UPSKILL MATH PLUS

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**1**. The diameter of an orange is \(7 \ cm\). Find the surface area of \(50\) oranges.

**Solution**:

Diameter of an orange, \(d\) \(=\) \(7 \ cm\)

Radius of an orange, \(r\) \(=\) $\frac{d}{2}=\frac{7}{2}=3.5\phantom{\rule{0.147em}{0ex}}\mathit{cm}$

Surface area of an orange \(=\) Surface area of a sphere

\(=\) \(4 \pi r^2\) sq. units

\(=\) $4\times \frac{22}{7}\times {\left(3.5\right)}^{2}$

\(=\) $4\times \frac{22}{7}\times 12.25$

\(=\) \(154\)

Surface area of an orange \(=\) \(154\) \(cm^2\).

Surface area of \(50\) oranges:

\(=\) \(154 \times 50\)

\(=\) \(7700\)

**Therefore, the surface area of**\(50\)

**oranges is**\(7700 \ cm^2\).

**2**. The radii of the frustum of a cone are \(6 \ cm\) and \(2 \ cm\), and the height of the cone is \(5 \ cm\). Find the total surface area of a frustum of a cone.

**Solution**:

Let \(R = 6 \ cm\), \(r = 2 \ cm\) and \(h = 5 \ cm\).

Slant height, \(l = \sqrt{h^2 + (R - r)^2}\)

\(l = \sqrt{5^2 + (6 - 2)^2}\)

\(l = \sqrt{25 + 16}\)

\(l = \sqrt{41}\)

\(l = 6.4 \ cm\) (approximately)

Total surface area of frustum of a cone \(=\) \(\pi l(R + r)\) \(+\) \(\pi R^2 + \pi r^2\) sq. units

T. S. A. \(=\) \(\frac{22}{7} \times 6.4 (6 + 2)\) \(+\) \(\frac{22}{7} (6^2 + 2^2)\)

\(=\) \(\frac{22}{7} \times 6.4 \times 8\) \(+\) \(\frac{22}{7} (36 + 4)\)

\(=\) \(\frac{22}{7} \times 51.2\) \(+\) \(\frac{22}{7} \times 40\)

\(=\) \(\frac{22}{7} \times (51.2 + 40)\)

\(=\) \(\frac{22}{7} \times 91.2\)

\(=\) \(286.63\) (approximately)

**Therefore, the total surface area of the frustum of a cone is**\(286.63 \ cm^2\).

Important!

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