CG SCERT (Chhattisgarh Board) Class 9 Mathematics Solutions | 2026-27 Official Edition
Complete 100% textbook-grounded CG Board Class 9 Mathematics Chapter 9 Solutions: Straight Line and Angle (सरल रेखा और कोण) in Bilingual Format (English & Hindi Medium).
Official Textbook Exercises
End-of-Chapter Exercises (Solutions)
अभ्यास के सभी प्रश्नों के हल
Question 1
Exercise Question
प्रश्न 1: संलग्न चित्र में ∠ POR तथा ∠ QOR एक रैखिक युग्म बन…
📝 Question:
- (i) 2x + 3x = 180°(Co-interior / Linear pair) ⇒ 5x = 180° ⇒ {x = 36°.
- y = 3x = 3(36°) = {108° (Vertically opposite / Corresponding).
- (ii) 4x – 23°+ 3x = 180° ⇒ 7x = 203° ⇒ {x = 29°.
- y = 3x = 3(29°) = {87°.
- (iii) 2y + 15°+ 3y – 10°= 180° ⇒ 5y + 5°= 180° ⇒ 5y = 175° ⇒ {y = 35°.
- x = 180°- (2y + 15°) = 180°- 85°= {95°.
- (i) x = {59°, y = {60°
- (ii) x = {40°, y = {40°
- (iii) x = {18°, y = {60°
- (iv) x = {20°, y = {63°
- (i) x = {140°
- (ii) x = {100°
- (iii) x = {250°
- (i) x = {40°, y = {70°
- (ii) x = {50°, y = {20°
- (iii) x = {51°, y = {35°
- (iv) x = {30°, y = {75°
💡 Solution:
- Solution / :
- Since ∠ POR and ∠ QOR form a linear pair:
- x + y = 180°—— (1)
- Given: x – y = 80°—— (2)
- Adding equations (1) and (2):
- 2x = 260° ⇒ x = {130°
- Substituting x = 130°in (1):
- 130°+ y = 180° ⇒ y = {50°
- Solution / :
- Lines PQ and RS intersect at O ⇒ ∠ POR = ∠ QOS = 40°(Vertically opposite angles).
- Given ∠ POR + ∠ QOT = 70° ⇒ 40°+ ∠ QOT = 70° ⇒ ∠ QOT = {30°.
- On line PQ: ∠ POR + ∠ ROT + ∠ QOT = 180°* 70°+ ∠ ROT = 180° ⇒ ∠ ROT = 110°.
- Reflex ∠ ROT = 360°- 110°= {250°.
- Solution / :
- Since ∠ NOB = 90°on line AB ⇒ ∠ NOA = 90° ⇒ x + y = 90°.
- Let x = 2k and y = 3k ⇒ 2k + 3k = 90° ⇒ 5k = 90° ⇒ k = 18°.
- Therefore, y = 3 × 18°= 54°.
- Since LM is a straight line, y + z = 180°(Linear pair).
- 54°+ z = 180° ⇒ z = 180°- 54°= {126°.
- Solution / :
- On the line AB:
- ∠ ECA + ∠ ECD = 180°(Linear pair) —— (1)
- ∠ EDB + ∠ EDC = 180°(Linear pair) —— (2)
- From (1) and (2): ∠ ECA + ∠ ECD = ∠ EDB + ∠ EDC.
- Since ∠ ECD = ∠ EDC, subtracting this equal angle from both sides:
- ∠ ECA = ∠ EDB. Hence proved.
- Solution / :
- Sum of all angles around point O is 360°:
- a + b + c + d = 360°* Since c + d = a + b: (a + b) + (a + b) = 360° ⇒ 2(a + b) = 360° ⇒ a + b = 180°.
- Since the adjacent angles a and b sum to 180°, their non-common arms form a straight line.
- Therefore, LOM is a straight line. Hence proved.
- Solution / :
- Since OC ⊥ AB ⇒ ∠ AOC = ∠ BOC = 90°.
- ∠ BOD = ∠ BOC + ∠ COD = 90°+ ∠ COD —— (1)
- ∠ AOD = ∠ AOC – ∠ COD = 90°- ∠ COD —— (2)
- Subtracting equation (2) from (1):
- ∠ BOD – ∠ AOD = (90°+ ∠ COD) – (90°- ∠ COD) = 2∠ COD
- ∠ COD = 1/2(∠ BOD – ∠ AOD). Hence proved.
- Solution / :
- Since AX is a straight line, ∠ ABC + ∠ CBX = 180° ⇒ 64°+ ∠ CBX = 180° ⇒ ∠ CBX = 116°.
- Ray BY bisects ∠ CBX ⇒ ∠ CBY = ∠ YBX = frac{116°{2} = 58°.
- ∠ ABY = ∠ ABC + ∠ CBY = 64°+ 58°= {122°.
- Reflex ∠ YBX = 360°- ∠ YBX = 360°- 58°= {302°.
- Solution / :
- Since AB ∥ CD ⇒ x + y = 180°(Co-interior angles).
- Since AB ∥ EF ⇒ x = z (Alternate interior angles).
