Question

Find the total area of 14 squares whose sides are 11 cm, 12 cm, 13 cm, … , 24 cm

Answer 1

Hint: We have been given the side of all the 14 squares and all are in a consecutive order. The total area will be the sum of areas of these squares given by the square of their respective sides. We can use the formula for sum of squares of n natural numbers to make the required calculation easier. Notice that 11 to 24 lies between the difference of the sum of numbers: 1 to 24 and 1 to 11 with respect to number 1.
Formula to be used:
 $ {1^2} + {2^2} + {3^2} + .... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6} $

Complete step-by-step answer:
We can see the sides of all the squares consecutive from 11 to 24 cm.
The area of a square is given as the square of side. The total area of 14 squares is the sum of squares side of the respective squares. So, the required area is given as:
 $ A = {11^2} + {12^2} + {13^2} + .... + {24^2}\_\_\_\_(1) $
We can find this value using the formula for sum of squares of n natural numbers. But, we need that only from numbers 11 to 24, which can be calculated by the difference between the sum of natural numbers from 1 to 24 and 1 to 10.
\[{11^2} + {12^2} + {13^2} + .... + {24^2} = \left[ {{1^2} + {2^2} + {3^2} + .... + {{24}^2}} \right] - \left[ {{1^2} + {2^2} + {3^2} + .... + {{10}^2}} \right]\_\_\_\_\_(2)\]
The sum of squares of n natural numbers is given as:
 $ \dfrac{{n(n + 1)(2n + 1)}}{6} $
Using this, we get:
\[
\Rightarrow {1^2} + {2^2} + {3^2} + .... + {24^2} = \dfrac{{24 \times \left( {24 + 1} \right)\left( {2 \times 24 + 1} \right)}}{6} \\
\Rightarrow {1^2} + {2^2} + {3^2} + .... + {24^2} = \dfrac{{24 \times 25 \times 49}}{6} \\
\Rightarrow {1^2} + {2^2} + {3^2} + .... + {24^2} = 4900 \\
 \]
And
\[
\Rightarrow {1^2} + {2^2} + {3^2} + .... + {11^2} = \dfrac{{10 \times \left( {10 + 1} \right)\left( {2 \times 10 + 1} \right)}}{6} \\
\Rightarrow {1^2} + {2^2} + {3^2} + .... + {11^2} = \dfrac{{10 \times 11 \times 21}}{6} \\
\Rightarrow {1^2} + {2^2} + {3^2} + .... + {11^2} = 385 \\
 \]
Substituting these values in (2), we get:
\[
\Rightarrow {11^2} + {12^2} + {13^2} + .... + {24^2} = 4900 - 385 \\
\Rightarrow {11^2} + {12^2} + {13^2} + .... + {24^2} = 4515 \\
 \]
Therefore, the total area of 14 squares whose sides are 11 cm, 12 cm, 13 cm, … , 24 cm is $ 4515c{m^2} $

Note: As the given units of sides were in cm, thus the area of the given square is also in the units $ c{m^2} $ . Like this general formula of sum of squares of n natural numbers, other known and important general formulas include the sum of all the natural numbers and their cubes. Note that the numbers greater than 1 comes under the category of natural numbers. Instead of applying the formula we could also have calculated the squares of the respective sides and added them but that would be a long and cumbersome process and thus we use the formulas to reduce the hard work and calculation errors.