Question

check your understanding

1. draw a cone with a height of 12 cm and a radius of 5 cm. find the volume.
2. draw a sphere with a radius of 10 ft. find the volume.
3. a. draw a cylinder with a height of 9 in. and a diameter of 4 in. find the volume.

b. draw a cone with a height of 9 in. and a diameter of 4 in. find the volume.
c. how many times the volume of the cone is the volume of the cylinder?
d. state a rule for the relating the volumes of a cylinder and cone that have the same height and diameter.
e. would your rule also apply to the volumes of a cylinder and cone that have the same height and radius?

algebra, unit 3 • geometry

Explanation:

Step1: Solve Q6: Cone volume

Formula: $V_{cone}=\frac{1}{3}\pi r^2 h$
$r=5$ cm, $h=12$ cm
$\frac{1}{3} \times \pi \times 5^2 \times 12 = \frac{1}{3} \times \pi \times 25 \times 12 = 100\pi \approx 314.16$ cm³

Step2: Solve Q7: Sphere volume

Formula: $V_{sphere}=\frac{4}{3}\pi r^3$
$r=10$ ft
$\frac{4}{3} \times \pi \times 10^3 = \frac{4000}{3}\pi \approx 4188.79$ ft³

Step3: Solve Q8a: Cylinder volume

Radius $r=\frac{4}{2}=2$ in, $h=9$ in
Formula: $V_{cylinder}=\pi r^2 h$
$\pi \times 2^2 \times 9 = 36\pi \approx 113.10$ in³

Step4: Solve Q8b: Cone volume

Same $r=2$ in, $h=9$ in
$V_{cone}=\frac{1}{3}\pi r^2 h$
$\frac{1}{3} \times \pi \times 2^2 \times 9 = 12\pi \approx 37.70$ in³

Step5: Solve Q8c: Volume ratio

$\frac{V_{cylinder}}{V_{cone}}=\frac{36\pi}{12\pi}=3$

Step6: Solve Q8d: State volume rule

Cylinder volume = 3 × cone volume (same height/diameter)

Step7: Solve Q8e: Rule applicability

Yes, same radius = same diameter, so rule holds.

Answer:

1. $\boldsymbol{100\pi \approx 314.16}$ cubic centimeters
2. $\boldsymbol{\frac{4000}{3}\pi \approx 4188.79}$ cubic feet
3. a. $\boldsymbol{36\pi \approx 113.10}$ cubic inches

b. $\boldsymbol{12\pi \approx 37.70}$ cubic inches
c. $\boldsymbol{3}$
d. The volume of a cylinder is 3 times the volume of a cone with the same height and diameter.
e. Yes, the rule applies, since same radius means same diameter, so the relationship between the volumes remains the same.