
Each angle of a regular polygon of n sides contains
A) 4n right angles
B) $\dfrac{{2\left( {n + 1} \right)}}{n}$ right angles
C) $\dfrac{{2\left( {n - 1} \right)}}{n}$ right angles
D) $\dfrac{{2\left( {n - 2} \right)}}{n}$ right angles
Answer
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Hint: A polygon is a two- dimensional figure that has at least three sides and three vertices. The polygon contains many angles. The area, as well as the perimeter of a polygon, depends on the type of polygon.
Complete step-by-step answer:
The sum of interior angles of a pentagon can be calculated as,
$A = \left( {2n - 4} \right) \times 90^\circ $
Here, n is the number of sides.
Substitute the number of sides of a polygon, we get,
$\begin{array}{l}
\Rightarrow A = \left( {2 \times 5 - 4} \right) \times 90^\circ \\
\Rightarrow A = 6 \times 90^\circ \\
\Rightarrow A = 540^\circ
\end{array}$
The sum of interior angles of a polygon is $540^\circ $
If a polygon has n sides, then it is divided into $\left( {n - 2} \right)$ triangles as the triangle has 3 sides which is less than the sides of a polygon.
The sum of interior angles of a triangle is $180^\circ $
So, now we can calculate the sum of the angles of $\left( {n - 2} \right)$ triangles,
Therefore,
$S = 180^\circ \times \left( {n - 2} \right)$
Here,
n is the number of sides
$180^\circ $ is the two right triangles
Now, on further solving the above expression we get,
$\begin{array}{l}
\Rightarrow S = 180^\circ \times \left( {n - 2} \right)\\
\Rightarrow S = 2 \times {\rm{right}}\;{\rm{angles}} \times \left( {n - 2} \right)
\end{array}$
The sum of the angles of $\left( {n - 2} \right)$ triangles = $2\left( {n - 2} \right){\rm{right}}\;{\rm{angles}}$
So, if polygon has n sides then,
Therefore,
$\Rightarrow S = \dfrac{{2\left( {n - 2} \right)}}{n}$ right angles
Therefore, the sum of interior angles of a polygon having n sides is $\dfrac{{\left( {2n - 2} \right)}}{n}$ right angles.
So, the correct answer is “Option D”.
Note: In this question, the students must have the knowledge about the polygon. The angle of the polygon and sides of a polygon must be known. The triangle has 3 sides and the total sides of the pentagon are 5.
Complete step-by-step answer:
The sum of interior angles of a pentagon can be calculated as,
$A = \left( {2n - 4} \right) \times 90^\circ $
Here, n is the number of sides.
Substitute the number of sides of a polygon, we get,
$\begin{array}{l}
\Rightarrow A = \left( {2 \times 5 - 4} \right) \times 90^\circ \\
\Rightarrow A = 6 \times 90^\circ \\
\Rightarrow A = 540^\circ
\end{array}$
The sum of interior angles of a polygon is $540^\circ $
If a polygon has n sides, then it is divided into $\left( {n - 2} \right)$ triangles as the triangle has 3 sides which is less than the sides of a polygon.
The sum of interior angles of a triangle is $180^\circ $
So, now we can calculate the sum of the angles of $\left( {n - 2} \right)$ triangles,
Therefore,
$S = 180^\circ \times \left( {n - 2} \right)$
Here,
n is the number of sides
$180^\circ $ is the two right triangles
Now, on further solving the above expression we get,
$\begin{array}{l}
\Rightarrow S = 180^\circ \times \left( {n - 2} \right)\\
\Rightarrow S = 2 \times {\rm{right}}\;{\rm{angles}} \times \left( {n - 2} \right)
\end{array}$
The sum of the angles of $\left( {n - 2} \right)$ triangles = $2\left( {n - 2} \right){\rm{right}}\;{\rm{angles}}$
So, if polygon has n sides then,
Therefore,
$\Rightarrow S = \dfrac{{2\left( {n - 2} \right)}}{n}$ right angles
Therefore, the sum of interior angles of a polygon having n sides is $\dfrac{{\left( {2n - 2} \right)}}{n}$ right angles.
So, the correct answer is “Option D”.
Note: In this question, the students must have the knowledge about the polygon. The angle of the polygon and sides of a polygon must be known. The triangle has 3 sides and the total sides of the pentagon are 5.
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