Question

Given: R=2m

Answer 1

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

Calculating Volume

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 -  $\frac{1}{4}L$²

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

Calculating Surface Area

Surface Area is calculated using the following formula

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

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

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

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

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

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

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

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

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

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