Three cylinders each of height $16$ cm and radius of base $4$ cm are placed on a plane, so that each cylinder touches the other two. Then the volume of region enclosed between the three cylinders in $cm^3$ is
The region enclosed between three cylinders of radius r forms an equilateral triangle of side 2r with three circular sectors of 60 degrees each. The area of the triangle is (sqrt(3)/4) * (2r)^2 = sqrt(3) * r^2, and the area of the three sectors is (3 * 60/360) * pi * r^2 = (1/2) * pi * r^2. The volume is height * (area of triangle - area of sectors) = 16 * (sqrt(3) * 16 - (pi * 16 / 2)) = 256 * sqrt(3) - 128 * pi = 128 * (2 * sqrt(3) - pi).
When the three cylinders of radius 4 cm are placed touching each other, their centers form an equilateral triangle with a side length of 8 cm. The volume of the enclosed region is the height multiplied by the area between the three circular bases and the triangle, which is found by subtracting the area of the three 60-degree sectors from the triangle's area. The area of the equilateral triangle is root three over four times 8 squared, equaling 16 root three, and the combined sector area is half of the circle's area, equating to half of pi times 4 squared, which is 8 pi. Multiplying the difference, 16 root three minus 8 pi, by the height of 16 cm gives the volume of the region as 128 times the quantity 2 root 3 minus pi.