Question

a heap of wheat is in the form of a cone whose diameter is 10.5 M and height is 3 M find its volume the hip is to be covered by Canvas to protect it from rain find the area of the Canvas required

Answer 1

Solutions :-

Diameter of a cone = 10.5 m
Radius = r = 10.5/2 = 5.25 m
Height of cone = h = 3 m

Find the volume of heap of wheat :-

Volume of cone = ⅓ π r²h
= ⅓ × 22/7 × 5.25 × 5.25 × 3
= 86.625 m³

Find the slant height (l) :-

l² = h² + r²
l² = 3² + 5.25²
l² = 36.5625
l = √36.5625 = 6.05 m

Find the area of the Canvas required :-

Curved surface area of canvas = πrl sq. units
= 22/7 × 5.25 × 6.05 m²
= 99.825 m²


Hence,
Volume of heap of wheat = 86.625 m³
Area of Canvas required = 99.825 m²

Answer 2

$\underline{\underline{\Huge\mathfrak{Answer ;}}}$

As Stated , that the Diameter of a cone is 10.5 m , than it's Radius will be = 10.5/2 = 5.5m and the height of the cons is 3m.

Now ,
We have to find the volume of the heap of wheat , So let's Find out ;-

We know that ;
Volume of Cone = 1/3π r^2h

So ,

$= > \frac{1}{3} \times \frac{22}{7} \times 5.25 \times 5.25 \times 3$

$= > 86.625 \: m^{3}$

Now ,
Finding the slanted height ;-

We know that ;-

$= > l^{2} = h^{2} + r^{2}$

So ,

$= > l^{2} = 3^{2} + 5.25^{2}$

$= > l^{2} = 36.5625$

$= > l = \sqrt{36.5625} = 6.05 \: m$

Now ,
At the end , we have to find the area of the stated Canvas ;-

We know that ;-
• Canvas's Curved surface area = πrl sq. units

So ,

$= > \frac{22}{7} \times 5.25 \times 6.05 \: m^{2} = 99.825 \: m^{2}$

Therefore ,

$<b>86.625 m^3 is the volume of heap of the wheat</b>$, $<b> 99.825 m^2 is the Area of the canvas needed.</b>$

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