Question #147373

A right circular cone is inscribed in a cube having a diagonal which

measures 20 square root of 3 cm.


3. What is the surface area of the cone?

4. What is the volume of the space between the cone and the cube?

Expert's answer

The diagonal dd of the cube with the side aa is


d2=a2+a2+a2=3a2d^2=a^2+a^2+a^2=3a^2

d=3ad=\sqrt{3}a

Given d=203 cm.d=20\sqrt{3}\ cm.

Then


a=d3=2033=20(cm)a=\dfrac{d}{\sqrt{3}}=\dfrac{20\sqrt{3}}{\sqrt{3}}=20(cm)

The radius rr of the circle inscribed in the square with the side aa is


r=a2r=\dfrac{a}{2}

Then we have the right circular cone inscribed in a cube with the side aa


radius=r=a2, height=h=aradius=r=\dfrac{a}{2}, \ height=h=a

Slant height LL is


L=r2+h2=(a2)2+a2=5a2L=\sqrt{r^2+h^2}=\sqrt{(\dfrac{a}{2})^2+a^2}=\dfrac{\sqrt{5}a}{2}

3. Surface area of a cone = Base Area + Curved Surface Area of a cone


A=πr2+πrLA=\pi r^2+\pi rL

A=π((a2)2+a2(5a2))=πa24(1+5)A=\pi((\dfrac{a}{2})^2+\dfrac{a}{2}(\dfrac{\sqrt{5}a}{2}))=\dfrac{\pi a^2}{4}(1+\sqrt{5})

A=π(20)24(1+5)=100π(1+5)(cm2)A=\dfrac{\pi (20)^2}{4}(1+\sqrt{5})=100\pi(1+\sqrt{5}) (cm^2)

1016.64(cm2)\approx1016.64(cm^2)

The surface area of the cone is 100π(1+5) cm21016.64 cm2.100\pi(1+\sqrt{5})\ cm^2\approx1016.64\ cm^2.


4. The volume of the cube is


Vcube=a3V_{cube}=a^3

The volume of the cone is


Vcone=13πr2h=13π(a2)2(a)=πa312V_{cone}=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi (\dfrac{a}{2})^2 (a)=\dfrac{\pi a^3}{12}

The volume of the space between the cone and the cube is


Vspace=VcubeVconeV_{space}=V_{cube}-V_{cone}

=a3πa312=a312(12π)=a^3-\dfrac{\pi a^3}{12}=\dfrac{ a^3}{12}(12-\pi)

Vspace=(20)312(12π)=2000(12π)3(cm3)V_{space}=\dfrac{(20)^3}{12}(12-\pi)=\dfrac{2000(12-\pi)}{3}(cm^3)

5905.605(cm3)\approx5905.605(cm^3)

The volume of the space between the cone and the cube is

2000(12π)3 cm35905.605 cm3.\dfrac{2000(12-\pi)}{3}\ cm^3\approx 5905.605 \ cm^3.



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