does Pythagoras theorem work with circles?

Question

if so could u give me an example. if u cant it dont matter.

Answer 1

If you draw a diameter of a circle and then have a triangle from the ends of the diameter which touches the circle - this triangle is a right angled triangle.

You can then apply Pythagoras to the triangle because it is a right angled triangle.

Alternatively, you could draw a circle with centre (0,0) and radius r with a point P(x,y) on the circle. You can then drop horizontal and vertical lines to the x and y axes and have another right angled triangle where you can apply Pythagoras: r^2 = x^2 + y^2

But to the circle itself ... well, the circle is not a triangle. You have to generate those right angled triangles.

Answer 2

Pythagorean theorem you mean?

I'm not sure how old you are, so I cannot predict how you'll comprehend. The answer is YES.

Ideally, there are two ways to graph a function, using Polar Coordinates, and Cartesian Coordinates. With Cartesian coordinates it may seem like the only use for Pythagorean's Theorem is for planar motion, but when converting to Polar Coordinates the idea is quite different.

Aside from the other 100's of times you will use Pythagorean's Theorem in Polar Coordinate math, which is based on the theory of circles; to convert from Cartesian coordinates to Polar coordinates you need an x=rcos(theta), and a y=rsin(theta) where, using none other than Pythagorean's Theorem and basic trigonometry, r = (sqr_root_of:((x^2) + (y^2))) which simplifies out to r^2 = (x^2) + (y^2)

Polar coordinates are used in a lot of post secondary, first and second year algebra classes. If you haven't gotten there yet and are still interested, I'd recommend you do take those classes when you get the chance.
There is SO much more to learn regarding what I've briefly touched on.

I Hope that was legible to you and helped!

Answer 3

Not generally (of course it will if you have a right triangle in there--and any isosceles triangle can be broken into two right triangles).

But, if you do calculate the linear distance between two points (as oppose to the arc length between them), you can calculate the perimeter of an n-gon this way. Archimedes first approximated pi using a similar technique of n-gons which were inscribed and circumscribed around a circle. (This is called the Method of Exhaustion.)

I'm sure Wikipedia or some math websites will give you more details, if you're so inclined.

Also, you could use the Law of Cosines (which, shall we call it a slightly more general version of the Pythaogrean Theorem, where you do not need a right angle) to calculate the linear distance as opposed to arc length between two points of an ellipse, instead of just a circle.

Answer 4

right that's my concept...on the 1st one, the quarter circle on the left isn't needed. It comes all the way down to the shaded section is comparable to the shaded semi minus the section between the arc and chord. the section between the arc and chord is the quarter circle minus the triangle. to locate the part of the shaded semicircle you may desire to be responsive to the radius, by fact the triangle is an isosceles suited, the hypotenuse (diameter of the shaded semi) is 6*sqrt2 so the radius is 3sqrt2 and the section could be (3sqrt2)^2 * pi /2 or 9pi. The quarter circle is 6^2 * pi / 4 or 9pi and the triangle is 6*6/2 or 18. the full element feels like: 9pi - (9pi - 18) or basically 18. the 2nd is easy in case you draw 2 segments from the factor of intersection of the two semi circles back perpendicular to the sides of the quarter circles (it is additionally the middle of the semi circles. you may desire to have what feels like the super quarter circle, a sq. interior the backside left corner and 1 / 4 circle on perfect of the sq. and on the splendid side of the sq.. At this factor, it would be sparkling that the section is massive Quarter Circle - (2 small quarter circles + sq.). you need to combine the two small quarters to a semicircle. super quarter is 10^2 pi / 4 or 25 pi. The small semi (or 2 quarters) could be 5^2 pi /2 or 12.5 pi and the sq. is 5^2 or 25. the super photograph could be 25pi - (12.5pi + 25) or 12.5pi -25. bear in mind, you won't be able to basically subtract the two semicircles by overlap...making the sq. and a couple of quarters is the least complicated physique of suggestions desire this helps.

Answer 5

Yes it does.

r^2 = [r sin(theta)]^2 + [r cos(theta)]^2 is the Pythagorean theorem for a circle.

r is the radius of the circle, and theta is some angle measured from a reference direction (like from the X axis) and theta = 2 pi = 360 deg for one full sweep around the circle.

If you draw a right triangle from the center inside a circle of radius r, you will see that r is the hypotenuse of that triangle. Further, you will see that x = r cos(theta) and y = r sin(theta); where x and y are the sides of the triangle, so that r^2 = x^2 + y^2, which is the standard theorem in Cartesian coordinates.

Answer 6

No. It only applies to right triangles. (Right triangles have one 90 degree angle.)

I suppose that theoretically you could have a right triangle inside a circle & use it to find a chord lengh that was one of the sides of the triangle or the radius if you had the lengths of sides c & b, but not a & a was also the radius of the circle. Once you know the radius, you could easily find the area & circumference.

Answer 7

No, but sometimes you would use the phytagoras theorem with circles if you have a triangle in a circle, for example, a radius and a tangent form a triangle within the circle, you can use phytagoras for such question

Answer 8

No but the Tomsick Theorum Does

A1 x A2 = 2pix3

where a is an angle

Answer 9

No. This applies only to a right angle triangle.

Answer 10

No, circles do not have a hypotenus.