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lys-0071 [83]
3 years ago
14

Given: R=2m

Mathematics
1 answer:
mr_godi [17]3 years ago
6 0

Answer:

V = \frac{2L\sqrt{3}}{3} \pi}

A = 4\pi +  \sqrt{3L^{2} + 16}

Step-by-step explanation:

Figure of cone is missing. See attachment

Given

Radius, R = 2m

Let L = KL=LM=KM

Required:

Volume, V and Surface Area, A

<u><em>Calculating Volume</em></u>

Volume is calculated using the following formula

V = \frac{1}{3} \pi R^{2} H

Where R is the radius of the cone and H is the height

First, we need to determine the height of the cone

The height is represented by length OL

It is given that KL=LM=KM in triangle KLM

This means that this triangle is an equilateral triangle

where OM = OK = \frac{1}{2} KL

OK = \frac{1}{2}L

Applying pythagoras theorem in triangle LOM,

|LM|² =  |OL|² + |OM|²

By substitution

L² = H² + ( \frac{1}{2}L)²

H² = L² -  \frac{1}{4}L²

H² = L² (1 -  \frac{1}{4})

H² = L² \frac{3}{4}

H² = \frac{3L^{2} }{4}

Take square root of bot sides

H = \sqrt{\frac{3L^{2} }{4}}

H = \frac{L\sqrt{3}}{2}

Recall that V = \frac{1}{3} \pi R^{2} H

V = \frac{1}{3} \pi 2^{2} * \frac{L\sqrt{3}}{2}

V = \frac{1}{3} \pi * 4} * \frac{L\sqrt{3}}{2}

V = \frac{1}{3} \pi} * {2L\sqrt{3}}

V = \frac{2L\sqrt{3}}{3} \pi}

in terms of \pi an d L where L = KL = LM = KM

<u><em>Calculating Surface Area</em></u>

Surface Area is calculated using the following formula

H = \frac{L\sqrt{3}}{4}

A=\pi r(r+\sqrt{h^{2} +r^{2} } )

A=\pi * 2(2+\sqrt{((\frac{L\sqrt{3}}{2})^{2} +2^{2} } ))

A=\pi * 2(2+\sqrt{{\frac{3L^{2}}{4} } + 4 } )

A=\pi * 2(2+\sqrt{{\frac{3L^{2}+16}{4} } })

A=2\pi(2+\sqrt{{\frac{3L^{2}+16}{4} } })

A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{\sqrt{4}} )

A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{2} )

A = 2\pi (2 + {\frac{1}{2} \sqrt{3L^{2} + 16})

A = 4\pi +  \sqrt{3L^{2} + 16}

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