Question : $A_1$ and $A_2$ are two regular polygons. The sum of all the interior angles of $A_1$ is $1080^{\circ}$. Each interior angle of $A_2$ exceeds its exterior angle by $132^{\circ}$. The sum of the number of sides $A_1$ and $A_2$ is:
Option 1: 21
Option 2: 22
Option 3: 23
Option 4: 24
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Correct Answer: 23
Solution :
Let the sides of polygon $A_1$ be $n$.
The sum of all the interior angles $A_1 = 1080^{\circ}$
$\Rightarrow (n-2)\times 180^{\circ} = 1080^{\circ}$
$\Rightarrow (n-2) =6$
$\Rightarrow n=8$
Now,
Let $I$ be the interior angle and $E$ be the exterior angle.
$\Rightarrow I-E = 132^{\circ}$.....(1)
$\Rightarrow I+E = 180^{\circ}$....(2)
Adding equation (1) and (2),
$\Rightarrow 2I = 180^{\circ} + 132^{\circ}$
$\Rightarrow 2I = 312^{\circ}$
$\Rightarrow I = 156^{\circ}$
Putting the value of $I$ in equation (1), we get,
$\Rightarrow 156^{\circ} - E = 132^{\circ}$
$\Rightarrow E = 24^{\circ}$
Number of sides of polygon $ A_2=\frac{360}{24} = 15$
Now,
The sum of the number of sides of polygon $A_1$ and $A_2 = (8 + 15) = 23$
Hence, the correct answer is 23.
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