower triangle:



We know that r%2Ba+=+h, so a=h-r
We know that h=6sqrt%283%29
So a=6sqrt%283%29-r

By the Pythagorean theorem, 

6%5E2%2Ba%5E2=r%5E2
36%2B%286sqrt%283%29-r%29%5E2=r%5E2
36%2B%2836%2A3-12sqrt%283%29%2Ar%2Br%5E2%29=r%5E2
36%2B108-12sqrt%283%29%2Ar%2Br%5E2=r%5E2
144-12sqrt%283%29%2Ar%2Br%5E2=r%5E2

Subtract r%5E2 from both sides:

144-12sqrt%283%29%2Ar=0

Solve for r:

144=12sqrt%283%29%2Ar

144%2F%2812sqrt%283%29%29=r

12%2Fsqrt%283%29=r

The area of the circle is given by

A=pi%2Ar%5E2

A=pi%2A%2812%2Fsqrt%283%29%29%5E2

A=pi%2A%28144%2F3%29

A=pi%2A48

A=48pi

So the area of the other part of the circle besides the
equilateral triangle is found by subtracting the area of 
the equilateral triangle from the area of the circle:

Answer: 48pi-36sqrt%283%29, or about 88.4

If we like, we can factor out 12

Answer:  12%284pi-3sqrt%283%29%29 but that isn't necessary.

Edwin