Question

Two regular polygons of the same number of sides have sides $40cm$ and $9cm$ in length. The lengths of the side of another regular polygon of the same number of sides and whose area is equal to the sum of the area of the given polygon is
$\left( 1 \right)\text{ 49}$
$\left( 2 \right)\text{ 31}$
$\left( 3 \right)\text{ 41}$
$\left( 4 \right)\text{ 360}$

Answer 1

Hint: In this question we have been given that the sum of two polygons of the same number of sides is equal to the area of a third polygon. We will solve this question by using the area formula of a polygon and we will equate the sum of the areas of the given polygon with the third polygon. We will use the formula of the area of a polygon which is given by $A=\dfrac{a\times p}{2}$, where $a=\dfrac{s}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}$, $s$ is the length of the side and $n$ is the number of sides and $p=s\times n$.

Complete step by step answer:
We have been given that the polygons have the same number of sides. Consider the number of sides in all the polygons be $n$.
We know that the sum of the area of the polygon with length of the side as $40cm$ and $9cm$ is equal to the area of the third polygon.
Let the length of the side of the third polygon be $x$.
On using the formula of the area of a polygon, from the given condition, we get:
$\Rightarrow \dfrac{xn\times \dfrac{x}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}}{2}=\dfrac{40n\times \dfrac{40}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}}{2}+\dfrac{9n\times \dfrac{9}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}}{2}$
On multiplying the numerators of the fractions, we get:
$\Rightarrow \dfrac{\dfrac{{{x}^{2}}n}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}}{2}=\dfrac{\dfrac{1600n}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}}{2}+\dfrac{\dfrac{81n}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}}{2}$
Now on calling the denominator from both the sides, we get:
$\Rightarrow \dfrac{{{x}^{2}}n}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}=\dfrac{1600n}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}+\dfrac{81n}{2\tan \left( \dfrac{{{180}^{\circ }}}{n} \right)}$
Now on calling the denominator from both the sides, we get:
$\Rightarrow {{x}^{2}}n=1600n+81n$
On taking the term $n$ common, we get:
$\Rightarrow {{x}^{2}}n=\left( 1600+81 \right)n$
On cancelling $n$ from both the sides, we get:
$\Rightarrow {{x}^{2}}=1600+81$
On adding the terms, we get:
$\Rightarrow {{x}^{2}}=1681$
Now we know that ${{41}^{2}}=1681$ therefore, on substituting, we get:
$\Rightarrow {{x}^{2}}={{41}^{2}}$
On taking the square root on both the sides, we get:
$\Rightarrow x=41cm$, which is the required length of the side of the third polygon for its area to be the sum of the polygons with length of side as $40cm$ and $9cm$.

So, the correct answer is “Option 3”.

Note: It is to be noted that in this question when taking the square root of the length only the positive value was considered because length can never be negative. The formula for the area of the polygon should be remembered for doing these types of questions. It is to be noted that the apothem is a line from the center of the polygon to its side at right angles.