Question

A heap of rice is in the form of a cone of diameter 12m and height 8m .Find it csa ,tsa, volumr .Hoe much canvas of clot is required to cover the heap ?

Answer 1

Answer:

Step-by-step explanation:

Answer 2

$\mathfrak{\huge{Answer:}}$

$\mathbb{GIVEN}$

♧ Diameter of the heap of rice = 12 m

♧ Height of the heap of rice = 8 m

$\mathbb{TO\:FIND}$

◇ The Total Surface Area of the heap of rice ( TSA )

◇ The Curved Surface Area of the heap of rice ( CSA )

◇ The Volume of the heap of rice

$\mathbb{METHOD}$

We will just put the values in the formulae we know. But first, we need to find the radius of the heap of rice:-

Diameter = 2 × Radius

=》 $\boxed{\tt{Radius = 6 m}}$

♤ CSA = $\sf{\pi r l}$

Length of the heap of rice = $\sf{\sqrt{h^{2} + r^{2}}}$

Length of the heap of rice = $\sf{\sqrt{8^{2} + 6^{2}}}$

Length of the heap of rice = $\sf{\sqrt{64+36}}$

Length of the heap of rice = $\sf{10 m}$

=》 CSA = $\sf{3.14 \times 6 \times 10}$

=》 $\boxed{\tt{CSA = 188.4 m^{2}}}$

♤ TSA = $\sf{\pi r^{2} + \pi r l}$

We've found the value of $\sf{\pi r l}$ that's why we can simply put that value and calculate the rest part.

=》 TSA = $\sf{\pi r^{2} + 188.4}$

=》 TSA = $\sf{3.14 \times 6^{2} + 188.4}$

=》 TSA = $\sf{113.04 + 188.4}$

=》 $\boxed{\tt{TSA = 301.44 m^{2}}}$

♤ Volume = $\sf{\frac{1}{3} \pi r^{2} h}$

=》 Volume = $\sf{\frac{1}{3} \times 3.14 \times 6^{2} \times 8}$

=》 $\boxed{\tt{Volume = 301.44 m^{3}}}$

They're your answers ! :)

$\bf{Note:}$ I've taken the value of $\sf{\pi\:as\:3.14}$ just to simplify the calculation part. You can even take the value as $\sf{\frac{22}{7}}$