Question

A right triangle ABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 5 cm, then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained.

Hint:

r = 5 cm
l = 13 cm
h = 12 cm
Volume = \(\frac 13 \pi r^2 h\)
= \(\frac 13 \pi \times 5^2 \times 12\)
= \(100\pi\) cm³

r = 12 cm
l = 13 cm
h = 5 cm
Volume = \(\frac 13 \pi r^2 h\)
= \(\frac 13 \pi \times {12}^2 \times 5\)
= \(240\pi\) cm³

Ratio = \(\frac {100\pi}{240\pi}\) = \(\frac {10}{24}\) =\( \frac {5}{12}\)= \(5:12\)

Answer 1

Step 1: Volume of the Solid Obtained by Revolving the Triangle About \( AB \) (5 cm)

When the right triangle is revolved about the side \( AB \), the solid formed is a cone. The radius of the cone is \( BC = 12 \, \text{cm} \), and the height of the cone is \( AB = 5 \, \text{cm} \). The formula for the volume of a cone is: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] where \( r \) is the radius and \( h \) is the height of the cone. Substituting the values \( r = 12 \, \text{cm} \) and \( h = 5 \, \text{cm} \): \[ V_{\text{cone}} = \frac{1}{3} \pi (12)^2 \times 5 = \frac{1}{3} \pi \times 144 \times 5 = \frac{1}{3} \pi \times 720 = 240 \pi \, \text{cm}^3 \] Therefore, the volume of the cone is \( 240 \pi \, \text{cm}^3 \).

Step 2: Volume of the Solid Obtained by Revolving the Triangle About \( BC \) (12 cm)

When the right triangle is revolved about the side \( BC \), the solid formed is a cone again. The radius of the cone is \( AB = 5 \, \text{cm} \), and the height of the cone is \( BC = 12 \, \text{cm} \). Using the same formula for the volume of a cone: \[ V_{\text{cone}} = \frac{1}{3} \pi r^2 h \] Substituting the values \( r = 5 \, \text{cm} \) and \( h = 12 \, \text{cm} \): \[ V_{\text{cone}} = \frac{1}{3} \pi (5)^2 \times 12 = \frac{1}{3} \pi \times 25 \times 12 = \frac{1}{3} \pi \times 300 = 100 \pi \, \text{cm}^3 \] Therefore, the volume of the cone is \( 100 \pi \, \text{cm}^3 \).

Step 3: Ratio of the Volumes

The ratio of the volumes of the two solids is: \[ \frac{V_{\text{cone about AB}}}{V_{\text{cone about BC}}} = \frac{240 \pi}{100 \pi} = \frac{240}{100} = \frac{12}{5} \] Therefore, the ratio of the volumes is \( \frac{12}{5} \).

Conclusion:

The volumes of the solids obtained are:

• Volume when revolved about \( AB = 240 \pi \, \text{cm}^3 \)
• Volume when revolved about \( BC = 100 \pi \, \text{cm}^3 \)

The ratio of the volumes is \( \frac{12}{5} \).