The diagonal d d d of the cube with the side a a a is
d 2 = a 2 + a 2 + a 2 = 3 a 2 d^2=a^2+a^2+a^2=3a^2 d 2 = a 2 + a 2 + a 2 = 3 a 2
d = 3 a d=\sqrt{3}a d = 3 a Given d = 20 3 c m . d=20\sqrt{3}\ cm. d = 20 3 c m .
Then
a = d 3 = 20 3 3 = 20 ( c m ) a=\dfrac{d}{\sqrt{3}}=\dfrac{20\sqrt{3}}{\sqrt{3}}=20(cm) a = 3 d = 3 20 3 = 20 ( c m ) The radius r r r of the circle inscribed in the square with the side a a a is
r = a 2 r=\dfrac{a}{2} r = 2 a Then we have the right circular cone inscribed in a cube with the side a a a
r a d i u s = r = a 2 , h e i g h t = h = a radius=r=\dfrac{a}{2}, \ height=h=a r a d i u s = r = 2 a , h e i g h t = h = a Slant height L L L is
L = r 2 + h 2 = ( a 2 ) 2 + a 2 = 5 a 2 L=\sqrt{r^2+h^2}=\sqrt{(\dfrac{a}{2})^2+a^2}=\dfrac{\sqrt{5}a}{2} L = r 2 + h 2 = ( 2 a ) 2 + a 2 = 2 5 a 3. Surface area of a cone = Base Area + Curved Surface Area of a cone
A = π r 2 + π r L A=\pi r^2+\pi rL A = π r 2 + π r L
A = π ( ( a 2 ) 2 + a 2 ( 5 a 2 ) ) = π a 2 4 ( 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 (( 2 a ) 2 + 2 a ( 2 5 a )) = 4 π a 2 ( 1 + 5 )
A = π ( 20 ) 2 4 ( 1 + 5 ) = 100 π ( 1 + 5 ) ( c m 2 ) A=\dfrac{\pi (20)^2}{4}(1+\sqrt{5})=100\pi(1+\sqrt{5}) (cm^2) A = 4 π ( 20 ) 2 ( 1 + 5 ) = 100 π ( 1 + 5 ) ( c m 2 )
≈ 1016.64 ( c m 2 ) \approx1016.64(cm^2) ≈ 1016.64 ( c m 2 ) The surface area of the cone is 100 π ( 1 + 5 ) c m 2 ≈ 1016.64 c m 2 . 100\pi(1+\sqrt{5})\ cm^2\approx1016.64\ cm^2. 100 π ( 1 + 5 ) c m 2 ≈ 1016.64 c m 2 .
4. The volume of the cube is
V c u b e = a 3 V_{cube}=a^3 V c u b e = a 3 The volume of the cone is
V c o n e = 1 3 π r 2 h = 1 3 π ( a 2 ) 2 ( a ) = π a 3 12 V_{cone}=\dfrac{1}{3}\pi r^2h=\dfrac{1}{3}\pi (\dfrac{a}{2})^2 (a)=\dfrac{\pi a^3}{12} V co n e = 3 1 π r 2 h = 3 1 π ( 2 a ) 2 ( a ) = 12 π a 3 The volume of the space between the cone and the cube is
V s p a c e = V c u b e − V c o n e V_{space}=V_{cube}-V_{cone} V s p a ce = V c u b e − V co n e
= a 3 − π a 3 12 = a 3 12 ( 12 − π ) =a^3-\dfrac{\pi a^3}{12}=\dfrac{ a^3}{12}(12-\pi) = a 3 − 12 π a 3 = 12 a 3 ( 12 − π )
V s p a c e = ( 20 ) 3 12 ( 12 − π ) = 2000 ( 12 − π ) 3 ( c m 3 ) V_{space}=\dfrac{(20)^3}{12}(12-\pi)=\dfrac{2000(12-\pi)}{3}(cm^3) V s p a ce = 12 ( 20 ) 3 ( 12 − π ) = 3 2000 ( 12 − π ) ( c m 3 )
≈ 5905.605 ( c m 3 ) \approx5905.605(cm^3) ≈ 5905.605 ( c m 3 ) The volume of the space between the cone and the cube is
2000 ( 12 − π ) 3 c m 3 ≈ 5905.605 c m 3 . \dfrac{2000(12-\pi)}{3}\ cm^3\approx 5905.605 \ cm^3. 3 2000 ( 12 − π ) c m 3 ≈ 5905.605 c m 3 .
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