The formula for the circumference of a circle is c= 2pi r, where r is the radius. Rearrange the formula to solve for our and sel
1 answer:
Answer: r = C/(2pi)
This is the same as writing
=========================================================
Explanation:
pi is some number (approximately 3.14) which means 2pi is also some number (roughly 6.28)
Saying c = 2pi*r means we have 2pi times r, and the result is the circumference c. To isolate r, we'll undo the multiplication. We'll undo it by dividing both sides by 2pi like so
C = 2pi*r
C/(2pi) = r
r = C/(2pi)
in which we can write it like
So whatever C is, we divide it over 2pi (aka roughly 6.28) to get the radius.
-----------------
Extra Info (Optional Section):
As an example, let's say the circumference of the circle is 628 feet. This means the distance around the circle is 628 feet. So C = 628 would lead to...
r = C/(2pi)
r = C/(6.28)
r = 628/(6.28)
r = 100
So the radius would be roughly 100 feet.
You might be interested in
Check the picture below.
now, we know that the slanted legs are congruent, since it's an isosceles trapezoid, we also know that the bases are the parallel sides, so, the "altitude" or distance from those bases are the same length, for each of those triangles in the picture.
now, the bases are parallel, that means the altitude segment is perpendicular to the base, the longest side at the bottom, so, we end up with a right-triangle that has a Hypotenuse and a Leg, equal to the other triangle's.
thus, by the HL theorem for right triangles, both of those triangles are congruent, and if the triangles are congruent, all their sides are also, including the ones on the base.
now, we know that the slanted legs are congruent, since it's an isosceles trapezoid, we also know that the bases are the parallel sides, so, the "altitude" or distance from those bases are the same length, for each of those triangles in the picture.
now, the bases are parallel, that means the altitude segment is perpendicular to the base, the longest side at the bottom, so, we end up with a right-triangle that has a Hypotenuse and a Leg, equal to the other triangle's.
thus, by the HL theorem for right triangles, both of those triangles are congruent, and if the triangles are congruent, all their sides are also, including the ones on the base.