Answer to Question #147373 in Geometry for solid mensuration

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?
1
Expert's answer
2020-11-30T12:57:13-0500

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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