Question

A heap of wheat is in the form of a cone whose diameter is 10.5 m and height is 3 m.

Find it volume. If 1cm3

wheat cost is Rs 10, then find total cost​

Answer 1

Answer:

Diameter of cone = 10.5 m

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 m

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 mVolume of cone = 1 / 3 πr2h

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 mVolume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 mVolume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3= 86.625m3

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 mVolume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3= 86.625m3Cost of 1m3 of wheat = 10

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 mVolume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3= 86.625m3Cost of 1m3 of wheat = 10∴ Cost of 86.625 m3 of wheat = 10 × 86.625

Diameter of cone = 10.5 mRadius of cone (r) = 5.25 m Height of cone (h) = 3 mVolume of cone = 1 / 3 πr2h = 1 / 3 × 22 / 7 × 5.25 × 5.25 × 3= 86.625m3Cost of 1m3 of wheat = 10∴ Cost of 86.625 m3 of wheat = 10 × 86.625= 86.625

Answer 2

$\begin{gathered} \sf{To \: Find :}\end{gathered}\\\begin{gathered} \sf{ Find \: the \: Volume \: and \: the \: Cost \: of \: wheat}\end{gathered} \\ \\\sf {Solution :} \\\dag† \: Formula \: Used : \\ {\underline{\boxed{\pmb{\sf{ Volume{\small_{(Cone)}} = \dfrac{1}{3} \pi {r}^{2} h }}}}}\\\sf{Where : }\\ \sf{ \pi = \dfrac{22}{7} } \\ \begin{gathered} \sf{π=r = Radius; h = Height }\end{gathered} \\\begin{gathered} \sf \dag† Calculating \: the \: Volume :\end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \pi {r}^{2} h } \end{gathered} \end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7}\times{ \bigg( \dfrac{Diameter}{2} \bigg) }^{2} \times 3 } \end{gathered} \end{gathered} \\ \begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \cancel\dfrac{10.5}{2} \bigg) }^{2} \times 3 }\end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( 5.25 \bigg) }^{2} \times 3 } \end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \dfrac{525}{100} \bigg) }^{2} \times 3 }\end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{525}{100} \times \dfrac{525}{100} \times 3 }\end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \cancel\dfrac{525}{100} \times \cancel\dfrac{525}{100} \times 3 }\end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{105}{20} \times \dfrac{105}{20} \times 3 } \end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \cancel\dfrac{105}{20} \times \cancel\dfrac{105}{20} \times 3 }\end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{\cancel7} \times \dfrac{\cancel{21}}{4} \times \dfrac{21}{4} \times 3 } \end{gathered} \end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times 22 \times \dfrac{3}{4} \times \dfrac{21}{4} \times 3 }\end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{\cancel3} \times 22 \times \dfrac{3}{4} \times \dfrac{\cancel{21}}{4} \times 3 } \end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = 1 \times 22 \times \dfrac{3}{4} \times \dfrac{7}{4} \times 3 } \end{gathered} \end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{66}{4} \times \dfrac{21}{4} } \end{gathered} \end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1386}{16} } \end{gathered} \end{gathered} \\\begin{gathered}\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \cancel\dfrac{1386}{16} }\end{gathered} \end{gathered} \\\begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Volume \; of \; Wheat = 86.625 \; {m}^{2} }}}}} \; {\pink{\bigstar}} \end{gathered}\\\begin{gathered} \sf\dag† Calculating \: the \: Cost :\end{gathered}\\\begin{gathered} \begin{gathered} \qquad \; \implies \; \; \sf { Cost = Volume \times Rate } \end{gathered} \end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \implies \; \; \sf { Cost = 86.625 \times Rate } \qquad \; \bigg\lgroup {\red{\sf{ 1 \; m = 100 \; cm }}} \bigg\rgroup \end{gathered} \end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \implies \; \; \sf { Cost = \bigg( 86.625 \times 100 \bigg) \times 10 }\end{gathered} \end{gathered} \\\begin{gathered} \begin{gathered} \qquad \; \implies \; \; \sf { Cost = 8662.5 \times 10 } \end{gathered} \end{gathered} \\\begin{gathered} \qquad \; \implies \; \; {\underline{\boxed{\pmb{\sf{ Cost \; of \; Wheat = Rs. \; 86625 }}}}} \;{ \pink{\bigstar}} \end{gathered} \\\sf\therefore \ Cost \: of \: the \: Wheat \: is \: Rs.86625 .$