一个纸杯的形状是一个倒置的圆锥。
纸杯顶部的直径是 8 cm。
圆锥的高度是 6 cm。
纸的厚度忽略不计。

1. Find the radius and slant height of the cone.

Top diameter $=8\text{ cm}$, so
$r=\frac{8}{2}=4\text{ cm}.$
Using Pythagoras' theorem,
$l=\sqrt{4^2+6^2}=\sqrt{52}=2\sqrt{13}\text{ cm}.$

2. Use the curved surface area formula for a cone.

Curved surface area
$=\pi r l$
$=\pi(4)(2\sqrt{13}).$

3. Calculate and round the area appropriately.

$\pi(4)(2\sqrt{13})=90.609\ldots$
$\approx 90.6\text{ cm}^2$

1. Explain why half the height does not give half the capacity in a cone.

The water surface is $3\text{ cm}$ below the top, so the height of the water is $6-3=3\text{ cm}$, which is half the full height. However, the cup is a cone, so the radius also changes with the height. Hence the volume is not proportional just to the height; a cone with half the height has less than half the volume.

1. Use the linear scale factor to find the percentage of volume filled.

The water forms a cone similar to the full cup.

Linear scale factor of water cone to full cone:
$\frac{3}{6}=\frac{1}{2}$

Volume scale factor:
$\left(\frac{1}{2}\right)^3=\frac{1}{8}$

Percentage filled:
$\frac{1}{8}\times 100\%=12.5\%$

1. Find the volume of water in the conical cup.

The water surface is $3\text{ cm}$ below the top, so the height of the water-filled cone is

$6-3=3\text{ cm}.$

Since the water-filled cone is similar to the full cone, its radius is half the top radius:

$r=\frac{1}{2}\times 4=2\text{ cm}.$

Volume of water:

$V=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi(2)^2(3)=4\pi\text{ cm}^3.$

2. Equate the water volume to the volume of the hemisphere.

Let the radius of the hemispherical bowl be $R\text{ cm}$.

Volume of hemisphere:

$\frac{2}{3}\pi R^3.$

Since the water fills the hemisphere completely,

$\frac{2}{3}\pi R^3=4\pi.$

3. Solve for the radius and calculate the diameter.

$\frac{2}{3}\pi R^3=4\pi$

Cancel $\pi$:

$\frac{2}{3}R^3=4$

$R^3=6$

$R=\sqrt[3]{6}=1.817\ldots$

Diameter:

$2R=2(1.817\ldots)=3.634\ldots$

$\text{Diameter}=3.63\text{ cm}$