Help on tough geometry / trig. math question?

Question

I have two circles. One has a radius of 6 inches, and the other has a radius of 8 inches. Their diameters are overlapping 2 inches. Don't use calculus to solve it. What is the area of the overlapped region?

Answer 1

I did originally goof. Here are the correct calculations. This uses a "brute force" method, ...

Center the 6" circle at the origin, so its equation is x^2 + y^2 = 36. Center the 8" circle at (0,12), so its equation is (x-12)^2 + y^2 = 64. When you solve these equations you find the points of intersection between the two circles--(29/6, -√455/6) and (29/6, √455/6). So if I draw a straight line connecting the two points of intersection--which is a chord in both circles--its length is √455/3.

Now in the 6" circle draw a radius to each point of intersection with the other circle. You have an isosceles triangle with two sides 6" and the third side √455/3". I can use the Law of Cosines to find the central angle. (I'll calculate it in radians instead of degrees.) This angle measures 1.268367682. Now I can use this to calculate the area of the triangle--17.18308726 sq in. I can also use the central angle to calculate the area of the circular segment--22.83061827 sq in. The difference between these will give the area of the portion of the 6" circle to the right (in our drawing) of the chord--5.64753101 sq in.

Now do the same thing in the 8" circle. Draw the two radii and find the central angle--0.92098685 radians. So the area of the triangle must be 25.47837076 sq in and the area of the sector must be 29.4715792 sq in. Thus, the portion of the 8" circle to the left of the chord must be 3.993208441 sq in.

Adding the two results, the overlapping region must have an area of 9.640739451 sq in or approximately 9.64 sq in.

Thanks to anonymous for catching my original mistake.

Answer 2

Form a triangle using the two radius's and the sum of the radiuses (or radii, whatever you prefer . . . I'm a computer scientist, not an English major ;-) less the overlap. So, the sides of that triangle are 6, 8, and 12. Use Heron's area formula to calculate the area of the triangle, and then it's height with the 12" side as a base. I won't type all the work (too much typing and I wrote a program to do it anyway; you can find the formula here: http://en.wikipedia.org/wiki/Heron%27s_formula ), but the area is sqrt(455). The area of a triangle is also bh/2, so h = 2*sqrt(455)/12 = sqrt(455)/6. Draw a line from the top corner perpendicular to the 12" side. That line has length sqrt(455)/6. The angle between the points of intersection in the 6" circle is therefore 2*arcsin(sqrt(455)/(6*6)), and the angle between the points of intersection in the 8" circle is therefore 2*arcsin(sqrt(455)/(6*8)). Notice that you can save some work by just adding the areas of the two sectors formed (I got 22.83 and 29.47 square inches as the areas of those sectors using the formula here: http://en.wikipedia.org/wiki/Circular_sector ) and subtracting twice the area of the first triangle we used (2*sqrt(455)). I got 9.639 (rounded up) square inches as the final result, but you should check to make sure I did it right. On the bright side, even if I made some error, you still know how to solve the problem now.

TDW: He gave the radii, not the diameters. Your calculations look sound, but they are based on erroneous data.

Answer 3

So, the diameters overlap by 2 inches. Therefore the distance from one centre to the other is 6+8-2=12. Orient the circles so that they are side to side (the centres are at the same height). Label the centre of the circle with radius 8 by A, the other centre by B and the upper intersection of the two circles by C. Now create a triangle ABC.

AB=12, AC= 8, and BC=6.

Using the law of cosines (http://hyperphysics.phy-astr.gsu.edu/hbase/lcos.html ) we can see that A≈ 26.38432974940796º ≈ 0.1465796097•π radians
and
B≈ 36.336057514613934º ≈ 0.2018669862•π radians
Now Sin(A) ≈ 0.4443901877
Then using the formula for area of a triangle
Area(ABC)=1/2(AB)(AC)sin(A) ≈ 21.33072901

using the formula for area of a circular cone
Area(A-cone)=r^2•(A/2) ≈ 14.73578960
Area(B-cone)=r^2•(B/2) ≈ 11.41530913

Therefore the overlapping area is Area(A-cone)+ Area(B-cone)-Area(ABC) ≈ 14.73578960 + 11.41530913 - 21.33072901 ≈ 4.82036972

This was only the top half of the overlap so the real total of overlap is 2•4.82036972 ≈ 9.64073944 sq in

Answer 4

i got here upon a way that incorporates the exterior attitude theorem of a triangle, the undeniable actuality that there are one hundred eighty degrees in a immediately line, and utilising a matrix to remedy 4 equations with 4 unknowns. the following that's: a million. make higher BD previous the triangle and call the recent endpoint G. Repeat with CE; call the recent endpoint H. 2. attitude (<) ADG = 40 degrees

Answer 5

Easy one: subtract the area of the smaller circle from the area of the larger one.

pi*8^2 - pi*6^2

pi(64-36) = 28pi

Answer 6

nope it is a very brute force problem.if u use calculus it will be easy and using calculus it is a kids problem.

Answer 7

Arggghhh!!! Why can't I use calculus? I should be able to use calculus whenever I want. - lol