UPSKILL MATH PLUS

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**1**. The diameter of an orange is \(8 \ cm\). Calculate the total surface area of the half orange.

**Solution**:

Diameter of an orange, \(d\) \(=\) \(8 \ cm\)

Radius of an orange, \(r\) \(=\) $\frac{d}{2}=\frac{8}{2}=4\phantom{\rule{0.147em}{0ex}}\mathit{cm}$

A half orange is in the shape of a hemisphere.

Total surface area of a hemisphere \(=\) \(3 \pi r^2\) sq. units

\(=\) $3\times \frac{22}{7}\times {4}^{2}$

\(=\) $3\times \frac{22}{7}\times 16$

\(=\) \(150.86\)

**The total surface area of the half orange is**\(150.86 \ cm^2\).

**2**. If the inner and outer radius of the hemispherical shell is \(3 \ cm\) and \(5 \ cm\), find the thickness and the curved surface area of the shell.

**Solution**:

Inner radius, \(r\) \(=\) \(3 \ cm\)

Outer radius, \(R\) \(=\) \(5 \ cm\)

Thickness \(=\) \(R - r\)

\(=\) \(5 - 3 = 2\)

**The thickness of the shell is**\(2 \ cm\).

Curved surface area \(=\) \(2 \pi (R^2 + r^2)\) sq. units

\(=\) $2\times \frac{22}{7}\times \left({5}^{2}+{3}^{2}\right)$

\(=\) $2\times \frac{22}{7}\times \left(25+9\right)$

\(=\) $2\times \frac{22}{7}\times 34$

\(=\) \(213.7\)

**The curved surface area of the hemispherical shell is**\(213.7\) \(cm^2\).

Important!

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