Finding the Area of a Circumscribed circle?

Question

My math teacher said we could use any resource to solve his problems. so here we go.

This is teh problem.

There is a square with a side of 5cm. then there is a circle circumscribed (outside) around the square. So now theres a circle drawn around the outside of a 5x5 square. How do I find the area of the circl expressed in a fraction in terms of pi

Answer 1

You picked a good resource. Draw a diagonal across the square. It will measure 5 x sqrt2 , which is about 1.414, so the diagional will be 7.07cm. The diagional is a diameter of the circle, and the circle area = (pi/4)d^2. d^2 is about 50, so your fraction is about 25/2 pi

Answer 2

Well lets point out a few things that we do know:

We know the area of the circle can be found with the formula: A =πr^2, where r is the radius of the circle. We also know that the radius is simply half of the diameter.

Now draw out the circle around the square and notice where the corners touch. Corner to opposite corner [diagonally] of the square end up being the circle's diameter. So we need to find what that length is. If we split the square into triangles, we end up with 2 equal sized triangles with the base being 5cm and the height being 5cm. Using the Pythagorean formula, we can find the missing side [as well as the diameter of the circle]

5^2 + 5^2 = D^2 [where D is the diameter, and the hypotenuse] 50cm = D^2, D = √50 = √2√25 = 5√2

Now we need to reduce the diameter to radius, which can be found by multiplying by 1/2 ---> 1/2*5√2 = 5√2/2

Ok, now we can plug our radius into our area of a circle formula: A = π (5√2/2)^2 = 25π/2

So our area is 25π/2cm^2

Good luck!

Answer 3

The radius of the circumscribed circle id 1/2 the diagonal of the square . The diagomnal is sqrt(5^2 + 5^2) = sqrt(50) = 5sqrt(2) by application of Pythagoras' theorem to the square.
Thus R = 5/2sqrt(2)

The area of the circle is now pi R^2 = pi x (50/4) = 25/2 pi

Answer 4

Draw a line from centre of circle/pentagon to each and each vertex of pentagon. each and each of those strains is a radius (r) of circle. we've 5 isosceles triangles, each and each with substantial perspective = 360°/5 = seventy two° and opposite area = 50 inches. Base angles = (a hundred and eighty?seventy two)/2 = fifty 4°. utilising regulation of sines, we get r/sin(fifty 4) = 50/sin(seventy two) r = 50*sin(fifty 4)/sin(seventy two) area of circle = ? * (50*sin(fifty 4)/sin(seventy two))² = 5683.1945 in²