Challenging Geometry problem, new?

Question

If I have a circle, diameter 10cm, and an equilateral triangleis formed from this diameter, jutting out of the "top" of the circle. what is the area of the two sections cut "off" the circle by the equilateral triangle? In other words, the area of the circle not enclosed by the triangle, and not in the "lower" half of the circle?

If you could explain your solution I would be appreciative. My gut reaction is that the answer is 4.08cm^2.

The center of the circle is the midpoint of the triangle. Two of the vertices of the triangle lie on the circle, at the endpoints of the diameter. the third vertex extends up and out of the circle.

Answer 1

This is a very nice calculus problem! :)

First draw your circle so that it is centered at the origin. It has equation x^2+y^2=5^2.

Now draw your equilateral triangle. Note that by symmetry, we can find the area in question by find the area on the right half and multiply the result by two.

The right vertex of your triangle is (5,0). The top vertex of your triangle is (use the Pythagorean theorem) is (0,5 \sqrt{3}), so an equation of the line the creates the right side of your triangle is
(use point-point form and solve for y) y=-\sqrt{3} (x-5).

To find where the line intersects the circle we solve the system
{x^2+y^2=5^2,y=-y=-\sqrt{3} (x-5)} to obtain
(5,0) and (5/2,5 \sqrt{3}/2).

Then, the area of the right half of the desired region is
\int_{5/2}^{5}[ \sqrt{25-x^2}-(-\sqrt{3} (x-5))]dx

and the total area is
2\int_{5/2}^{5}[ \sqrt{25-x^2}-(-\sqrt{3} (x-5))]dx

This is a nasty integral requiring a trig substitution. After some work, we get

25/6 (2 \pi -3 \sqrt{3}) approx 4.5293

Answer 2

The trick to this problem is figuring out where the triangle intersects the circle.

One way to do this, is to draw an equilateral hexagon within the circle. The hexagon can then be cut into six triangles, like you would cut a pizza. Each triangle is exactly the same size, and is also exactly 1/4 of the area of the of the equilateral triangle that you are describing.

Now you just have to calculate the area of a arc sector, and subtract the area of the little triangle to calculate the area of a single section, then multiply by 2 to get both sections.

(2)((1/6)(pi)(5^2) - (5)(2.5)(3^(1/2))/2)
(25/3)(pi) - (25/2)(3^1/2)
4.52930369

Or you can do it this way.

a = area of 1/2 of circle = (pi)(5^2)/2 = 39.26990817
t = 3/4 * area of triangle = (3/4)5(3)^(1/2)(10)/2 = 32.47595264
area of sections cut off = (2/3)(a - t) = 4.529303689

Answer 3

Here's how to do it. Let the diameter be AB; the center of the circle O; the third (top) vertex of the triangle C (so the triangle is ABC). Let D be the point where BC intersects the circle. Draw OD.

Now, OD = OB = 5 (radii of the circle). Triangle BOD is isosceles. so base angles ODB and OBD are equal. But angle OBD is 60 degrees (from equilateral triangle ABC). So triangle DOB is also equilateral.

The area of the sector of the circle DOB is (60/360)(25 PI) = 25 PI/6. The height of triangle ODB is (5/2) sqrt(3). The area of triangle ODB is 5(5/2) sqrt(3)/2 = (25/4) sqrt(3).

Subtract the area of the triangle from the area of the segment to get the portion of the circle cut off by the chord. This is

(25/6) pi - (25/4) sqrt(3)

Double that to get your answer:

(25/3) pi - (25/2) sqrt(3) = (25/6)(2 pi - 3 sqrt(3))

On the calculator, this is 4.529 sq. cm.

Answer 4

Very interesting question. The key is to find the angle of elevation from the horizontal (diameter) to the point of intersection. By inspection this looks like close to 70 deg. Once the angle is found, find the area of the sector, subtract the area of the triangle within.

Find the angle:

The two shapes can be represented by the equations
y = sqrt (25 - x^2)
y = -x*sqrt(3) + 5*sqrt(3)
(i assume you know how to get these, if not, ask)

25 - x^2 = 3*(5-x)^2
25 - x^2 = 3*(x^2-10x+25)
4x^2-30x+50=0
2x^2-15x+25=0
(2x-5)(x-5)=0
x=5/2 or 5
y(5/2)=5*sqrt(3)/2
theta=60 deg

Now it seems obvious that this could be found much more quickly by construction, but nevertheless...

A=sector-triangle= (25/6)[pi-(3sqrt(3)/2)] (standard stuff)
2A=4.53cm^2

doing this by integration requires trig sub for sqrt (25 - x^2), which is too hard for the grade level

Answer 5

I'll give you my thoughts, but I haven't worked the answer to a conclusion:

In an xy plane, draw circle (x-5)^2 + y^2 = 25, and the equilateral triangle with vertices at (0,0), (10,0), (5,5*SQUAREROOT3).

The side of the triangle connecting (0,0) and (5,5*SQUAREROOT3) is straight line y=(SQUAREROOT3)*x.

Find the point where the circle intersects y=(SQUAREROOT3)*x. Call this point P.

Using calculus, integrate the area from 0 to P of the area between the circle and y=(SQUAREROOT3)*x.

Multiply by two, because there are two such areas.

Answer 6

get the area of the triangle. and the area of the circle half = 2*pi*r and divide by 2 then subtract the area of the triangle.

Answer 7

If you add both sides, then you get 4.53 cm^2.
I solved this graphically through geometric construction, so I did not use a specific expression or formula to arrive at the answer. I can assure you that the answer is correct.