Example problems of a sphere and frustum of a cone — lesson. Mathematics State Board, Class 10.

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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,

$r$

$=$

$\frac{d}{2} = \frac{7}{2} = 3.5 cm$

Surface area of an orange

$=$ Surface area of a sphere

$=$

$4 \pi r^2$ sq. units

$=$

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

$=$

$4 \times \frac{22}{7} \times 12.25$

$=$

$154$

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.

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4. Example problems of a sphere and frustum of a cone

Theory: