Question

Ridhi drew a polygon with n sides. The smallest exterior angle is \(8^\circ\) and each subsequent exterior angle is \(4^\circ\) more than the previous exterior angle. Find the number of sides of the polygon that Ridhi had drawn.

Hint:

The exterior angles form an A.P. with \(a=8\), \(d=4\). Put \(S_n = 360\) and solve for \(n\).

Answer 1

Step 1: Setting up:
Let the polygon have \(n\) sides. Its exterior angles go \(8^\circ, 12^\circ, 16^\circ, \ldots\) in steps of \(4^\circ\), so they are in A.P. The last (largest) exterior angle is
\[ l = 8 + (n-1)\times 4 = 4n + 4 \]

Step 2: Use the sum in the first-plus-last form:
For an A.P., the sum is \(\dfrac{n}{2}(a + l)\). Since all exterior angles of a polygon add up to \(360^\circ\),
\[ \frac{n}{2}\left(8 + 4n + 4\right) = 360 \]
\[ \frac{n}{2}(4n + 12) = 360 \implies n(2n + 6) = 360 \]

Step 3: Reduce and use the quadratic formula:
\[ 2n^2 + 6n - 360 = 0 \implies n^2 + 3n - 180 = 0 \]
The discriminant is \(3^2 + 4 \times 180 = 729 = 27^2\). So
\[ n = \frac{-3 \pm 27}{2} \]
This gives \(n = 12\) or \(n = -15\).

Step 4: Choose the valid root:
A polygon cannot have a negative number of sides, so only \(n = 12\) is allowed. The largest exterior angle is then \(4 \times 12 + 4 = 52^\circ\), which is less than \(180^\circ\), as an exterior angle of a convex polygon should be.

Final Answer:
The polygon has 12 sides. \[ \boxed{n = 12} \]