Answer:
[tex]V = \frac{2L\sqrt{3}}{3} \pi}[/tex]
[tex]A = 4\pi + \sqrt{3L^{2} + 16}[/tex]
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
[tex]V = \frac{1}{3} \pi R^{2} H[/tex]
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 = [tex]\frac{1}{2} KL[/tex]
OK = [tex]\frac{1}{2}L[/tex]
Applying pythagoras theorem in triangle LOM,
|LM|² = |OL|² + |OM|²
By substitution
L² = H² + ( [tex]\frac{1}{2}L[/tex])²
H² = L² - [tex]\frac{1}{4}L[/tex]²
H² = L² (1 - [tex]\frac{1}{4}[/tex])
H² = L² [tex]\frac{3}{4}[/tex]
H² = [tex]\frac{3L^{2} }{4}[/tex]
Take square root of bot sides
[tex]H = \sqrt{\frac{3L^{2} }{4}}[/tex]
[tex]H = \frac{L\sqrt{3}}{2}[/tex]
Recall that [tex]V = \frac{1}{3} \pi R^{2} H[/tex]
[tex]V = \frac{1}{3} \pi 2^{2} * \frac{L\sqrt{3}}{2}[/tex]
[tex]V = \frac{1}{3} \pi * 4} * \frac{L\sqrt{3}}{2}[/tex]
[tex]V = \frac{1}{3} \pi} * {2L\sqrt{3}}[/tex]
[tex]V = \frac{2L\sqrt{3}}{3} \pi}[/tex]
in terms of [tex]\pi[/tex] an d L where L = KL = LM = KM
Calculating Surface Area
Surface Area is calculated using the following formula
[tex]H = \frac{L\sqrt{3}}{4}[/tex]
[tex]A=\pi r(r+\sqrt{h^{2} +r^{2} } )[/tex]
[tex]A=\pi * 2(2+\sqrt{((\frac{L\sqrt{3}}{2})^{2} +2^{2} } )[/tex])
[tex]A=\pi * 2(2+\sqrt{{\frac{3L^{2}}{4} } + 4 }[/tex] )
[tex]A=\pi * 2(2+\sqrt{{\frac{3L^{2}+16}{4} } })[/tex]
[tex]A=2\pi(2+\sqrt{{\frac{3L^{2}+16}{4} } })[/tex]
[tex]A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{\sqrt{4}} )[/tex]
[tex]A = 2\pi (2 + \frac{\sqrt{3L^{2} + 16}}{2} )[/tex]
[tex]A = 2\pi (2 + {\frac{1}{2} \sqrt{3L^{2} + 16})[/tex]
[tex]A = 4\pi + \sqrt{3L^{2} + 16}[/tex]
