The shaded area consists of the triangle and the rectangle. r2 1 2 314 × 25 785 cm2 Area of rectangle l × b Area of shaded region 785. The corresponding lengths and breadths of the rectangles are equal, and the radius of the quarter. 1 Area of each of two quadrant 4 r2 (i) Radius of the circle the.
We let the top edges of the two shaded triangles. We could also solve this question algebraically. Therefore the total shaded area is half the total area of the original rectangle. If (circleDistance.x > (rect.width/2 + circle. The shaded area is then calculated by subtracting the area of the circle from the. Besides, we can use the Pythagorean Theorem to relate a and b with the radius R of the semi-circle. Areas of the shaded regions in figures 1 and 2 are equal. It is then evident that the shaded area is the sum of exactly half the area of each of the smaller rectangles.
Here is how I would do it: bool intersects(CircleType circle, RectType rect)ĬircleDistance.x = abs(circle.x - rect.x) ĬircleDistance.y = abs(circle.y - rect.y)
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