Geometric Measurement and Dimension: Solving Problems with Volume Formulas (CCSS.G-GMD.3)

Cylinder volume V=πr²h=π(4)²(12)=192π≈603.2 m³. Choice B used πrh (forgot to square r), D used the diameter instead of the radius, and C mixes units (treating meters as centimeters).

Cylinder: V_cyl=πr²h=π(5)²(14)=350π≈1099.6 m³. Cone: V_cone=⅓πr²h=⅓π(5)²(6)=50π≈157.1 m³. Total =400π≈1256.6 m³. Choice B forgot the ⅓ in the cone, A omits the cone, and C is an arithmetic slip in the sum.

Cylinder: V_cyl=π(9)²(8)=648π≈2036.3 m³. Hemisphere: V_hemi=⅔π(9)³=486π≈1527.2 m³. Total =1134π≈3563.4 m³. Choice A used a full sphere (⁴⁄₃πr³) instead of a hemisphere, C omits the dome, and D mixes units.

Pyramid volume V=⅓Bh with B=6×6=36. So V=⅓(36)(4)=48 m³. Choice A forgot the ⅓ factor, B failed to square the base side, and D uses the wrong units.

$V_{\text{cone}}=\tfrac{1}{3}\pi r^2 h=\tfrac{1}{3}\pi(3)^2(12)=36\pi\approx113.1\ \text{cm}^3.$ $V_{\text{hemi}}=\tfrac{2}{3}\pi r^3=\tfrac{2}{3}\pi(3)^3=18\pi\approx56.5\ \text{cm}^3.$ Total $=36\pi+18\pi=54\pi\approx169.6\ \text{cm}^3.$ Choice A forgot the $\tfrac{1}{3}$ in the cone formula, and Choice B omits the hemisphere.

$V_{\text{cyl}}=\pi r^2 h=\pi(4)^2(10)=160\pi\approx502.7\ \text{m}^3.$ Choice B forgets $\pi$, Choice C uses the diameter as the radius ($r=8$), and Choice D is an arithmetic slip using $150\pi$ instead of $160\pi.$

$V_{\text{cyl}}=\pi r^2 h=\pi(5)^2(12)=300\pi\approx942.5\ \text{m}^3.$ $V_{\text{cone}}=\tfrac{1}{3}\pi r^2 h=\tfrac{1}{3}\pi(5)^2(4)=\tfrac{100}{3}\pi\approx104.7\ \text{m}^3.$ Total $=300\pi+\tfrac{100}{3}\pi=\tfrac{1000}{3}\pi\approx1047.2\ \text{m}^3.$ Choice C forgot the $\tfrac{1}{3}$ in the cone formula, and Choice A omits the cone entirely.

$V_{\text{pyr}}=\tfrac{1}{3}Bh=\tfrac{1}{3}(8^2)(9)=\tfrac{1}{3}(64)(9)=192\ \text{m}^3.$ Choice C forgot the $\tfrac{1}{3}$, and Choice A used $\tfrac{1}{2}Bh$ instead of $\tfrac{1}{3}Bh.$

$V_{\text{sphere}}=\tfrac{4}{3}\pi r^3=\tfrac{4}{3}\pi(6)^3=\tfrac{4}{3}\pi(216)=288\pi\approx904.8\ \text{m}^3.$ Choice A uses $\pi r^3$ (missing the $\tfrac{4}{3}$), Choice D uses the hemisphere formula $\tfrac{2}{3}\pi r^3$, and Choice B incorrectly uses the diameter as the radius.