Question

$\sf \red{Question:}$
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. 

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Answer 1

Answer:

since the heap of wheat is in the form of a cone and the canvas required to cover the heap will be equal to the curved surface area of the cone.

volume of a cone of base radius, 'r' and height, 'h' = 1/3πr²h

curved surface area of the cone having a base radius, 'r' and slant height, 'l' = πrl

slant height of the cone, l = √r² + h²

diameter of the conical heap, d = 10.5 m

radius of the conical heap, r = 10.5/2 m = 5.25 m

height of the conical heap, h = 3 m

volume of the conical heap = 1/3πr²h

= 1/3 × 22/7 × 5.25 m × 5.25 m × 3 m

= 86.625 m³

slant height, l = √r² + h²

= √(5.25)² + (3)²

= √27.5625 + 9

= √36.5625

= 6.046 m (approx.)

the area of the canvas required to cover the heap of wheat = πrl

= 22/7 × 5.25 m × 6.046 m

= 99.759 m²

the volume of the conical heap is 86.625 m³ and the area of the canvas required is 99.759 m².

Step-by-step explanation:

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Answer 2

Given : Diameter of the heap is 10.5 m and the height is 3 m . Cost of 1 m³ wheat is Rs.10 .

To Find : Find the Volume and the Cost of wheat

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SolutioN :

$\dag$ Formula Used :

• ${\underline{\boxed{\pmb{\sf{ Volume{\small_{(Cone)}} = \dfrac{1}{3} \pi {r}^{2} h }}}}}$

Where :

• $\sf{ \pi = \dfrac{22}{7} }$

• r = Radius
• h = Height

$\dag$ Calculating the Volume :

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \pi {r}^{2} h } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \dfrac{Diameter}{2} \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \cancel\dfrac{10.5}{2} \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( 5.25 \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \dfrac{525}{100} \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{525}{100} \times \dfrac{525}{100} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \cancel\dfrac{525}{100} \times \cancel\dfrac{525}{100} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \dfrac{105}{20} \times \dfrac{105}{20} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times \cancel\dfrac{105}{20} \times \cancel\dfrac{105}{20} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{\cancel7} \times \dfrac{\cancel{21}}{4} \times \dfrac{21}{4} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times 22 \times \dfrac{3}{4} \times \dfrac{21}{4} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{\cancel3} \times 22 \times \dfrac{3}{4} \times \dfrac{\cancel{21}}{4} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = 1 \times 22 \times \dfrac{3}{4} \times \dfrac{7}{4} \times 3 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{66}{4} \times \dfrac{21}{4} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1386}{16} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \cancel\dfrac{1386}{16} } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Volume \; of \; Wheat = 86.625 \; {m}^{2} }}}}} \; {\blue{\bigstar}} \end{gathered}$

$\dag$ Calculating the Cost :

$\begin{gathered} \qquad \; \implies \; \; \sf { Cost = Volume \times Rate } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; \sf { Cost = 86.625 \times Rate } \qquad \; \bigg\lgroup {\purple{\sf{ 1 \; m = 100 \; cm }}} \bigg\rgroup \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; \sf { Cost = \bigg( 86.625 \times 100 \bigg) \times 10 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; \sf { Cost = 8662.5 \times 10 } \\ \\ \\ \end{gathered}$

$\begin{gathered} \qquad \; \implies \; \; {\underline{\boxed{\pmb{\sf{ Cost \; of \; Wheat = Rs. \; 86625 }}}}} \; {\orange{\bigstar}} \end{gathered}$

$\therefore \;$ Cost of the Wheat is Rs.86625 .

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