Geometry Quiz: Informal Arguments For Circle Solid Formulas

What this quiz covers

This quiz focuses on Informal Arguments For Circle Solid Formulas, giving you a quick way to practice the rules, question types, and explanations that matter most for Geometry.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Both solids have identical cross-sectional areas at every height level parallel to their bases
2. Both solids have proportionally similar cross-sectional areas that decrease linearly from base to apex (correct answer)
3. Both solids have cross-sectional areas that follow the same quadratic relationship with height
4. Both solids have bases with equal areas and identical rates of volume change per unit height

Question 2

1. $\lim_{n \to \infty} P_n = \lim_{n \to \infty} n \cdot 2r \sin\left(\frac{\pi}{n}\right) = 2\pi r$ (correct answer)
2. $\lim_{n \to \infty} P_n = \lim_{n \to \infty} n \cdot 2r \cos\left(\frac{\pi}{n}\right) = 2\pi r$
3. $\lim_{n \to \infty} P_n = \lim_{n \to \infty} n \cdot r \tan\left(\frac{\pi}{n}\right) = \pi r$
4. $\lim_{n \to \infty} P_n = \lim_{n \to \infty} n \cdot 2r \tan\left(\frac{\pi}{n}\right) = 2\pi r$

Question 3

1. The sectors become increasingly triangular, and $n$ triangles with base $\frac{2\pi r}{n}$ and height $r$ give area $\pi r^2$
2. The parallelogram's area exactly equals $\pi r^2$ regardless of $n$, but the shape looks more rectangular as $n$ increases
3. The parallelogram's base approaches $\pi r$ and height approaches $r$, while the "steps" become negligible (correct answer)
4. The rearranged sectors form a shape with perimeter $2\pi r$ and average width $r$, giving area $\pi r^2$

Question 4

1. Because the rearranged shape has base about $\pi r$ and height $r$, so its area is about $(\pi r)(r)=\pi r^2$. (correct answer)
2. Because the circle's area must be $\pi r^2$ since $\pi$ is defined using circles.
3. Because the rearranged shape has base $r$ and height $\pi r$, so its area is $r+\pi r$.
4. Because the perimeter of the circle is $2\pi r$, the area is also $2\pi r$.

Question 5

1. At each height the pyramid slice is similar to the base, and the collection of shrinking slice areas accounts for one-third the prism's volume with the same B and h. (correct answer)
2. Since both solids have the same base area B, they must have the same volume, so the pyramid volume is Bh.
3. Since the pyramid has four triangular faces, its volume is four times the prism's volume, so $V=4Bh$.
4. Since the pyramid's edges slope, its volume equals its lateral surface area, so $V=\tfrac13Bh$.

Question 6

1. The base of the rearranged shape approaches $\frac{1}{2}C$, so its area is $\left(\frac{1}{2}C\right)r=\pi r^2$. (correct answer)
2. The base of the rearranged shape equals $C$, so its area is $Cr=2\pi r^2$.
3. The height of the rearranged shape equals the diameter, so its area is $\left(\frac{1}{2}C\right)(2r)=2\pi r^2$.
4. The rearranged shape has the same perimeter as the circle, so its area must be $\pi r^2$.

Question 7

1. The circumference $2\pi r$ cannot be accurately measured when the cylinder is unrolled into a flat rectangle
2. The unrolled shape loses the circular cross-section information needed to calculate the $\pi r^2$ base area
3. The unrolling process changes the height dimension, making the volume calculation impossible using this method
4. The unrolled shape is a rectangle with dimensions $2\pi r \times h$, but this represents surface area, not volume (correct answer)

Question 8

1. As $n \to \infty$, both $A_n \to A_c$ and the prism volume $A_n h \to A_c h$, justifying $V = \pi r^2 h$ (correct answer)
2. As $n \to \infty$, the ratio $\frac{A_n}{A_c} \to 1$ while the volume ratio approaches $\frac{2}{3}$
3. As $n \to \infty$, the perimeter approaches the circumference but the area relationship remains constant
4. As $n \to \infty$, both the surface area and volume formulas converge to their circular analogs simultaneously

Question 9

1. The cylinder and prism have equal volumes because their cross-sectional areas match at every height.
2. The cylinder's volume equals the prism's volume, so the cylinder volume is $\pi r^2 h$.
3. The cylinder's lateral surface area must equal the prism's lateral surface area because the slice areas match. (correct answer)
4. If the common height is doubled while slice areas stay the same, the volume doubles.

Question 10

1. Radius is $R\left(\frac{y}{h}\right)$, and integrating $\pi r^2$ from $0$ to $h$ gives $\frac{\pi R^2 h}{3}$
2. Radius is $R\left(1 - \frac{y}{h}\right)$, and integrating $\pi r^2$ from $0$ to $h$ gives $\frac{\pi R^2 h}{3}$ (correct answer)
3. Radius is $R\left(1 - \frac{y}{h}\right)$, and integrating $\pi r^2$ from $0$ to $h$ gives $\frac{2\pi R^2 h}{3}$
4. Radius is $\frac{R}{h}(h - y)$, and integrating $2\pi r$ from $0$ to $h$ gives $\pi R^2 h$

Question 11

1. Since both solids have the same surface area at each height, their volumes are equal, so $V=\pi r^2 h$.
2. Since cross-sections at every height have equal area and the heights match, the volumes match, so $V=(\pi r^2)h$. (correct answer)
3. Since the cylinder's base area is $2\pi r$, multiplying by height gives $V=2\pi r h$.
4. Since both solids look about the same from the side, their volumes must be equal, so $V=\pi r^2 h$.

