Question

Find two regular polygons such that the number of their sides may be as 3 to 4 and the number of degrees in an angle of the first to the number of grades in an angle of the second as 4 to 5.

Answer 1

Hint: We will apply the formula for interior angle which is given by $\dfrac{{{180}^{\text{o}}}\left( n-2 \right)}{n}$ where n is called the number of sides of a regular polygon. We will also use the relation between degree and grade which is given by ${{1}^{g}}={{\dfrac{90}{100}}^{\text{o}}}$.

Complete step-by-step answer:
A polygon is any shape made by n number of sides. The interior angle is an angle which is made by the sides inside that polygon. Numerically, the formula for interior angle is given by $\dfrac{{{180}^{\text{o}}}\left( n-2 \right)}{n}$ where n is called the number of sides of a regular polygon. And since, the polygon is regular therefore all its interior angles will be the same.
In this question we are not given the number of sides of the two polygons. But the ratio of the number of their sides may be as 3 to 4.
So, we can let that one side be ${{n}_{1}}=3x$ and the other side be ${{n}_{2}}=4x$...(i).
 As we know that the formula for the number of sides of any polygon is given by $\dfrac{{{180}^{\text{o}}}\left( n-2 \right)}{n}$. Therefore, we have the ratio according to the statements as,
$\dfrac{\dfrac{{{180}^{\text{o}}}\left( {{n}_{1}}-2 \right)}{{{n}_{1}}}}{\dfrac{{{180}^{\text{o}}}\left( {{n}_{2}}-2 \right)}{{{n}_{2}}}}=\dfrac{\dfrac{{{180}^{\text{o}}}\left( 3x-2 \right)}{3x}}{\dfrac{{{180}^{\text{o}}}\left( 4x-2 \right)}{4x}}$
Also, we have that the number of degrees in an angle of the first to the number of grades in an angle of the second is 4 to 5. Therefore, we actually have $\dfrac{{{4}^{\text{o}}}}{{{5}^{g}}}$. Now, we will apply the relation between grade and degree which is given by ${{90}^{\text{o}}}={{100}^{g}}$ or ${{1}^{g}}={{\dfrac{90}{100}}^{\text{o}}}$. Thus, we get
$\begin{align}
  & {{5}^{g}}=5\times {{\left( 1 \right)}^{g}} \\
 & \Rightarrow {{5}^{g}}=5\times {{\dfrac{90}{100}}^{\text{o}}} \\
 & \Rightarrow {{5}^{g}}=1\times {{\dfrac{9}{2}}^{\text{o}}} \\
 & \Rightarrow {{5}^{g}}={{4.5}^{\text{o}}} \\
\end{align}$
By substituting this value in $\dfrac{{{4}^{\text{o}}}}{{{5}^{g}}}$ we have
$\begin{align}
  & \dfrac{{{4}^{\text{o}}}}{{{5}^{g}}}=\dfrac{{{4}^{\text{o}}}}{{{4.5}^{\text{o}}}} \\
 & \Rightarrow \dfrac{{{4}^{\text{o}}}}{{{5}^{g}}}={{\left( \dfrac{4}{4.5} \right)}^{\text{o}}} \\
\end{align}$
Now according to question now we have,
 $\begin{align}
  & \dfrac{\dfrac{{{180}^{\text{o}}}\left( {{n}_{1}}-2 \right)}{{{n}_{1}}}}{\dfrac{{{180}^{\text{o}}}\left( {{n}_{2}}-2 \right)}{{{n}_{2}}}}=\dfrac{\dfrac{{{180}^{\text{o}}}\left( 3x-2 \right)}{3x}}{\dfrac{{{180}^{\text{o}}}\left( 4x-2 \right)}{4x}} \\
 & \Rightarrow \dfrac{\dfrac{{{180}^{\text{o}}}\left( 3x-2 \right)}{3x}}{\dfrac{{{180}^{\text{o}}}\left( 4x-2 \right)}{4x}}={{\left( \dfrac{4}{4.5} \right)}^{\text{o}}} \\
 & \Rightarrow \dfrac{{{180}^{\text{o}}}\left( 3x-2 \right)}{{{180}^{\text{o}}}\left( 4x-2 \right)}={{\left( \dfrac{4}{4.5}\times \dfrac{3}{4} \right)}^{\text{o}}} \\
 & \Rightarrow \dfrac{{{180}^{\text{o}}}\times 3x-{{180}^{\text{o}}}\times 2}{{{180}^{\text{o}}}\times 4x-{{180}^{\text{o}}}\times 2}={{\left( \dfrac{2}{3} \right)}^{\text{o}}} \\
 & \Rightarrow \dfrac{540{{x}^{\text{o}}}-{{360}^{\text{o}}}}{720{{x}^{\text{o}}}-{{360}^{\text{o}}}}={{\left( \dfrac{2}{3} \right)}^{\text{o}}} \\
 & \Rightarrow 3\left( 540x-360 \right)=2\left( 720x-360 \right) \\
 & \Rightarrow 1620x-1080=1440x-720 \\
 & \Rightarrow 1620x-1440x=1080-720 \\
 & \Rightarrow 180x=360 \\
 & \Rightarrow x=\dfrac{360}{180} \\
 & \Rightarrow x=2 \\
\end{align}$
After substituting the value of x = 2 in the statement (i). Therefore, we can convert one side be ${{n}_{1}}=3x$ or ${{n}_{1}}=3\left( 2 \right)$ or ${{n}_{1}}=6$ and other side be ${{n}_{2}}=4x$ or ${{n}_{2}}=4\left( 2 \right)$ or ${{n}_{2}}=8$.
Hence, one polygon is a hexagon and another polygon is an octagon.

Note: While forming the ratio check about its power or the form in which it should be written. As in this case we can see that and the number of degrees in an angle of the first to the number of grades in an angle of the second as 4 to 5 which is clearly written as $\dfrac{{{4}^{\text{o}}}}{{{5}^{g}}}$ instead of $\dfrac{{{4}^{\text{o}}}}{{{5}^{\text{o}}}}$. The names of the polygons according to the number of sides are discussed in note. The three sides of polygon is called triangle, the 4 sides polygon is called quadrilateral (it can be square or rectangle or rhombus), polygon with 5 sides is pentagon, 6 sides polygon is hexagon, and then heptagon with 7 sides, octagon with 8 sides.