Question

Q no.2
3
A 21m deep well with diameter 6m is dig and the
earth from the digging is evenly spread to from a
platform 27m x 11m. Find the height of the
platform​

Answer 1

Answer:

2m

Step-by-step explanation:

volume of well=πr^2h

volume=22/7*3*3*21=594

volume of earth=lbh

594=27*11*h

h=2m

Answer 2

Given :-

Dept of the platform = 21 m

Diameter of the platform = 6 m

Dimension of the platform = 27 m × 11 m

To Find :-

The height of the platform.

Solution :-

We know that,

• h = Height
• d = Diameter
• r = Radius

Given that,

Diameter (d) = 6 m

Dept = 21 m

According to the question,

It is given that the shape of the well is in the shape of a cylinder with a diameter of 7 m

So, radius = $\sf \dfrac{Diameter}{2}$

Radius = $\sf \dfrac{7}{2} \: m$

Volume of the earth dug out will be equal to the volume of the cylinder

$\underline{\boxed{\sf Volume \ of \ cylinder= \pi r^{2}h}}$

Substituting them, we get

$\longrightarrow \sf 22 \times 7 \times 5 \: m^{3}$

Let the height of the platform = H

Volume of soil from well (cylinder) = Volume of soil used to make such platform

πr²h = Area of platform × Height of the platform

We know that the dimension of the platform is = 22 × 14

So, Area of platform = $\sf 22 \times 14 \: m^{2}$

∴ $\sf \pi r^{2}h=22 \times 14 \times H$

$\implies \sf Height=2.5 \: m$

Therefore, the height of the platform is 2.5 m

To Note :-

Two end faces of right circular cylinder are circles having each area = πr²

Rectangular solids and cylinders are somewhat similar because they both have two bases and a height.