Example problem of a right circular cone — lesson. Mathematics CBSE, Class 9.

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A diameter of the wafer cone is \(10 \ cm\) and, its height is \(12 \ cm\). Calculate the curved surface area of \(20\) such wafer cones.

Solution:

Diameter of the cone \((d)\) \(=\) \(10 \ cm\)

Radius of the cone \((r)\) \(=\)

\frac{d}{2} = \frac{10}{2} = 5

\(cm\)

Height of the cone \((h)\) \(=\) \(12 \ cm\)

Let us first find the slant height of the cone.

\(l = \sqrt{r^2 + h^2}\)

\(l = \sqrt{5^2 + 12^2}\)

\(l = \sqrt{25 + 144}\)

\(l = \sqrt{169}\)

\(l = 13\) \(cm\)

Curved surface area of the cone \(=\) \(\pi r l\) sq. units

\(=\)

\frac{22}{7} \times 5 \times 13

\(=\)

\frac{1430}{7}

\(=\) \(204.28\) \(cm^2\)

Curved surface area of a cone is \(204.28 \ cm^2\).

Curved surface area of \(20\) cones:

\(=\) \(20 \times 204.28\)

\(=\) \(4085.6\)

Therefore, the curved surface area of \(20\) wafer cones is \(4085.6 \ cm^2\).

Important!

The value of \(\pi\) should be taken as \(\frac{22}{7}\) unless its value shared in the problem.

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2. Example problem of a right circular cone