Question

A cistern $6$ m long and $4$ m wide contains water up to a depth of $1$ m $25$ cm.
The total area of the wet surface is:
$\left( A \right)\text{ 49}{{\text{m}}^{2}}$
$\left( B \right)\text{ 50}{{\text{m}}^{2}}$
$\left( C \right)\text{ 53}\text{.5}{{\text{m}}^{2}}$
$\left( D \right)\text{ 55}{{\text{m}}^{2}}$

Answer 1

Hint: In this question we have been given with a cistern which has a length of $6$ meters which is $4$ meters wide. We also know the depth of the cistern which is $1$ m $25$ cm. We will consider the structure of the cistern as a cuboid of the given length, breadth and height. We will consider the depth of the cistern as its height. We have to find the total surface which is wet when the cistern is full therefore, we will add the surface area of the $4$ walls and the bottom of the surface to get the required solution.

Complete step by step solution:
We know that:
$l=6$ m
$b=4$ m
$h=1.25$ m
now since each side represents a rectangle, we will use the formula area of a rectangle to find the surface area which is given by $l\times b$.
Now since the depth is same for all the $4$ walls, but the length is different, we will have $2$ walls which have length as $6$ and $2$ walls which have length as $4$.
On adding, we get:
$\Rightarrow \left( 6\times 1.25 \right)+\left( 4\times 1.25 \right)+\left( 6\times 1.25 \right)+\left( 4\times 1.25 \right)$
On simplifying, we get:
$\Rightarrow 7.5+5+7.5+5$
On adding all the terms, we get:
$\Rightarrow 25{{m}^{2}}$, which is the total surface area of all the $4$ walls.
Now the surface area of the bottom will be as:
$\Rightarrow 6\times 4$
On simplifying, we get:
$\Rightarrow 24{{m}^{2}}$
And the total surface area which will be wet will be the summation of both therefore, we get:
$\Rightarrow sa=25{{m}^{2}}+24{{m}^{2}}$
On adding, we get:
$\Rightarrow sa=49{{m}^{2}}$, which is the required solution.

So, the correct answer is “Option A”.

Note: It is to be remembered that when calculating areas, the units should be handled with utmost care. It is to be noted that the depth was given to us as $1$ meter and $25$ centimeters. We converted the $25$ centimeters to meters and added it to get $1.25$ meters so that all the units of the cistern become the same.