The perimeters of the ends of a frustum of a right circular

Question:

The perimeters of the ends of a frustum of a right circular cone are 44 cm and 33 cm. If the height of the frustum be 16 cm, find its volume, the slant surface and the total surface.

Solution:

The height of the frustum of the cone is = 16 cm. The perimeters of the circular ends are 44 cm and 33 cm. Let the radii of the bottom and top circles are r1 cm and r2 cm respectively. Then, we have

$2 \pi r_{1}=44$

$\Rightarrow \pi r_{1}=22$

$\Rightarrow \quad r_{1}=\frac{22 \times 7}{22}$

$\Rightarrow \quad r_{1}=7$

$2 \pi r_{2}=33$

$\Rightarrow \pi r_{2}=\frac{33}{2}$

$\Rightarrow r_{2}=\frac{33}{2} \times \frac{7}{22}$

$\Rightarrow r_{2}=\frac{21}{4}$

The slant height of the bucket is

$l=\sqrt{\left(r_{1}-r_{2}\right)^{2}+h^{2}}$

$=\sqrt{\left(7-\frac{21}{4}\right)^{2}+16^{2}}$

$=16.1 \mathrm{~cm}$

The curved/slant surface area of the frustum cone is

$=\pi\left(r_{1}+r_{2}\right) \times l$

$=\left(\pi r_{1}+\pi r_{2}\right) \times l$

$=(22+16.5) \times 16.1$

$=619.85 \mathrm{~cm}^{2}$

Hence Curved surface area $=619.85 \mathrm{~cm}^{2}$

The volume of the frustum of the cone is

$V=\frac{1}{3} \pi\left(r_{1}^{2}+r_{1} r_{2}+r_{2}^{2}\right) \times h$

$=\frac{1}{3} \pi\left(7^{2}+7 \times 5.25+5.25^{2}\right) \times 16$

$=1900 \mathrm{~cm}^{3}$

Hence Volume of frustum $=1900 \mathrm{~cm}^{3}$

The total surface area of the frustum cone is

$=\pi\left(r_{1}+r_{2}\right) \times l+\pi r_{1}^{2}+\pi r_{2}^{2}$

$=619.85+\frac{22}{7} \times 7^{2}+\frac{22}{7} \times 5.25^{2}$

$=860.25$ Square $\mathrm{cm}$

Hence Total surface area $=860.25 \mathrm{~cm}^{2}$

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