I mean I solved it, but not very mathematically X3: You only gave us the information that the length is 8, and that there are two semicircles. So no matter how big the semicircles are, the solution would have to be the same. So I just made the semicircle so small that the area is just the green thing, so the diameter is 8, so the area obviusly is 8pi.

@featureboxx

Ай бұрын

yes, I followed the same method, thinking about a missing value, then concluding that the problem implies that the result doesn't depend on the size of any circle, and subsequently making the calculation where the smallest circle is just a point.

@luiscostacarlos

24 күн бұрын

A observação foi muito boa, mas nada tira a beleza de álgebra. observation was very good, but nothing takes away from the beauty of algebra.

I believe the problem needs an additional constraint that was not mentioned... the chord is parallel to the diameter. Without this constraint only a general expression for the area is possible. That seems like a much trickier problem.

Shaded Area A = ⅛ π c² A = ⅛ π 8² A = 8π cm² ( Solved √ ) Just apply formula of half circular ring !!!, respect to the chord

Solution: R = radius of the large semicircle, r = radius of the small semicircle. Pythagoras: R²-r² = (8/2)² = 16 Green area = area of the large semicircle - area of the small semicircle = π*R²/2-π*r²/2 = π/2*(R²-r²) = π/2*16 = 8π ≈ 25.1327

My way of solution is ▶ O is the center of the larger circle. The radius of this semicircle is equal to R, and the radius of the small circle is equal to r. If we draw a radius from the center of the larger circle to the point where the tangent of the small circle touches the larger circle, let's call this point A. The vertical length formed by combining the radius R and the tangent of the circle produces a right triangle: ΔOAB OA= R AB= 8/2 AB= 4 length units BO= r According to the Pythagorean theorem, we get: OA²= AB²+BO² R²= 4²+r² R²-r²= 16 Agreen= (π/2)*(R²-r²) = (π/2)*16 Agreen= 8π square units Agreen ≈ 25,13 square units

@Qwertyuiopasdfghjklzxcvbnmlwjd

2 ай бұрын

Was just about to comment this but saw yours. I didn’t realise this at first and spent so long finding out small radius is either 0 or 4, and at the end when working out the area I found out I could’ve just subbed in 16 as it was difference of two squares 😅

El problema se hace trivial si consideramos dos círculos completos de radios "r" y "R", centrados de forma que el área verde es la de una corona circular. En la figura equivalente obtenida se comprueba que: R²-r²= (8/2)²=16 → Área verde =(πR²-πr²)/2 =π(R²-r²)/2 =16π/2=8π. Gracias y un saludo.

Green area=1/2(π)R^2-1/2(π)(r^2 π/2(R^2-r^2) R^2=r^2+4^2 R^2-r^2=16 So green area=π/2(16)=8π=25.13 square units.

In a circle with a radius 5. There are values of y to be plugged in the equation x^2+y^2=5^2 which produces exact values for x. Here are they: if y=0, x=5 if y=1.4, x=4.8 if y=3, x=4

I knew it is 8 pi instantly. The radius of the 2 circles do not matter so set the big semicircle to r=4 and the small semicircle to r=0. Therefore the green area is 16pi/2=8pi