Convert €65.90 to pounds.
Give your answer rounded to 2 DP.
Answers
Answer:
\large\underline{\sf{Solution-}}
Solution−
Given that,
Diameter of cone = 3.5 cm = 7/2 cm
So, radius of cone, r = 7/4 cm
Height of cone, h = 3 cm
We know, Volume of cone of radius r and height h is given by
\begin{gathered}\boxed{\rm{ \:Volume_{(cone)} \: = \: \frac{\pi}{3} \: {r}^{2} \: h \: \: }} \\ \end{gathered}
Volume
(cone)
=
3
π
r
2
h
So, using this result, we have
\begin{gathered}\rm \: \:Volume_{(1 \: cone)} \: = \: \frac{\pi}{3} \: {r}^{2} \: h \: \: \\ \end{gathered}
Volume
(1cone)
=
3
π
r
2
h
So,
\begin{gathered}\rm \: \:Volume_{(504 \: cones)} \: = \: 504 \times \frac{\pi}{3} \: {r}^{2} \: h \: \: \\ \end{gathered}
Volume
(504cones)
=504×
3
π
r
2
h
\begin{gathered}\rm \: \:Volume_{(504 \: cones)} \: = \: 168 \:\pi \: {r}^{2} \: h \: - - - (1) \: \\ \end{gathered}
Volume
(504cones)
=168πr
2
h−−−(1)
Now, it is given that,
504 cones, each of diameter 3.5 cm and height 3 cm are melted and recast into a metallic sphere.
Let assume that radius of sphere be R cm.
So, as 504 cones melted and recast in to a metallic sphere.
\begin{gathered}\rm \: Volume_{(504 \: cones)} = Volume_{(sphere)} \\ \end{gathered}
Volume
(504cones)
=Volume
(sphere)
\begin{gathered}\rm \: 168\pi \: {r}^{2}h \: = \: \dfrac{4}{3}\pi \: {R}^{3} \\ \end{gathered}
168πr
2
h=
3
4
πR
3
\begin{gathered}\rm \: 42\pi \: {r}^{2}h \: = \: \dfrac{1}{3}\pi \: {R}^{3} \\ \end{gathered}
42πr
2
h=
3
1
πR
3
\begin{gathered}\rm \:3 \times 42\: \times \frac{7}{4} \times \frac{7}{4} \times 3 \: = \: {R}^{3} \\ \end{gathered}
3×42×
4
7
×
4
7
×3=R
3
\begin{gathered}\rm \:3 \times 21\: \times \frac{7}{2} \times \frac{7}{4} \times 3 \: = \: {R}^{3} \\ \end{gathered}
3×21×
2
7
×
4
7
×3=R
3
\begin{gathered}\rm \:3 \times 7 \times 3\: \times \frac{7}{2} \times \frac{7}{2 \times 2} \times 3 \: = \: {R}^{3} \\ \end{gathered}
3×7×3×
2
7
×
2×2
7
×3=R
3
\begin{gathered}\rm \: {\bigg(\dfrac{21}{2} \bigg) }^{3} \: = \: {R}^{3} \\ \end{gathered}
(
2
21
)
3
=R
3
\begin{gathered}\rm\implies \:R = \dfrac{21}{2} \: cm \\ \end{gathered}
⟹R=
2
21
cm
Hence,
\begin{gathered}\rm\implies \:Diameter \: of \: sphere \: = \: 21 \: cm \\ \end{gathered}
⟹Diameterofsphere=21cm
Now,
\begin{gathered}\rm \: Surface \: Area_{(sphere)} \\ \end{gathered}
SurfaceArea
(sphere)
\begin{gathered}\rm \: = \:4 \: \pi \: {r}^{2} \\ \end{gathered}
= 4πr
2
\begin{gathered}\rm \: = \:4 \times \dfrac{22}{7} \times \dfrac{21}{2} \times \dfrac{21}{2} \\ \end{gathered}
= 4×
7
22
×
2
21
×
2
21
\begin{gathered}\rm \: = \:22 \times 3 \times 21 \\ \end{gathered}
= 22×3×21
\begin{gathered}\rm \: = \:1386 \: {cm}^{2} \\ \end{gathered}
= 1386cm
2
\begin{gathered}\rm\implies \:\rm \: Surface \: Area_{(sphere)} \: = \: 1386 \: {cm}^{2} \\ \end{gathered}
⟹SurfaceArea
(sphere)
=1386cm
2
\rule{190pt}{2pt}
Additional Information
\begin{gathered}\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{ \red{More \: Formulae}}}} \\ \\ \bigstar \: \bf{CSA_{(cylinder)} = 2\pi \: rh}\\ \\ \bigstar \: \bf{Volume_{(cylinder)} = \pi {r}^{2} h}\\ \\ \bigstar \: \bf{TSA_{(cylinder)} = 2\pi \: r(r + h)}\\ \\ \bigstar \: \bf{CSA_{(cone)} = \pi \: r \: l}\\ \\ \bigstar \: \bf{TSA_{(cone)} = \pi \: r \: (l + r)}\\ \\ \bigstar \: \bf{Volume_{(sphere)} = \dfrac{4}{3}\pi {r}^{3} }\\ \\ \bigstar \: \bf{Volume_{(cube)} = {(side)}^{3} }\\ \\ \bigstar \: \bf{CSA_{(cube)} = 4 {(side)}^{2} }\\ \\ \bigstar \: \bf{TSA_{(cube)} = 6 {(side)}^{2} }\\ \\ \bigstar \: \bf{Volume_{(cuboid)} = lbh}\\ \\ \bigstar \: \bf{CSA_{(cuboid)} = 2(l + b)h}\\ \\ \bigstar \: \bf{TSA_{(cuboid)} = 2(lb +bh+hl )}\\ \: \end{array} }}\end{gathered}\end{gathered}\end{gathered}\end{gathered}
MoreFormulae
MoreFormulae
★CSA
(cylinder)
=2πrh
★Volume
(cylinder)
=πr
2
h
★TSA
(cylinder)
=2πr(r+h)
★CSA
(cone)
=πrl
★TSA
(cone)
=πr(l+r)
★Volume
(sphere)
=
3
4
πr
3
★Volume
(cube)
=(side)
3
★CSA
(cube)
=4(side)
2
★TSA
(cube)
=6(side)
2
★Volume
(cuboid)
=lbh
★CSA
(cuboid)
=2(l+b)h
★TSA
(cuboid)
=2(lb+bh+hl)