[tex] \sf \red{Question:}[/tex]
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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Answers & Comments
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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Given : Diameter of the heap is 10.5 m and the height is 3 m . Cost of 1 m³ wheat is Rs.10 .
[tex] \\ \\ [/tex]
To Find : Find the Volume and the Cost of wheat
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
SolutioN :
[tex] \dag [/tex] Formula Used :
Where :
[tex] \\ \\ [/tex]
[tex] \dag [/tex] Calculating the Volume :
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \pi {r}^{2} h } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \dfrac{Diameter}{2} \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered} [/tex]
[tex] \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} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( 5.25 \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times \dfrac{22}{7} \times { \bigg( \dfrac{525}{100} \bigg) }^{2} \times 3 } \\ \\ \\ \end{gathered} [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \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} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{3} \times 22 \times \dfrac{3}{4} \times \dfrac{21}{4} \times 3 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1}{\cancel3} \times 22 \times \dfrac{3}{4} \times \dfrac{\cancel{21}}{4} \times 3 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = 1 \times 22 \times \dfrac{3}{4} \times \dfrac{7}{4} \times 3 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{66}{4} \times \dfrac{21}{4} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \dfrac{1386}{16} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; \sf { Volume = \cancel\dfrac{1386}{16} } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \dashrightarrow \; \; {\underline{\boxed{\pmb{\sf{ Volume \; of \; Wheat = 86.625 \; {m}^{2} }}}}} \; {\blue{\bigstar}} \end{gathered} [/tex]
[tex] \\ \\ [/tex]
[tex] \dag [/tex] Calculating the Cost :
[tex] \begin{gathered} \qquad \; \implies \; \; \sf { Cost = Volume \times Rate } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \implies \; \; \sf { Cost = 86.625 \times Rate } \qquad \; \bigg\lgroup {\purple{\sf{ 1 \; m = 100 \; cm }}} \bigg\rgroup \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \implies \; \; \sf { Cost = \bigg( 86.625 \times 100 \bigg) \times 10 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \implies \; \; \sf { Cost = 8662.5 \times 10 } \\ \\ \\ \end{gathered} [/tex]
[tex] \begin{gathered} \qquad \; \implies \; \; {\underline{\boxed{\pmb{\sf{ Cost \; of \; Wheat = Rs. \; 86625 }}}}} \; {\orange{\bigstar}} \end{gathered} [/tex]
[tex] \\ \\ [/tex]
[tex] \therefore \; [/tex] Cost of the Wheat is Rs.86625 .
[tex] \\ \qquad{\rule{200pt}{2pt}} [/tex]
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