CG Board Class 8 Mathematics Chapter 11 Solutions: Construction of Quadrilateral (चतुर्भुज की रचना) – 2026-27

CG SCERT (Chhattisgarh Board) Class 8 Mathematics Solutions | 2026-27 Official Edition

Complete 100% textbook-grounded CG Board Class 8 Mathematics Chapter 11 Solutions: Construction of Quadrilateral (चतुर्भुज की रचना) in Bilingual Format (English & Hindi Medium).


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Question 1

Find the sum of interior angles of a…


📝 Question:

Find the sum of interior angles of a polygon having

💡 Solution:

(i) 5 sides**

(ii) 6 sides

(iii) 8 sides

(iv) 12 sides

(v) n sides

  • (i) For 5 sides:
  • (ii) For 6 sides:
  • (iii) For 8 sides:
  • (iv) For 12 sides:
  • (v) For n sides:

Question 2

Find the number of sides of a polygon…


📝 Question:

Find the number of sides of a polygon if the sum of its interior angles is

💡 Solution:

(i) 900°**

(ii) 1620°

(iii) 1080°

(iv) 2340°

  • (i) When Sum = 900°:
  • (ii) When Sum = 1620°:

(n – 2) × 180°= 1620°n – 2 = (1620)/(180) = 9 ⇒ n = 9 + 2 = 11 sides

  • (iii) When Sum = 1080°:
  • (iv) When Sum = 2340°:

(n – 2) × 180°= 2340°n – 2 = (2340)/(180) = 13 ⇒ n = 13 + 2 = 15 sides

Question 3

Find the measure of each interior angle of…


📝 Question:

Find the measure of each interior angle of a regular polygon having

💡 Solution:

(i) 6 sides**

(ii) 8 sides

(iii) 10 sides

(iv) 12 sides

  • (i) For n = 6: frac{(6 – 2) × 180°{6} = frac{720°{6} = 120° (ii) For n = 8: frac{(8 – 2) × 180°{8} = frac{1080°{8} = 135° (iii) For n = 10: frac{(10 – 2) × 180°{10} = frac{1440°{10} = 144° (iv) For n = 12: frac{(12 – 2) × 180°{12} = frac{1800°{12} = 150° Answer: (i) 120°, (ii) 135°, (iii) 144°, (iv) 150°

Question 4

Find the measure of each exterior angle of…


📝 Question:

Find the measure of each exterior angle of a regular polygon having

💡 Solution:

(i) 5 sides**

(ii) 8 sides

(iii) 9 sides

(iv) 15 sides

  • (i) For n = 5: frac{360°{5} = 72° (ii) For n = 8: frac{360°{8} = 45° (iii) For n = 9: frac{360°{9} = 40° (iv) For n = 15: frac{360°{15} = 24° Answer: (i) 72°, (ii) 45°, (iii) 40°, (iv) 24°

Question 5

Find the number of diagonals in a polygon…


📝 Question:

Find the number of diagonals in a polygon having

💡 Solution:

(i) 4 sides**

(ii) 5 sides

(iii) 6 sides

(iv) 8 sides

(v) 12 sides

  • (i) For n = 4: (4(4 – 3))/(2) = (4 × 1)/(2) = 2
  • (ii) For n = 5: (5(5 – 3))/(2) = (5 × 2)/(2) = 5
  • (iii) For n = 6: (6(6 – 3))/(2) = (6 × 3)/(2) = 9
  • (iv) For n = 8: (8(8 – 3))/(2) = (8 × 5)/(2) = 20
  • (v) For n = 12: (12(12 – 3))/(2) = (12 × 9)/(2) = 54

(i) 2, (ii) 5, (iii) 9, (iv) 20, (v) 54

Question 6

If four angles of a pentagon are 100…


📝 Question:

If four angles of a pentagon are 100°, 110°, 120°, 85°, find the fifth angle

💡 Solution:

  • Sum of angles of a pentagon / = (5 – 2) × 180°= 540°.
  • Let fifth angle be x.
  • 100°+ 110°+ 120°+ 85°+ x = 540° 415°+ x = 540° ⇒ x = 540°- 415°= 125° Answer: 125°

Question 1

Construct a quadrilateral ABCD in which AB 4…


📝 Question:

Construct a quadrilateral ABCD in which AB = 4 cm, BC = 3 cm, CD = 4.5 cm, AD = 5 cm, and diagonal AC = 6 cm

💡 Solution:

  1. Draw line segment AB = 4 cm.
  2. With center A and radius 6 cm, draw an arc.
  3. With center B and radius 3 cm, draw another arc intersecting the previous arc at point C.
  4. Join AC and BC.
  5. With center A and radius 5 cm, draw an arc on the other side of AC.
  6. With center C and radius 4.5 cm, draw an arc intersecting the previous arc at point D.
  7. Join AD and CD. ABCD is the required quadrilateral.

Question 2

Construct a quadrilateral PQRS in which PQ 42…


📝 Question:

Construct a quadrilateral PQRS in which PQ = 4.2 cm, QR = 3.6 cm, RS = 4.8 cm, PS = 5 cm, and diagonal PR = 6.4 cm

💡 Solution:

  1. Draw line segment PQ = 4.2 cm.
  2. With center P, radius 6.4 cm, and center Q, radius 3.6 cm, draw arcs intersecting at R.
  3. With center P, radius 5 cm, and center R, radius 4.8 cm, draw arcs intersecting at S.
  4. Join QR, RS, SP, and PR. PQRS is the required quadrilateral.

### *EXERCISE 11.2

Question 1

Construct a quadrilateral ABCD in which AB 35…


📝 Question:

Construct a quadrilateral ABCD in which AB = 3.5 cm, BC = 4 cm, CD = 5 cm, diagonal AC = 5.5 cm, and diagonal BD = 6 cm

💡 Solution:

  1. Draw line segment AB = 3.5 cm.
  2. With center A, radius 5.5 cm, and center B, radius 4 cm, draw arcs intersecting at C.
  3. With center B, radius 6 cm, and center C, radius 5 cm, draw arcs intersecting at D.
  4. Join AD, CD, BC, and BD. ABCD is the required quadrilateral.

### *EXERCISE 11.3

Question 1

Construct a quadrilateral ABCD in which AB 4…


📝 Question:

Construct a quadrilateral ABCD in which AB = 4 cm, BC = 3.5 cm, ∠ A = 60°, ∠ B = 100°, and ∠ C = 80°

💡 Solution:

  1. Draw line segment AB = 4 cm.
  2. At point A, construct angle ∠ XAB = 60°.
  3. At point B, construct angle ∠ YBA = 100°.
  4. Cut an arc of radius 3.5 cm from B on ray BY to mark point C.
  5. At point C, construct angle ∠ BCZ = 80°intersecting ray AX at point D.
  6. ABCD is the required quadrilateral.


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