 UPSKILL MATH PLUS

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### Theory:

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

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

$$=$$ $4×\frac{22}{7}×12.25$

$$=$$ $$154$$

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.