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The measure of an angle in degrees, grades and radians be D, G and C respectively, then relation between them \[\dfrac{D}{90}=\dfrac{G}{100}=\dfrac{2C}{\pi }\] but \[{{1}^{\circ }}={{\left( \dfrac{180}{\pi } \right)}^{o}}\simeq {{57}^{\circ }},17',44.8''\] and sum of interior angles of a n-sided regular polygon is \[(2n-4)\dfrac{\pi }{2}\]. On the basis of above information, answer the following question: The number of sides of two regular polygons are as 5:4 and the difference between their angles is \[\dfrac{\pi }{20}\], the number of sides in the polygons respectively are –
A. 25, 20
B. 20, 16
C. 15, 12
D. 10, 8

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Last updated date: 22nd Feb 2024
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IVSAT 2024
Answer
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Hint: We are given relations between sides of a polygon and their interior angles. Firstly, we will deal with the ratio of the sides of the two polygons and then use it in the formula of the interior angles of a polygon. When we get the interior angle expression of each of the polygons, we will put it in the equation which stated their difference as \[\dfrac{\pi }{20}\] as per the question. Then, we will get the values of the number of sides of both the polygons.

Complete step by step solution:
Polygon is the general term given for figures with many sides. For example – square is a polygon.
According to the given question, the ratio of the sides given is 5:4
Let each side be $x$.
Then the number of sides in the first polygon be $5x$.
And the number of sides in the second polygon be $4x$.
And we are given that sum of interior angles of a polygon is \[(2n-4)\dfrac{\pi }{2}\],
So, the interior angle of a polynomial would be the ratio of sum of interior angles of a polygon and the number of sides of a polygon.
Interior angle of a polygon \[=\dfrac{(2n-4)\dfrac{\pi }{2}}{n}=\dfrac{(n-2)\pi }{n}\]
Let the interior angle of the first polygon be ‘a’
Then the interior angle of the second polygon be ‘b’
Therefore,
\[a=\dfrac{(5x-2)\pi }{5x}\]
\[b=\dfrac{(4x-2)\pi }{4x}\]
We are also given that the difference in the interior angle of the first and second polygon is \[\dfrac{\pi }{20}\].
 We get,
\[a-b=\dfrac{\pi }{20}\]
\[\Rightarrow \dfrac{(5x-2)\pi }{5x}-\dfrac{(4x-2)\pi }{4x}=\dfrac{\pi }{20}\]
We will take the LCM here, we then get
\[\Rightarrow \dfrac{(20{{x}^{2}}-8x-20{{x}^{2}}+10x)\pi }{20{{x}^{2}}}=\dfrac{\pi }{20}\]
\[\Rightarrow \dfrac{2x\pi }{20{{x}^{2}}}=\dfrac{\pi }{20}\]
Cancelling the common terms, we have
\[\Rightarrow \dfrac{2}{x}=1\]
\[\Rightarrow x=2\]
Putting the values of x in the sides of polygon, we get,
Therefore, the number of sides in the first polygon \[=5x=5(2)=10\]
And the number of sides in the second polygon be \[=4x=4(2)=8\]
Therefore, the answer is D 10,8.

Note: The sum of interior angles of a polygon given in the question should not be taken for the interior angle of a polygon, else the answer will be wrong. While substituting the value of x in the sides of the polygon, the calculation must be done neatly.