Answer
424.2k+ 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 \[V=\dfrac{1}{3}\pi {{r}^{2}}h\].
![seo images](https://www.vedantu.com/question-sets/1212fee7-c9be-481f-b015-6d2ae2ae350a7191093461723093159.png)
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
How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Why Are Noble Gases NonReactive class 11 chemistry CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let X and Y be the sets of all positive divisors of class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
At which age domestication of animals started A Neolithic class 11 social science CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Which are the Top 10 Largest Countries of the World?
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Give 10 examples for herbs , shrubs , climbers , creepers
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Difference Between Plant Cell and Animal Cell
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Write a letter to the principal requesting him to grant class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Change the following sentences into negative and interrogative class 10 english CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE
![arrow-right](/cdn/images/seo-templates/arrow-right.png)