Prisms and Cylinders — Hard Practice Quiz
A Geometry cheat sheet for Prisms and Cylinders — every key formula with its symbols defined — plus a hard-level practice quiz to test recall.
Formulas & key concepts
Prism Volume
Where: \(B\) = base area, \(h\) = height
Prism Surface Area
Where: \(P\) = base perimeter
Cylinder Volume
Cylinder Surface Area
Practice quiz
A cylinder has an initial volume $V_0$ and surface area $SA_0$. If its radius is doubled and its height is adjusted to maintain the same volume $V_0$, what is the new surface area $SA'$ in terms of $SA_0$ and the initial ratio of height to radius, $\frac{h}{r}$?
- $\frac{1}{2} SA_0 + 7\pi r^2$
- $\frac{1}{2} SA_0 + 3\pi r^2$
- $\frac{1}{2} SA_0 + 6\pi r^2$
- $\frac{1}{2} SA_0 + 5\pi r^2$
Answer: $\frac{1}{2} SA_0 + 6\pi r^2$
A rectangular prism has initial base dimensions $L$ and $W$, and height $H$. If its length is doubled, width is halved, and height is tripled, how does its surface area change? Let $SA_{initial}$ be the initial surface area.
- The new surface area is $2LW + 12LH + 3WH$.
- The new surface area is $2LW + 6LH + 3WH$.
- The new surface area is $2LW + 12LH + 6WH$.
- The new surface area is $2LW + 6LH + 6WH$.
Answer: The new surface area is $2LW + 12LH + 3WH$.
A prism has a square base. If its volume is $V$ and its height is $h$, express its surface area $SA$ in terms of $V$ and $h$.
- $\frac{2V}{h} + 4\sqrt{Vh}$
- $\frac{V}{h} + 4\sqrt{Vh}$
- $\frac{2V}{h} + 2\sqrt{Vh}$
- $\frac{V}{h} + 2\sqrt{Vh}$
Answer: $\frac{2V}{h} + 4\sqrt{Vh}$
A cylinder has a surface area $SA$ and its height is equal to its diameter. What is its volume in terms of $SA$?
- $\frac{SA}{3} \sqrt{\frac{SA}{6\pi}}$
- $\frac{SA}{6} \sqrt{\frac{SA}{3\pi}}$
- $\frac{SA}{2} \sqrt{\frac{SA}{6\pi}}$
- $\frac{SA}{3} \sqrt{\frac{SA}{3\pi}}$
Answer: $\frac{SA}{3} \sqrt{\frac{SA}{6\pi}}$
A rectangular prism has a square base of side $s$ and height $h_P$. A cylinder has radius $r$ and height $h_C$. If the prism's volume is equal to the cylinder's volume, and $h_P = 2s$ for the prism, while $h_C = 2r$ for the cylinder, what is the ratio of the prism's surface area to the cylinder's surface area?
- $\frac{5}{3\pi^{1/3}}$
- $\frac{10}{3\pi^{1/3}}$
- $\frac{5}{6\pi^{1/3}}$
- $\frac{10}{3\pi^{2/3}}$
Answer: $\frac{5}{3\pi^{1/3}}$
If a cylinder's radius is increased by $50\%$ and its height is decreased by $50\%$, which of the following statements is true regarding its volume and surface area?
- The volume increases by $12.5\%$, and the change in surface area depends on the initial ratio of radius to height.
- The volume increases by $25\%$, and the surface area decreases.
- The volume remains unchanged, and the surface area increases.
- The volume decreases by $12.5\%$, and the change in surface area depends on the initial ratio of radius to height.
Answer: The volume increases by $12.5\%$, and the change in surface area depends on the initial ratio of radius to height.
For a cylinder, if the ratio of its surface area to its volume is $k$, and its height is $h$, what is its radius $r$ in terms of $k$ and $h$?
- $\frac{2h}{kh - 2}$
- $\frac{h}{kh - 2}$
- $\frac{2h}{kh + 2}$
- $\frac{h}{kh + 2}$
Answer: $\frac{2h}{kh - 2}$
A rectangular prism has a base area $B_P$ and height $h_P$. A cylinder has a base area $B_C$ and height $h_C$. If $B_P = 2B_C$ and $h_C = 3h_P$, what is the ratio of the prism's volume to the cylinder's volume?
- $\frac{2}{3}$
- $\frac{3}{2}$
- $\frac{1}{6}$
- $6$
Answer: $\frac{2}{3}$
A prism has a base that is a right-angled triangle with legs $a$ and $b$. If its volume is $V$ and its height is $h$, express its surface area $SA$ in terms of $V$, $h$, $a$, and $b$.
- $\frac{2V}{h} + (a+b+\sqrt{a^2+b^2})h$
- $\frac{V}{h} + (a+b+\sqrt{a^2+b^2})h$
- $\frac{2V}{h} + (a+b+\sqrt{a^2-b^2})h$
- $\frac{V}{h} + (a+b+\sqrt{a^2-b^2})h$
Answer: $\frac{2V}{h} + (a+b+\sqrt{a^2+b^2})h$
A cylinder with radius $r$ and height $h_C$ is placed centrally on top of a rectangular prism with base dimensions $L \times W$ and height $h_P$. The cylinder's base is entirely within the prism's top face. Which expression correctly represents the total surface area of the composite solid?
- $2LW + 2(L+W)h_P + 2\pi r h_C$
- $LW + 2(L+W)h_P + 2\pi r h_C + \pi r^2$
- $2LW + 2(L+W)h_P + 2\pi r h_C + 2\pi r^2$
- $LW + 2(L+W)h_P + 2\pi r h_C$
Answer: $2LW + 2(L+W)h_P + 2\pi r h_C$
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