- Therefore, z + y = 180°.
- Let y = 3k and z = 7k ⇒ 3k + 7k = 180° ⇒ 10k = 180° ⇒ k = 18°.
- z = 7 × 18°= 126° ⇒ {x = 126°.
- Solution / :
- Since AB ∥ CD ⇒ ∠ AGE = ∠ GED = {126° (Alternate interior angle).
- ∠ GEF = ∠ GED – ∠ FED = 126°- 90°= {36°.
- Since AB is a straight line, ∠ AGE + ∠ FGE = 180° ⇒ 126°+ ∠ FGE = 180° ⇒ ∠ FGE = {54°.
- Solution / :
- Draw a line RU ∥ ST passing through R.
- ∠ SRU + ∠ RST = 180° ⇒ ∠ SRU = 180°- 130°= 50°.
- Since PQ ∥ RU ⇒ ∠ QRU = ∠ PQR = 110°(Alternate interior angle).
- ∠ QRS = ∠ QRU – ∠ SRU = 110°- 50°= {60°.
- Solution / :
- Since AB ∥ CD and PQ is transversal ⇒ x = ∠ APQ = {50° (Alternate interior angle).
- Since AB ∥ CD and PR is transversal ⇒ ∠ APR = ∠ PRD = 127°.
- 50°+ y = 127° ⇒ y = 127°- 50°= {77°.
- Solution / :
- Corresponding angle x = {80°.
- Co-interior angle y = 180°- 80°= {100°.
- Solution / :
- Draw normals BN ⊥ PQ and CM ⊥ RS.
- Since PQ ∥ RS ⇒ BN ∥ CM.
- By laws of reflection: ∠ 1 = ∠ 2 and ∠ 3 = ∠ 4.
- Since BN ∥ CM with transversal BC ⇒ ∠ 2 = ∠ 3 (Alternate interior angles).
- Multiplying by 2: 2∠ 2 = 2∠ 3 ⇒ ∠ 1 + ∠ 2 = ∠ 3 + ∠ 4 ⇒ ∠ ABC = ∠ BCD.
- Since alternate interior angles ∠ ABC and ∠ BCD are equal, AB ∥ CD. Hence proved.
- Solution / :
- ∠ ACB + ∠ BCE = 180°(Linear pair) ⇒ ∠ ACB = 180°- 163°= {17°.
- Exterior angle ∠ BCE = ∠ BAC + ∠ ABC ⇒ 163°= 34°+ ∠ ABC ⇒ ∠ ABC = {129°.
- ∠ DBC + ∠ ABC = 180°(Linear pair) ⇒ ∠ DBC = 180°- 129°= {51°.
- Solution / :
- Using angle relationships in parallel lines and angle bisector property:
- {∠ 1 = 60°.
- Solution / :
- ∠ QPR = 180°- 135°= 45°(Linear pair)
- ∠ PQR = 180°- 110°= 70°(Linear pair)
- In △ PQR: ∠ PRQ = 180°- (45°+ 70°) = 180°- 115°= {65°.
- Solution / :
- In △ XYZ: ∠ XZY = 180°- (62°+ 54°) = 180°- 116°= 64°.
- ZO bisects ∠ XZY ⇒ ∠ OZY = frac{64°{2} = {32°.
- YO bisects ∠ XYZ ⇒ ∠ OYZ = frac{54°{2} = 27°.
- In △ OYZ: ∠ YOZ = 180°- (27°+ 32°) = 180°- 59°= {121°.
- Solution / :
- Since AB ∥ DE ⇒ ∠ CED = ∠ BAC = 35°(Alternate interior angle).
- In △ CDE: ∠ DCE = 180°- (53°+ 35°) = 180°- 88°= {92°.
- Solution / :
- In △ PRT: ∠ PTR = 180°- (95°+ 40°) = 180°- 135°= 45°.
- Vertically opposite angle ∠ STQ = ∠ PTR = 45°.
- In △ STQ: ∠ SQT = 180°- (75°+ 45°) = 180°- 120°= {60°.
- Solution / :
- Exterior angle property for △ BCD: x + 28°= 65° ⇒ {x = 37°? (Or alternate angle x + 28°= 65°).
- In right triangle ABC: y = 90°- x = 90°- 37°= {53°.
- Solution / :
- In △ ABC, exterior ∠ ACD = ∠ BAC + ∠ ABC.
- Dividing by 2: 1/2∠ ACD = 1/2∠ BAC + 1/2∠ ABC —— (1)
- In △ BEC, exterior ∠ ECD = ∠ BEC + ∠ EBC.
- Since EC bisects ∠ ACD ⇒ ∠ ECD = 1/2∠ ACD.
- Since EB bisects ∠ ABC ⇒ ∠ EBC = 1/2∠ ABC.
- So 1/2∠ ACD = ∠ BEC + 1/2∠ ABC —— (2)
- Comparing (1) and (2):
- 1/2∠ BAC + 1/2∠ ABC = ∠ BEC + 1/2∠ ABC
- ∠ BEC = 1/2∠ BAC. Hence proved.