Question

The area of a square field is 8 hectares. How long would a man take to cross it diagonally by walking at the rate of 4 kmph?

Answer 1

Hint: Area of a square is equal to the square of its side and diagonal of the square is $\sqrt 2 $ times the side. We will calculate the side of the square field using its area and then the length of diagonal using the side’s length. Now, we have the distance travelled by the man and his speed, we can easily calculate the time taken by the formula: $t = \dfrac{d}{s}$.

Complete step by step solution:

The shape of the field is square, and for a square all its sides are equal.
We know that area of the square is equal to the square of its side.
Let $a$ be the side of the square field and $d$ be it is diagonal
Now, the formula for area of the square field will be ${a^2}$
And, given area of a square field is 8 hectares.
Now, we need to convert the given area from hectares to square kilometers
We know that:
1 hectare is equal to $0.01$ square kilometers
Thus, 8 hectares is equal to $8 \times 0.01 = 0.08$ square kilometers
Now, for the given square field, we have
${a^2} = 0.08k{m^2}$
${a^2} = \dfrac{8}{{100}}k{m^2}$
${a^2} = 2 \times \dfrac{4}{{100}}k{m^2}$
Taking square root both sides:
$a = \sqrt {2 \times \dfrac{4}{{100}}} km$
$a = \sqrt 2 \dfrac{2}{{10}}km$
$a = \dfrac{{\sqrt 2 }}{5}km$
Thus, the side of square field is $\dfrac{{\sqrt 2 }}{5}km$
Now, we know that diagonal of a square is $\sqrt 2 $ times its side,
Therefore, the diagonal of the square field is $\sqrt 2 $ times the side of the square field.
Let $d$ be the diagonal of a square field,
$d = \sqrt 2 a$
$d = \sqrt 2 \dfrac{{\sqrt 2 }}{5}km$
$d = \dfrac{2}{5}km$
Now, to calculate the time a man would take to cross the field diagonally by walking at the rate of 4 kmph, we need to know the relation between speed, time and distance, covered by an object.
Speed is equal to distance divided by time,
Time is equal to distance divided by the speed
Let $t$ be the time the man would take to cross the field diagonally
And the speed of man as given is 4 kmph, the distance he has to cover is the diagonal of the field.
Distance is $\dfrac{2}{5}km$
Therefore, according to the formula, $t = \dfrac{d}{s}$
$t = \dfrac{{\dfrac{2}{5}km}}{{4{\text{ }}km{h^{ - 1}}}}$
$t = \dfrac{{0.4km \times hr}}{{4{\text{ }}km}}$
$t = \dfrac{1}{{{\text{10}}}}hr$
Now, we know that in an hour, there are 60 minutes
Thus, $t = \dfrac{1}{{{\text{10}}}}hr = \dfrac{1}{{{\text{10}}}} \times 60\min $
$t = 6\min $

Hence, a man would take $6\min $ to cross the field diagonally by walking at the rate of 4 kmph

Note: Make use of the speed – distance time relationship. In case of any confusion remembering the following formula: $t = \dfrac{d}{s}$
Where t is the time, d is the distance and s is the speed. It can be remembered by using units.
We know, the unit of distance is km,
Unit of time is the hour.
And, the unit of speed in kmph.
We have speed as: $\dfrac{{km}}{h}$
Km can be written as distance and time can be written as the hour
So, we have a speed equal to distance divided by time.
Hence, by this method, there is no need for cramming the formula.
We can also solve the above question using Pythagoras theorem as \[{\text{d = }}\sqrt {{{\text{a}}^{\text{2}}}{\text{ + }}{{\text{a}}^{\text{2}}}} \], as there is a right angled triangle if we consider two adjacent sides and diagonal.