UPSKILL MATH PLUS

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The 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

\(=\) $\frac{22}{7}\times 5\times 13$

\(=\) $\frac{1430}{7}$

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

**The 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 is shared in the problem.