Geometry: Solving Problems with Volumes of Cones, Cylinders, and Spheres (CCSS.8.G.9)
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Common Core 8th Grade Math › Geometry: Solving Problems with Volumes of Cones, Cylinders, and Spheres (CCSS.8.G.9)
A right cylinder has radius 3 cm and height 5 cm. What is the volume of the cylinder?
$30\pi\ \text{cm}^3$
$45\ \text{cm}^3$
$45\pi\ \text{cm}^3$
$180\pi\ \text{cm}^3$
Explanation
Use $V=\pi r^2 h$. Substitute $r=3$, $h=5$: $V=\pi(3)^2(5)=\pi\cdot 9\cdot 5=45\pi\ \text{cm}^3$. The $30\pi$ option uses $2\pi rh$ (surface area idea), $45$ omits $\pi$, and $180\pi$ treats the diameter (6) as the radius.
A right cone has radius 4 m and height 6 m. What is the volume of the cone?
$32\pi\ \text{m}^3$
$96\pi\ \text{m}^3$
$32\ \text{m}^3$
$128\pi\ \text{m}^3$
Explanation
Use $V=\tfrac{1}{3}\pi r^2 h$. Substitute $r=4$, $h=6$: $V=\tfrac{1}{3}\pi(4)^2(6)=\tfrac{1}{3}\pi\cdot 16\cdot 6=32\pi\ \text{m}^3$. The $96\pi$ option forgets the $\tfrac{1}{3}$, $32$ omits $\pi$, and $128\pi$ uses the diameter (8) as the radius.
A sphere has radius 6 inches. What is its volume?
\$144\pi\ \text{in}^3$
\$288\ \text{in}^3$
\$2304\pi\ \text{in}^3$
\$288\pi\ \text{in}^3$
Explanation
Use $V=\tfrac{4}{3}\pi r^3$. With $r=6$: $V=\tfrac{4}{3}\pi(6)^3=\tfrac{4}{3}\pi\cdot 216=288\pi\ \text{in}^3$. The $144\pi$ option confuses surface area $4\pi r^2$, $288$ omits $\pi$, and $2304\pi$ uses the diameter (12) in place of the radius.
A cylindrical can has a diameter of 8 m and a height of 3 m. What is the volume of the can?
$24\pi\ \text{m}^3$
$48\pi\ \text{m}^3$
$48\ \text{m}^3$
$192\pi\ \text{m}^3$
Explanation
First find the radius: $r=\tfrac{8}{2}=4$ m. Use $V=\pi r^2 h$: $V=\pi(4)^2(3)=\pi\cdot 16\cdot 3=48\pi\ \text{m}^3$. The $24\pi$ option uses $2\pi rh$ (surface-area-related), $48$ omits $\pi$, and $192\pi$ treats the diameter as the radius.
What is the volume of this cylinder with radius 3 cm and height 5 cm?
$48\pi\ \text{cm}^3$
$45\pi\ \text{cm}^3$
$180\pi\ \text{cm}^3$
$45\ \text{cm}^3$
Explanation
Use $V=\pi r^2 h$. With $r=3$ and $h=5$: $V=\pi(3)^2(5)=\pi\cdot9\cdot5=45\pi\ \text{cm}^3$. Choice A uses a surface area expression, C uses diameter as radius, and D omits $\pi$.
A cone has radius 4 m and height 6 m. What is its volume?
$96\pi\ \text{m}^3$
$48\pi\ \text{m}^3$
$32\ \text{m}^3$
$32\pi\ \text{m}^3$
Explanation
Use $V=\tfrac{1}{3}\pi r^2 h$. With $r=4$, $h=6$: $V=\tfrac{1}{3}\pi(4)^2(6)=\tfrac{1}{3}\pi\cdot16\cdot6=32\pi\ \text{m}^3$. Choice A forgets the $\tfrac{1}{3}$, B uses $\tfrac{1}{2}$ instead of $\tfrac{1}{3}$, and C omits $\pi$.
What is the volume of a sphere with radius 6 inches?
\$288\pi\ \text{in}^3$
\$144\pi\ \text{in}^3$
\$2304\pi\ \text{in}^3$
\$288\ \text{in}^3$
Explanation
Use $V=\tfrac{4}{3}\pi r^3$. With $r=6$: $V=\tfrac{4}{3}\pi(6)^3=\tfrac{4}{3}\pi\cdot216=288\pi\ \text{in}^3$. Choice B is the surface area number reused as a volume, C uses the diameter as the radius, and D omits $\pi$.
A cylindrical can has diameter 8 cm and height 10 cm. What is its volume?
$112\pi\ \text{cm}^3$
$640\pi\ \text{cm}^3$
$160\pi\ \text{cm}^3$
$160\ \text{cm}^3$
Explanation
Convert diameter to radius: $r=4$ cm. Use $V=\pi r^2 h=\pi(4)^2(10)=\pi\cdot16\cdot10=160\pi\ \text{cm}^3$. Choice A is a surface area expression, B treats the diameter as the radius, and D omits $\pi$.
A cylinder has radius 3 cm and height 5 cm. What is the volume of this cylinder? Give your answer in cubic centimeters.
$45\pi$
$48\pi$
$45$
$180\pi$
Explanation
Use $V=\pi r^2 h$. Here $r=3$, $h=5$: $V=\pi(3)^2(5)=\pi\cdot9\cdot5=45\pi$. Distractors: $48\pi$ confuses surface area, $45$ omits $\pi$, $180\pi$ uses the diameter as the radius.
A cone has radius 3 m and height 8 m. What is the volume of the cone? Give your answer in cubic meters.
$72\pi$
$24$
$96\pi$
$24\pi$
Explanation
Use $V=\tfrac{1}{3}\pi r^2 h$. With $r=3$, $h=8$: $V=\tfrac{1}{3}\pi(3)^2(8)=\tfrac{1}{3}\pi\cdot9\cdot8=24\pi$. Distractors: $72\pi$ uses the cylinder formula (missing $\tfrac{1}{3}$), $24$ omits $\pi$, $96\pi$ incorrectly uses the diameter 6 as $r$.