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

MathsCBSE Class 10 · Standard · Sample paper, session 2026-27 · 3 marks · Short answer

CBSE Class 10 Mathematics Standard sample paper 2026-27: question 28(A)

From the official CBSE sample question paper for 2026-27, section C, question 28(A), for 3 marks. Try it first; the hints and our step-by-step solution are below.

Q28(A) · 3 marks · Short answerArithmetic Progressions

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.

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2026-27.

Hints

Open one hint at a time and try again before the next.

Hint 1

What is the sum of all exterior angles of any polygon?

Hint 2

The exterior angles 8°, 12°, 16°, … form an A.P. Set the sum of its first $n$ terms equal to 360 and solve the quadratic in $n$ .

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

The exterior angles form an A.P. with first term $a = 8$ and common difference $d = 4$ .
The exterior angles of any polygon add up to $36 0^{\circ}$ , so $S_{n} = 360$ :
$\frac{n}{2} [2 \times 8 + (n - 1) \times 4] = 360$ .
1 mark
Simplify:
$\frac{n}{2} (4 n + 12) = 360$
$\Rightarrow 2 n^{2} + 6 n - 360 = 0$
$\Rightarrow n^{2} + 3 n - 180 = 0$
$\Rightarrow (n + 15) (n - 12) = 0$ .
1 mark
So $n = - 15$ or $n = 12$ .
½ mark
The number of sides cannot be negative, so $n = 12$ .
The polygon has 12 sides.
½ mark

Answer: 12 sides

Where the marks are

Uses sum of exterior angles = 360° to form $\frac{n}{2} [16 + 4 (n - 1)] = 360$ (1 mark)
Reduces to $n^{2} + 3 n - 180 = 0$ and factorises (1 mark)
Gets $n = - 15$ or $12$ (½ mark)
Rejects the negative value and concludes 12 sides (½ mark)

Revise the chapter

Chapter: Arithmetic Progressions (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Arithmetic Progressions, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2026-27 (Standard): every question, by section.

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CBSE Class 10 Mathematics Standard sample paper 2026-27: question 28(A)

Hints

Step-by-step solution

Revise the chapter

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