Question

A contractor is entrusted to erect a tent for flood-victims. He is allowed a fixed amount for this task. He has two options :
(i) to erect a tent which is cylindrical upto a height of 3 m and conical above it. The diameter of the base is 105 m and slant height of the conical part is 53 m.
(ii) to erect the tent as described in option(i), only replacing the conical part as a hemispherical part.
The contractor chooses the option-(ii) and decides to donate the extra (difference) canvas to be used in this case.
How much canvas is donated by the contractor?

Answer 1

Answer:

in both option cylinderical shape part is samr

hence we will find canvas used in conical and hemispherical shape.

for that we need to find curved surface areas of both

conical shape

radius = 105/2 m

slant height 53 m

curved surface area = (22/7)×(105/2) × 53

= 8745 m^2

hemispherical shape

radius = 105/2 m

curved surface area = 2*(22/7)(105/2)^2

= 17325 m^2

he would not save canvas by making hemispherical type top of tent

Answer 2

The contractor would donate total canvas of "8576.55 $m^2$ ".

Step by step Explanation:

Given: radius of base, r = $\dfrac{diameter}{2}$

$r = \dfrac{105}{2} = 52.5 m$

Height of the cylinder, h = 3 m,

Slant height of cone, l  = 53 m,

Now we can calculate the surface area of cylinder as:

$\textrm{surface area of the cylinder} = 2\pi r^2+ 2\pi rh$

$\textrm{surface area of the cylinder} = 2\pi (52.5)^2+ 2\pi \times 52.5 \times 3$$\textrm{surface area of the cylinder} = 2\pi \times 52.5^2+ 2\pi \times 52.5 \times 3$

$\textrm{surface area of the cylinder} = 18307.62 m$

For case (i) for conical tent, we can calculate the total surface area of the tent as:

Surface area of conical tent = surface area of cylinder + surface area of cone

$\textrm{surface area of the conical tent} = \textrm{surface area of the cylinder} + (\pi r^2 + \pi rl)$

$\textrm{surface area of the conical tent} = \textrm{surface area of the cylinder} + (\pi \times 52.5^2 + \pi \times 52.5 \times 53)$$\textrm{surface area of the conical tent} = 18307.62 + 17400.49$

$\textrm{surface area of the conical tent} = 35708.11 m^2$

For case (ii) for Hemisphere tent, we can calculate the total surface area of the tent as:

$\textrm{surface area of the conical tent} = \textrm{surface area of the cylinder} + \textrm{surface area of hemisphere}$$\textrm{surface area of the hemisphere tent} = 18307.62 + 3\pi r^2\\\\\textrm{surface area of the hemisphere tent} = 18307.62 + 3\pi 52.5^2\\\textrm{surface area of the hemisphere tent} = 44284.66 m^2$

Now we can calculate the total donated canvas by the contractor = canvas used in hemisphere tent - canvas used in cylindrical tent

Donated canvass = 44284.66 - 35708.11

Donated canvas = 8576.55 $m^2$