Ex.13.7 Q8 Surface Areas and Volumes Solution - NCERT Maths Class 9


Question

If the triangle \(ABC\) in the Question 7 above is revolved about the side \(5 \rm\,rm \) then find the volume of the solid so obtained. Find also the ratio of the volumes of the two solids obtained in Questions 7 and 8.

 Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-7 | Question 8

Text Solution

Reasoning:

Volume of the right circular cone is \(\begin{align}\frac{1}{3} \end{align}\)time of the volume of a cylinder\(\begin{align} = \frac{1}{3}\pi {r^2}h \end{align}\).

What is  known?

Sides of the right triangle.

What is  unknown?

Volume of the cone and ratio between the volumes.

Solution:

Radius of the cone \(= 12\rm\, cm\)

Height of the cone \(= 5 \rm\,cm\)

Volume of cone 

\[\begin{align}&= \frac{1}{3}\pi {r^2}h \\ &= \frac{1}{3} \times \pi \times {(12)^2} \times 5 \\ & = 240\pi \,\,\,\rm\,c{m^3} \end{align}\]

Volume in question \(7\)\(\begin{align} = 100\pi \,\,\rm\,c{m^3} \end{align}\)

Ratio \(\begin{align}100:240 = 5:12 \end{align}\)

Answer:

Volume of the cone is \(\begin{align} = 240\pi \,\,\rm\,c{m^3} \end{align}\)

Ratio between the volume \(\begin{align}=5:12 \end{align}\)

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