Question

A bucket open at the top has top and bottom radii of circular ends as 40 cm and 20 cm respectively. Find the volume of the bucket if its depth is 21 cm. Also find the area of the tin sheet required for making the bucket

Answer 1

Given:-

Let the top radius be R and bottom radius be r

Now

Radius of top(R)= 40cm

Radius of bottom (r)= 20cm

And height (h)= 21cm

Now

To find:-

The volume of bucket

The area of sheet required for making the bucket

A.T.Q:-

$\boxed{\sf{\purple{Volume=\dfrac{1}{2}\pi h(R{}^{2}+r{}^{2})+R\times r}}}$

$\sf\implies Volume =\dfrac{1}{\cancel3}\times \dfrac{22}{7}\times \cancel{21}\times[ (40){}^{2}+(20){}^{2}+40\times 20)\\ \\ \sf\implies Volume=\dfrac{22}{7}\times \cancel{7}\times ( 1600+400+800)\\ \\ \sf\implies Volume =22\times 2800\\ \\ \sf\implies Volume =22\times 2800\\ \\ \sf\implies Volume =61600cm{}^{3}$

$\boxed{\tt{\blue{Volume\:of\: bucket=61600cm{}^{3}}}}$

Now the Area of sheet required to make the bucket

$\boxed{\sf{\purple{Area\:of\: sheet\: required=C.S.A\:of\: bucket+Area\:of\: base}}}$

$\sf\bullet C.S.A\:of\: bucket=\pi (R+r)\times slant height\\ \\ \sf \bullet Area of base =\pi R {}^{2}$

$\sf\implies Slant\: height(l)=\sqrt{H{}^{2}+(R-r){}^{2}}\\ \\ \sf\implies l=\sqrt{(21){}^{2}+(40-20){}^{2}}\\ \\ \sf\implies l=\sqrt{441+(20){}^{2}}\\ \\ \sf\implies l=\sqrt{441+400}\\ \\ \sf\implies l=\sqrt{841}\\ \\ \sf\implies slant\: height=29cm$

• Now the Area of sheet

$\sf\implies Area \:of\:sheet=\pi (R+r)l+\pi R{}^{2}\\ \\ \sf\implies Area \:of\:sheet= \dfrac{22}{7}\times (40+20)\times 29+\dfrac{22}{7}\times (40){}^{2}\\ \\ \sf\implies Area_{sheet}= \dfrac{22\times 60\times 29}{7} +\dfrac{22\times 1600}{7}\\ \\ \sf\implies Area\:of\:sheet =\dfrac{38280}{7}+\dfrac{35200}{7}\\ \\ \sf \implies Area_{sheet}=\dfrac{38280+35200}{7}\\ \\ \sf\implies Area\:of\:sheet =\cancel{\dfrac{73480}{7}}\\ \\ \sf\implies Area\:of\:sheet= 10497.14cm{}^{2}$

$\boxed{\sf{\blue{Area\:of\:sheet\: required=10497.14cm{}^{2}}}}$

$\boxed{\sf{Volume\:of \:bucket=61600cm{}^{3}}}$

Answer 2

$\huge{\mathtt{Solution⟹ }}$

$\mathtt{Given ⟹}$

$\mathtt{⟹\: Radius\: of \: the \: base \: of \: bucket\:=\: 20\: cm}$

$\mathtt{⟹ \: Radius\: of \: top\: of \: bucket\:=\: 40\: cm}$

$\mathtt{⟹\: Height\: of\: the \:bucket\:=\: 21\: cm}$

$\mathtt{ \: To \: Find \:⟹}$

$\mathtt{⟹ \: Volume\: of \: the \: bucket}$

$\mathtt{⟹ \: and\: Tin \:required\: to\: form\: the\: bucket }$

_______________________

$\mathtt{Formula\: of \:volume\:⟹ }$

$\frac{\pi.h}{3} ( {r}^{2} + {R}^{2} + r.R \: )$

$\mathtt{⟹\: Here \:r\: = \:Radius\: of \:base \:and \:R\: =\: radius\: of \:top }$

$⟹ \frac{\pi.21}{3} ( {20}^{2} + {40}^{2} + (20 \times 40))$

$⟹ \frac{22 \times 21}{7 \times 3} (400 + 1600 + 800)$

$⟹ \: \frac{22 \times 7 \times 3}{7 \times 3} (2800)$

$⟹ \: 22(2800) = 61600 {cm}^{3}$

${\mathtt{\red{⟹\: The \:volume \:of \:bucket\: is\: 61600\: {cm}^{3}}}}$

$\mathtt{Tin\: required\: ⟹}$

$\mathtt{⟹\: lateral\: surface\: area\: + \: base\: area}$

$\mathtt{Formula \:of\: lateral\: surface\: area\: and \:base \:area\: ⟹}$

$⟹ \: base \: area \: = \pi. {r}^{2}$

$⟹ \: \: lsa = \pi(R \: + r).l$

$\mathtt{⟹ \:Here\: l\: = \:slant \:height}$

$⟹ \: l \: = \sqrt{ {h}^{2} + {(R - r) }^{2} } \\ </p><p>$

$⟹ \: \sqrt{ {21}^{2} + {(40 - 20)}^{2} }$

$⟹ \: \sqrt{441 + 400} = \sqrt{841}$

$\mathtt{⟹\: l \:=\: 29}$

$⟹ \: lsa \: = \frac{22}{7} (40 + 20).29$

$⟹ \: \frac{22}{7} (60 \times 29)$

$⟹ \: base \: area \: = \frac{22}{7} (20 \times 20)$

$required \: area \: of \: tin \: = \frac{22}{7} (60(29) + 400)$

$⟹ \: \frac{22}{7}( 1740 + 400)$

$⟹ \: \frac{22}{7} (2140)$

$⟹ \: \frac{47080}{7} = 6725.71 {cm}^{2}$

$\mathtt \red{area \: of \: tin \: is \: 6725.71 {cm}^{2} }$