Question

Given: R=2m

KL = LM = KM

Find: V and

Surface Area of the cone

Answer 1

`V = (2L√(3))/(3) \pi}`

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

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 = (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 =
`(1)/(2) KL`

OK =
`(1)/(2)L`

Applying pythagoras theorem in triangle LOM,

|LM|² = |OL|² + |OM|²

By substitution

L² = H² + (
`(1)/(2)L`)²

H² = L² -
`(1)/(4)L`²

H² = L² (1 -
`(1)/(4)`)

H² = L²
`(3)/(4)`

H² =
`(3L^(2) )/(4)`

Take square root of bot sides

`H = \sqrt{(3L^(2) )/(4)}`

`H = (L√(3))/(2)`

Recall that
`V = (1)/(3) \pi R^(2) H`

`V = (1)/(3) \pi 2^(2) * (L√(3))/(2)`

`V = (1)/(3) \pi * 4} * (L√(3))/(2)`

`V = (1)/(3) \pi} * {2L√(3)}`

`V = (2L√(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 = (L√(3))/(4)`

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

`A=\pi * 2(2+\sqrt{(((L√(3))/(2))^(2) +2^(2) } )`)

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

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

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

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

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

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

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