Question

1. The lateral edge of a regular hexagonal pyramid is two times the length of the base edge. If the apothem of the base is 8 cm, find the altitude and the volume of a cone inscribed in the pyramid.

2. If the diameter of the base remains constant, by what factor should the altitude be multiplied to produce a cone with twice volume as the original.

3.If the altitude of a cone remains constant, by what factor should the diameter be multiplied in order to construct a cone with a volume that is triple the original.

Answer 1

Answer on Question #58134 – Math – Geometry

Question

1. The lateral edge of a regular hexagonal pyramid is two times the length of the base edge. If the apothem of the base is $8\mathrm{cm}$, find the altitude and the volume of a cone inscribed in the pyramid.

Solution

Base edge $a = \frac{apothem}{cos30^0} = \frac{16}{\sqrt{3}}$;

Lateral edge $l = \frac{16}{\sqrt{3}} * 2 = \frac{32}{\sqrt{3}}$;

radius of the cone $r = apothem = 8$;

altitude of the cone $h = \sqrt{l^2 - r^2} = \sqrt{\frac{4*16^2}{3} - \frac{16^2}{3}} = 16 \, \text{cm}$;

volume of a cone is

$$V = \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi * 8^2 * 16 = \frac{1024\pi}{3} \approx 1072.33 \, \text{cm}^3.$$

Answer: $16 \, \text{cm}, 1072.33 \, \text{cm}^3$.

Question

2. If the diameter of the base remains constant, by what factor should the altitude be multiplied to produce a cone with twice volume as the original.

Solution

$$V = \frac{1}{3} \pi r^2 h \rightarrow 2V = \frac{1}{3} \pi r^2 (ah) \rightarrow \frac{1}{3} \pi r^2 h = \frac{1}{6} \pi r^2 (ah) \rightarrow a = 2;$$

altitude should be doubled.

Answer: 2.

Question

3. If the altitude of a cone remains constant, by what factor should the diameter be multiplied in order to construct a cone with a volume that is triple the original.

Solution

$$V = \frac{1}{3} \pi r^2 h \rightarrow 3V = \frac{1}{3} \pi (ar)^2 h \rightarrow \frac{1}{3} \pi r^2 h = \frac{1}{9} \pi (ar)^2 h \rightarrow a = \sqrt{3}.$$

Radius (and diameter) should be multiplied by $\sqrt{3}$.

Answer: $\sqrt{3}$.

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