sat suite question viewer

Geometry and Trigonometry / Area and volume Difficulty: Hard

The circumference of the base of a right circular cylinder is $20 π$ $20 pi$ meters, and the height of the cylinder is $6$ $6$ meters. What is the volume, in cubic meters, of the cylinder?

Back question 69 of 86 Next

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86

Explanation

Choice C is correct. The volume, $V$ $V$ , of a right circular cylinder is given by the formula $V = π r^{2} h$ $V = π r^{2} h$ , where $r$ $r$ is the radius of the base of the cylinder and $h$ $h$ is the height of the cylinder. It’s given that a right circular cylinder has a height of $6$ $6$ meters. Therefore, $h = 6$ $h equals 6$ . It's also given that the right circular cylinder has a base with a circumference of $20 π$ $20 pi$ meters. The circumference, $C$ $C$ , of a circle is given by $C = 2 π r$ $upper C equals 2 pi r$ , where $r$ $r$ is the radius of the circle. Substituting $20 π$ $20 pi$ for $C$ $C$ in the formula $C = 2 π r$ $upper C equals 2 pi r$ yields $20 π = 2 π r$ $20 pi equals 2 pi r$ . Dividing each side of this equation by $2 π$ $2 pi$ yields $10 = r$ $10 equals r$ . Substituting $10$ $10$ for $r$ $r$ and $6$ $6$ for $h$ $h$ in the formula $V = π r^{2} h$ $V = π r^{2} h$ yields $V = π {(10)}^{2} (6)$ $upper V equals pi left parenthesis 10 right parenthesis squared left parenthesis 6 right parenthesis$ , or $V = 600 π$ $upper V equals 600 pi$ . Therefore, the volume, in cubic meters, of the cylinder is $600 π$ $600 pi$ .

Choice A is incorrect and may result from conceptual or calculation errors.

Choice B is incorrect. This is the lateral surface area, not the volume, of the cylinder.

Choice D is incorrect. This is the result of using the diameter, not the radius, for the value of $r$ $r$ in the formula $V = π r^{2} h$ $V = π r^{2} h$ .