Answer
Verified
387.9k+ views
Hint: First of all, we should find the radius of the cone. We know that the volume of the cone is equal to V if r is the radius of the cone and h is the height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}h\]. By using this formula, we can find the volume of the cone. We know that the curved surface area of the cone is equal to A if r is the radius of the cone, h is the height of the cone and l is the slant height of the cone, then \[A=\pi rl\] where \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\]. By using this formula, we can find the area of the canvas.
Complete step-by-step solution:
From the question, it is clear that a heap of wheat is in the form of a cone whose diameter is 10.5 m and the height is 3 m.
We know that the volume of the cone is equal to V if r is the radius of the cone and h is the height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}h\].
We were given that the diameter of the cone is equal to 10.5 cm.
Let us assume the diameter of the cone is equal to d.
\[\Rightarrow d=10.5....(1)\]
We know that if r is the radius of the cone and d is the diameter of the cone, then \[d=2r\].
Now let us assume the radius of the cone is equal to r.
\[\Rightarrow 10.5=2r\]
By using cross multiplication, we get
\[\begin{align}
& \Rightarrow r=\dfrac{10.5}{2} \\
& \Rightarrow r=5.25.....(2) \\
\end{align}\]
We were given that the height of the cone is equal to 3m.
Let us assume the height of the cone is equal to h.
\[\Rightarrow h=3......(3)\]
Let us assume the volume of the cone is equal to V.
We know that the volume of the cone is equal to V if r is the radius of the cone and h is the height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}h\].
\[\begin{align}
& \Rightarrow V=\dfrac{1}{3}\pi {{\left( 5.25 \right)}^{2}}\left( 3 \right) \\
& \Rightarrow V=86.59.....(4) \\
\end{align}\]
From equation (4), it is clear that the volume of the cone is equal to \[86.59c{{m}^{3}}\].
Now we should find the area of the canvas.
We know that the curved surface area of the cone is equal to A if r is the radius of the cone, h is the height of the cone and l is the slant height of the cone, then \[A=\pi rl\] where \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\].
So, let us assume l is the slant height of the cone.
\[\begin{align}
& \Rightarrow l=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 5.25 \right)}^{2}}} \\
& \Rightarrow l=\sqrt{36.5625} \\
& \Rightarrow l=6.04669.....(5) \\
\end{align}\]
From equation (5), it is clear that the slant height of the cone is equal to 6.04669 m.
Let us assume the curved surface area of the cone is equal to A.
\[\begin{align}
& \Rightarrow A=\pi \left( 5.25 \right)\left( 6.04669 \right) \\
& \Rightarrow A=99.756.......(6) \\
\end{align}\]
From equation (6), it is clear that the area of the canvas is equal to \[99.756{{m}^{2}}\].
Note: Students may assume that the curved surface area of the cone is equal to A if r is the radius of the cone, h is the height of the cone, then \[A=\pi rh\]. Students may also assume that that the volume of the cone is equal to V if r is the radius of the cone, h is the height of the cone and l is the slant height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}l\] where \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\]. But we know that these are incorrect. So, these misconceptions should be avoided.
Complete step-by-step solution:
From the question, it is clear that a heap of wheat is in the form of a cone whose diameter is 10.5 m and the height is 3 m.
We know that the volume of the cone is equal to V if r is the radius of the cone and h is the height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}h\].
We were given that the diameter of the cone is equal to 10.5 cm.
Let us assume the diameter of the cone is equal to d.
\[\Rightarrow d=10.5....(1)\]
We know that if r is the radius of the cone and d is the diameter of the cone, then \[d=2r\].
Now let us assume the radius of the cone is equal to r.
\[\Rightarrow 10.5=2r\]
By using cross multiplication, we get
\[\begin{align}
& \Rightarrow r=\dfrac{10.5}{2} \\
& \Rightarrow r=5.25.....(2) \\
\end{align}\]
We were given that the height of the cone is equal to 3m.
Let us assume the height of the cone is equal to h.
\[\Rightarrow h=3......(3)\]
Let us assume the volume of the cone is equal to V.
We know that the volume of the cone is equal to V if r is the radius of the cone and h is the height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}h\].
\[\begin{align}
& \Rightarrow V=\dfrac{1}{3}\pi {{\left( 5.25 \right)}^{2}}\left( 3 \right) \\
& \Rightarrow V=86.59.....(4) \\
\end{align}\]
From equation (4), it is clear that the volume of the cone is equal to \[86.59c{{m}^{3}}\].
Now we should find the area of the canvas.
We know that the curved surface area of the cone is equal to A if r is the radius of the cone, h is the height of the cone and l is the slant height of the cone, then \[A=\pi rl\] where \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\].
So, let us assume l is the slant height of the cone.
\[\begin{align}
& \Rightarrow l=\sqrt{{{\left( 3 \right)}^{2}}+{{\left( 5.25 \right)}^{2}}} \\
& \Rightarrow l=\sqrt{36.5625} \\
& \Rightarrow l=6.04669.....(5) \\
\end{align}\]
From equation (5), it is clear that the slant height of the cone is equal to 6.04669 m.
Let us assume the curved surface area of the cone is equal to A.
\[\begin{align}
& \Rightarrow A=\pi \left( 5.25 \right)\left( 6.04669 \right) \\
& \Rightarrow A=99.756.......(6) \\
\end{align}\]
From equation (6), it is clear that the area of the canvas is equal to \[99.756{{m}^{2}}\].
Note: Students may assume that the curved surface area of the cone is equal to A if r is the radius of the cone, h is the height of the cone, then \[A=\pi rh\]. Students may also assume that that the volume of the cone is equal to V if r is the radius of the cone, h is the height of the cone and l is the slant height of the cone, then \[V=\dfrac{1}{3}\pi {{r}^{2}}l\] where \[l=\sqrt{{{r}^{2}}+{{h}^{2}}}\]. But we know that these are incorrect. So, these misconceptions should be avoided.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Define absolute refractive index of a medium
Find out what do the algal bloom and redtides sign class 10 biology CBSE
Prove that the function fleft x right xn is continuous class 12 maths CBSE
Trending doubts
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Which type of bond is stronger ionic or covalent class 12 chemistry CBSE
What organs are located on the left side of your body class 11 biology CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
How fast is 60 miles per hour in kilometres per ho class 10 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
When people say No pun intended what does that mea class 8 english CBSE