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Pls help


A right circular cone has a lateral surface area of 188.50 square inches. If its slant height lis 10 inches, then what is the volume of the cone in cubic inches?


Sagot :

Answer:

169.04 in² (nearest hundredth)

Step-by-step explanation:

Surface area of a cone = [tex]\pi[/tex]r² + [tex]\pi[/tex]r[tex]l[/tex]

(where r = radius of the base and [tex]l[/tex] = slant height)

Given slant height [tex]l[/tex] = 10 and surface area = 188.5

Surface area  = [tex]\pi[/tex]r² + [tex]\pi[/tex]r[tex]l[/tex]

188.5 = [tex]\pi[/tex]r² + 10[tex]\pi[/tex]r

[tex]\pi[/tex]r² + 10[tex]\pi[/tex]r - 188.5 = 0

r = [tex]\frac{-10\pi +\sqrt{(10\pi )^2-(4\times\pi \times-188.5)} }{2\pi }[/tex] = 4.219621117...

Volume of a cone = (1/3)[tex]\pi[/tex]r²h

(where r = radius of the base and h = height)

We need to find an expression for h in terms of [tex]l[/tex] using Pythagoras' Theorem a² + b² = c², where a = radius, b = height and c = slant height

r² + h² = [tex]l[/tex]²

h² = [tex]l[/tex]² - r²

h = √([tex]l[/tex]² - r²)

Therefore, substituting found expression for h:

volume of a cone = (1/3)[tex]\pi[/tex]r²√([tex]l[/tex]² - r²)

Given slant height [tex]l[/tex] = 10 and r = 4.219621117...

volume = 169.0431969... = 169.04 in² (nearest hundredth)

Check the picture below.

notice, the slant-height is simply the hypotenuse from the pythagorean theorem.

[tex]188.5=\stackrel{\textit{lateral area}}{\pi r\sqrt{r^2+h^2}}\implies 188.5=\pi r10\implies \cfrac{188.5}{10\pi }=r\implies \boxed{\cfrac{18.85}{\pi }=r} \\\\[-0.35em] ~\dotfill\\\\ h^2=(slant-height)^2-r^2\implies h=\sqrt{(slant-height)^2-r^2} \\\\\\ h=\sqrt{10^2-\cfrac{18.85^2}{\pi^2}}\implies h=\sqrt{\cfrac{100\pi^2-18.85^2}{\pi^2}}\implies \boxed{h=\cfrac{\sqrt{100\pi^2-18.85^2}}{\pi }} \\\\[-0.35em] ~\dotfill[/tex]

[tex]V=\cfrac{\pi }{3}\left( \cfrac{18.85}{\pi } \right)^2\left( \cfrac{\sqrt{100\pi^2-18.85^2}}{\pi } \right)\implies h\approx 301.6031084547~in^3[/tex]

View image jdoe0001