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Simplest solution: I, Myself wrote a proof for the area of the circle. The idea is that a polygon is created by triangles. For example, a square is created by 4 triangles and its area can be calculated by summing up the area of the triangles. r = 1 2a then S = 4(1 2ra) = 2ra = a2 For hexagonal it is S = 6(1 2ra) and for n sided polygon it is S ...
A = 1 2l 4R2−l2√ 2 = l 4R2−l2√ 4 A = 1 2 l 4 R 2 − l 2 2 = l 4 R 2 − l 2 4. We can finally calculate the area of the regular inscribed polygon. We have n n triangles with equal area, so the total area will be n multiplied by the area of a single triangle. Atot = nl 4R2−l2√ 4 A t o t = n l 4 R 2 − l 2 4.
So, the area of one half of the intersection is the area of a circular segment with angle θ = 2π 3 and radius r, which gives an area of r2 2(θ − sinθ) = r2 2 (2π 3 − √3 2) and so the area of the entire intersection is twice this. This gives an area of r2(2π 3 − √3 2). @MhenniBenghorbal I'm a little confused by the edit.
The result follows almost immediately: ∫r − r2√r2 − y2dy = [r2arcsiny r + y√r2 − y2 ]r − r = r2arcsin(1) − r2arcsin(− 1) = = r2π 2 + r2π 2 = πr2. So, using the cartesian coordinates, the only important observation is simply to consider the right extreme values for each variable. Using the circumference equation x2 + y2 = r2 ...
For the radius of the sphere d d, the arc length id dθ d θ, where sin(θ) = r/d sin (θ) = r / d. The area of the "circle" is then πd2θ2 π d 2 θ 2. My mistake. Here is the solution: Suppose you call r r the length of the arc along the sphere, and x x the radius in the plane. At a position l l along the arc, x = d sin(θ) x = d sin (θ ...
the diameters differ by 9cm, so if the inner circle has radius r, the outer circle has radius r + 4.5. The area of the crescent is the difference in the areas of the circles: pi(r + 4.5)^2 - pi * r^2. All that's left is finding r. Define C as the point (0,0), then point E is at (0, r - 0.5) (because CE is 5cm less than the larger radius).
Right now, if you want to calculate the area of a circle of radius $4$, you're summing up the circumferences in the following picture, times the "thickness" of each skin, which is $1$: $\hskip1in$ However, this method has a problem: Each circumference represents the area of everything inside it, but before the next circle.
You know the area of a circular sector OPR O P R is given by r21a 2 r 1 2 a 2 where a a is in radians of course. 1) Area of triangle OPO′ O P O ′ minus area of circular sector O′PS O ′ P S = area of sector OPS O P S. 2)Requested area = 2 (r21a 2 r 1 2 a 2 minus area of sector OPS O P S) Share. Cite.
A formula for the area is worked out in Circle-Circle Intersection at Wolfram MathWorld: $$ A = r^{2}\cos^ ...
This is an ellipse: Note that it doesn't have a radius, but two "axis". The area of a circle is π ⋅r2 π ⋅ r 2, with r r the radius. The area of an ellipse if π ⋅ A ⋅ B π ⋅ A ⋅ B, which is a nice generalization. The drawing you made could be half an ellipse, so you have to divide the area by a factor of 2 2.