Question 12

1. Each small pyramid has volume $\frac{1}{3} \cdot \frac{s^3}{n^3} \cdot h = \frac{s^3 h}{3n^3}$, and $n^2$ of them give $\frac{s^3 h}{3n}$
2. Each small pyramid has volume $\frac{1}{3} \cdot \frac{s^2}{n^2} \cdot \frac{h}{n} = \frac{s^2 h}{3n^3}$, and $n^2$ of them give $\frac{s^2 h}{3n}$
3. Each small pyramid has volume $\frac{1}{3} \cdot \frac{s^2}{n^2} \cdot h = \frac{s^2 h}{3n^2}$, and $n^2$ of them give $\frac{s^2 h}{3}$ (correct answer)
4. Each small pyramid has volume $\frac{1}{3} \cdot \frac{s^2}{n} \cdot h = \frac{s^2 h}{3n}$, and $n^2$ of them give $\frac{s^2 h}{3} \cdot n$

Question 13

1. At height $y$, the hemisphere has area $\pi\sqrt{r^2 - y^2}$ and the cylinder-minus-cone has matching linear decrease
2. At height $y$, the hemisphere has area $\pi(r^2 - y^2)$ and the cylinder-minus-cone has area $\pi r^2 - \pi(r-y)^2$
3. At height $y$, both solids have circular cross-sections with the same radius $\sqrt{r^2 - y^2}$
4. At height $y$, the hemisphere has area $\pi(r^2 - y^2)$ and the cylinder-minus-cone has area $\pi r^2 - \pi y^2 = \pi(r^2 - y^2)$ (correct answer)

Question 14

1. Because each layer keeps the same area and the heights add to h, the rearranged solid has the same volume as the cone, matching one-third of the cylinder with base $\pi r^2$. (correct answer)
2. Because the cone's curved surface can be flattened into a sector, the volume must be $\tfrac13\pi r^2 h$.
3. Because the cone has circular slices, its volume equals the circumference times height, so $V=2\pi r h$.
4. Because the rearranged stepped solid looks like it has base area $\pi r^2$, its volume is $\pi r^2 h$.

Question 15

1. The base is $\pi r$ because it is half of $2\pi r$, so area is $(\pi r)\cdot r$. (correct answer)
2. The base is $2\pi r$ because it matches the full circumference, so area is $(2\pi r)\cdot r$.
3. The height is $\pi r$ because arcs become straight, so area is $r\cdot \pi r$.
4. The area equals the circumference times the diameter, so $A=(2\pi r)(2r)$.

Question 16

1. Stacking slices shows the cone uses one-third the volume of the matching cylinder because slice areas grow from 0 to $\pi r^2$ in the same way each time. (correct answer)
2. A cone has volume $\tfrac13\pi r^2 h$ because that is the standard formula for cones.
3. Because the cone's surface area is one-third the cylinder's surface area, the cone's volume is one-third the cylinder's volume.
4. Because the cone's cross-sections are rectangles, comparing them to the cylinder gives $V=\tfrac13\pi r^2 h$.

Question 17

1. Adding the areas of the unrolled rectangles is like adding circumferences times thickness, which builds up to half of $2\pi r$ times $r$, giving $\pi r^2$. (correct answer)
2. Since each ring has circumference $2\pi r$, the circle's area is $2\pi r$.
3. Because the unrolled rings make a triangle, the area is $\tfrac12 r^2$.
4. Because the diameter is $2r$, the area is $\pi(2r)^2=4\pi r^2$.

Question 18

1. Each disk has area $\pi r^2$, and stacking disks to height h makes volume equal to base area times height, so $V=(\pi r^2)h$. (correct answer)
2. Each disk has circumference $2\pi r$, so stacking to height h gives $V=2\pi r h$.
3. Because the cylinder's surface area is $2\pi r h$, the volume is also $2\pi r h$.
4. Because the disks are circles, the cylinder's volume is $\pi r^2+h$.

Question 19

1. At every height, each cross-section of the cylinder and prism has the same area $\pi r^2$, so the solids have equal volume, giving $V=\pi r^2h$. (correct answer)
2. The cylinder volume is $\pi r^2h$ because that is the memorized cylinder formula, and slicing does not change it.
3. At every height, the cylinder's cross-section is a circle but the prism's is a polygon, so their volumes cannot be compared.
4. Because the cylinder's lateral surface area is $2\pi rh$, its volume must be $2\pi rh$.

Question 20

1. Because each disk has area $\pi r^2$ and the stack reaches height $h$, the total volume matches the cylinder's volume, so $V=\pi r^2h$. (correct answer)
2. Because the cylinder's curved surface can be unrolled into a rectangle, the volume is the rectangle's area $2\pi rh$.
3. Because disks are flat and a cylinder is curved, stacking disks cannot model the cylinder's volume.
4. Because the cylinder formula is $\pi r^2h$, any slicing argument must lead to that formula.