Example problems of a hemisphere and hollow hemisphere — lesson. Mathematics State Board, Class 10.

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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,

$r$

$=$

$\frac{d}{2} = \frac{8}{2} = 4 cm$

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

$=$

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

$=$

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

$=$

$2 \times \frac{22}{7} \times 34$

$=$

$213.7$

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.

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5. Example problems of a hemisphere and hollow hemisphere

Theory: