Verified answer

[tex]\large\underline{\sf{Solution-}}[/tex]

Given that, A heap of wheat is in the form of a cone.

Diameter of cone = 9 m

So, Radius of cone, r = 4.5 m

Height of cone, h = 3.5 m

We know, Volume of cone of radius r and height h is given by

[tex]\boxed{ \sf{ \:Volume_{(Cone)} = \frac{1}{3} \times \pi \times {r}^{2} h \: }} \\ \\ [/tex]

So, on substituting the values, we get

[tex]\sf \: Volume_{(Cone)} = \frac{1}{3} \times 3.14 \times {(4.5)}^{2} \times 3.5 \\ \\ [/tex]

[tex]\sf \: Volume_{(Cone)} = \frac{1}{3} \times 3.14 \times \frac{9}{2} \times \frac{9}{2} \times \frac{7}{2} \\ \\ [/tex]

[tex]\sf \: Volume_{(Cone)} = 3.14 \times \frac{3}{2} \times \frac{9}{2} \times \frac{7}{2} \\ \\ [/tex]

[tex]\sf \: Volume_{(Cone)} = \frac{593.46}{8} \\ \\ [/tex]

[tex]\sf \:  \implies \: Volume_{(Cone)} = 74.18 \: {m}^{3} \: \{approx. \} \\ \\ [/tex]

Now, we have to how much canvas cloth is required to just cover the heap?

So, we have to find the curved surface area of cone.

We know, slant height (l), height (h) and radius (r) are connected by the relationship

[tex]\sf \: {l}^{2} = {r}^{2} + {h}^{2} \\ \\ [/tex]

[tex]\sf \: {l}^{2} = {(4.5)}^{2} + {(3.5)}^{2} \\ \\ [/tex]

[tex]\sf \: {l}^{2} = {\bigg(\dfrac{9}{2} \bigg) }^{2} + {\bigg( \dfrac{7}{2} \bigg) }^{2} \\ \\ [/tex]

[tex]\sf \: {l}^{2} = \frac{81}{4} + \frac{49}{4} \\ \\ [/tex]

[tex]\sf \: {l}^{2} = \frac{81 + 49}{4} \\ \\ [/tex]

[tex]\sf \: {l}^{2} = \frac{130}{4} \\ \\ [/tex]

[tex]\sf \: l = \frac{ \sqrt{130} }{2} \\ \\ [/tex]

[tex]\sf \:  \implies \: l = 5.70087 \\ \\ [/tex]

[tex]\sf \:  \implies \: l = 5.7 \: \: \: \{approx. \}\\ \\ [/tex]

Now,

Amount of canvas required = Curved Surface Area of cone

[tex]\sf \: Amount_{(Canvas\:required)} = \pi \: r \: l \\ \\ [/tex]

[tex]\sf \: Amount_{(Canvas\:required)} = 3.14 \times 4.5 \times 5.7 \\ \\ [/tex]

[tex]\sf \:\bf \:  \implies \: Amount_{(Canvas\:required)} = 80.541 \: {m}^{2} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{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}[/tex]

Answer:

Volume = 74.18 m³

Amount of cloth required = 80.541 m²

[tex]\rule{70mm}{2pt}[/tex]

Step-by-step explanation:

It's given that a heap of wheat is in the form of a cone of diameter 9 m and height 3.5 m. Need to find it's volume and the amount of canvas cloth; required to just cover the heap.

In short we have—

• Diameter of cone (D) = 9 m
• Radius of cone = D/2 = 9/2 = 4.5 m
• Height of cone = 3.5 m

We know that volume of the conical heap is given by—

‎ ‎‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎‎ ‎ ‎ 1/3 πr²h

[tex]\implies\:1/3\: \times 3.14\: \times (4.5)^{2} \:\times\:3.5[/tex]

[tex]\implies\:74.18\:m^{3}[/tex]

Volume of cone is 74.18 m³.

Now, to find the amount of canvas required to conver the heap of the cone. Use the formula of curved surface area of cone.

We know that curved surface area of cone is given by—

‎ ‎‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎‎ ‎πrl

Since the value of length is not given, find its length using this formula-

‎ ‎‎ ‎ ‎ ‎ ‎‎‎ ‎‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ l² = r² + h²

[tex]\implies\:l^{2}\:=\:(4.5)^{2}+(3.5)^{2}[/tex]

[tex]\implies\:l^{2}\:=\:20.25+12.25[/tex]

[tex]\implies\:l^{2}\:=\:32.5[/tex]

[tex]\implies\:l^{2}\:=\:(5.7)^{2}[/tex]

[tex]\implies\:l\:=\:5.7[/tex]

Therefore,

Curved surface area of conical heap = 3.14 × 4.5 × 5.7 = 80.541 m² (approx.)