Internet Stumped over Geometry Problem

After my class learns Chapter 10 of Geometry, they might be able to solve it.

After my class learns Chapter 10 of Geometry, they might be able to solve it.

Internet Stumped over Geometry Problem

After my class learns Chapter 10 of Geometry, they might be able to solve it.

After my class learns Chapter 10 of Geometry, they might be able to solve it.

- mr womp womp
- Official Cemetech Cat Manager (Posts: 1787)

- 08 Feb 2022 01:02:55 pm
- Last edited by mr womp womp on 11 Feb 2022 03:21:08 pm; edited 1 time in total

You can solve this geometrically.

The coordinates (-1/2,√3/2) means the left line of the triangle is the hypotenuse of a right angle triangle with side lengths of 1/2 and √3/2. With Pythagoras, you can figure out the length is 1. The right side is obviously the same length since it is also a radius of the same circle. I think you can convince yourself by looking at it that the top line is also of length 1 but you can also just look at the coordinates and check with the distance formula. So we have an equilateral triangle of side length 1.

You can then solve the area of that using the formula (√3/4)*a², which follows from the Pythagorean theorem again. Then you just have to subtract that from the area of the sector. Since it is formed by an equilateral triangle, the angle is 60°, so you can do (60/360)*πr², with a side length of 1, that gives you an area of π/6. You then subtract the area of the triangle, which is √3/4 to get the answer: π/6-√3/4 or approx. 0.0906.

Edit: That was fun! I haven't done this kind of problem since high school

The coordinates (-1/2,√3/2) means the left line of the triangle is the hypotenuse of a right angle triangle with side lengths of 1/2 and √3/2. With Pythagoras, you can figure out the length is 1. The right side is obviously the same length since it is also a radius of the same circle. I think you can convince yourself by looking at it that the top line is also of length 1 but you can also just look at the coordinates and check with the distance formula. So we have an equilateral triangle of side length 1.

You can then solve the area of that using the formula (√3/4)*a², which follows from the Pythagorean theorem again. Then you just have to subtract that from the area of the sector. Since it is formed by an equilateral triangle, the angle is 60°, so you can do (60/360)*πr², with a side length of 1, that gives you an area of π/6. You then subtract the area of the triangle, which is √3/4 to get the answer: π/6-√3/4 or approx. 0.0906.

Edit: That was fun! I haven't done this kind of problem since high school

- arusher999
- Advanced Member (Posts: 181)

- Re: I can already imagine this headline
- 08 Feb 2022 05:46:13 pm

After my class learns Chapter 10 of Geometry, they might be able to solve it.

What Geometry class r u in? I'm in Geo honors, and we are doing circles for chapter 10 as well, lol, we might have the same textbook.

For the record: I figured out how to solve that one pretty quickly (within a few seconds). Didn't solve it, though. Pretty sure womp is right

(There's a big difference between figuring out how to solve for a certain number and actually solving it. Solving it is trivial; it's how you go to figuring out how to solve it that's important...)

If you like these kinds of problems, go and take a look at AoPS. It's essentially Cemetech but math and also they actually have classes and are an accredited school. At least in California. The community is very friendly, and I suspect that Cemetech and AoPS (despite their different focuses... foci?) have communities that would mix very well.

(There's a big difference between figuring out how to solve for a certain number and actually solving it. Solving it is trivial; it's how you go to figuring out how to solve it that's important...)

If you like these kinds of problems, go and take a look at AoPS. It's essentially Cemetech but math and also they actually have classes and are an accredited school. At least in California. The community is very friendly, and I suspect that Cemetech and AoPS (despite their different focuses... foci?) have communities that would mix very well.

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