Question

Prove that the interior angle of a regular five-sided polygon (pentagon) is three times the exterior angle of a regular decagon.

Answer 1

Hint: We will first find an interior angle of a pentagon using the formula, $\dfrac{\left( n-2 \right)\left( 180 \right)}{n}$, here for pentagon n will be 5. We will also use the concept that the exterior angle of any polygon is equal to 360˚ and then we will find the exterior angle of decagon using the formula, $\dfrac{360}{n}$, for decagon n will be 10. Then we will compare both the angles obtained in the last step.

Complete step by step solution:
It is given in the question that we have to prove that the interior angle of a regular five-sided polygon (pentagon) is three times the exterior angle of a regular decagon.
We will consider the 5 sided regular polygon, that is, a regular pentagon. We can represent a regular pentagon as shown below.

Now, the interior angle of any polygon can be calculated using the formula, $\dfrac{\left( n-2 \right)\left( 180 \right)}{n}$, where n is the number of sides and for a regular pentagon, the value of n is equal to 5.
On putting the value of n = 5 in the formula to find the interior angle of a polygon, we get,
$\begin{align}
  & \dfrac{\left( 5-2 \right)\left( 180 \right)}{5} \\
 =& \dfrac{\left( 3 \right)\left( 180 \right)}{5} \\
 =& \dfrac{540}{5} \\
 =& {{108}^{\circ }} \\
\end{align}$
So, each interior angle of a regular pentagon is equal to 108˚.
Now, we know that the exterior angle of a regular polygon is equal to 360˚. So, for a decagon, the exterior angle is equal to 360˚. And we know that a regular decagon has 10 sides, so we can represent it as follows.

Now, each exterior angle of a decagon will be equal to $\dfrac{360}{n}$, here n = 10. So, we get an exterior angle of a regular decagon as, $\dfrac{360}{10}={{36}^{\circ }}$.
Now, if we compare the angle of interior angle of pentagon and the exterior angle of a decagon, we get,
108 = 3 (36)
So, we get that the interior angle of a regular five-sided polygon (pentagon) is three times the exterior angle of a regular decagon. Hence proved.

Note: The most common mistake that the students make while solving this question, is by writing the wrong formulas. They might write the formula of the exterior angles of a polygon as $\dfrac{\left( n-2 \right)\left( 180 \right)}{n}$ instead of $\dfrac{360}{n}$ and may write the formula of interior angles of a polygon as $\dfrac{360}{n}$ instead of $\dfrac{\left( n-2 \right)\left( 180 \right)}{n}$. So, the students must be careful while solving this question.