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| source: ncert_exercises |
| topic: geometry |
| class_level: class_9 |
| chapter: ch6 |
| exercise: ex6.2 |
| difficulty: intermediate |
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| Class 9 | Ch6: Lines and Angles | Exercise 6.2 |
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| Key Theorems (Parallel lines + Transversal): |
| - Corresponding angles are equal (and converse). |
| - Alternate interior angles are equal (and converse). |
| - Co-interior (same-side interior) angles are supplementary — sum=180° (and converse). |
| - Lines parallel to same line are parallel to each other. |
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| Q1. In the figure, find the values of x and y, then show that AB∥CD. |
| Answer: x=50° (vertically opposite). y=130° (linear pair with 50°). x+y=180° → co-interior angles sum=180° → AB∥CD. ✅ |
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| Q2. In the figure, if AB∥CD, CD∥EF and y:z=3:7, find x. |
| Answer: AB∥CD∥EF. x=z (alternate interior, AB∥EF). y+z=180° (co-interior, CD∥EF... checking). y:z=3:7 → y=54°, z=126°. x=z=126°. ✅ |
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| Q3. In the figure, if AB∥CD, EF⊥CD and ∠GED=126°, find ∠AGE, ∠GEF and ∠FGE. |
| Answer: ∠GEF=∠GED-∠FED=126°-90°=36°. ∠AGE=∠GED=126° (alternate interior, AB∥CD). ∠FGE=180°-∠AGE=54° (linear pair). ✅ |
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| Q4. In the figure, if PQ∥ST, ∠PQR=110° and ∠RST=130°, find ∠QRS. |
| Answer: Draw line through R parallel to PQ and ST. ∠QRP'=180°-110°=70° (co-interior with PQ). ∠P'RS=180°-130°=50° (co-interior with ST). ∠QRS=70°+50°=120°. ✅ |
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| Q5. In the figure, if AB∥CD, ∠APQ=50° and ∠PRD=127°, find x and y. |
| Answer: x=∠APQ=50° (alternate interior angles, AB∥CD). ∠PRD=∠QPR+x → 127°=y+50° → y=77°. ✅ |
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| Q6. In the figure, PQ and RS are two mirrors placed parallel to each other. An incident ray AB strikes mirror PQ at B, reflected ray moves along BC and strikes mirror RS at C, again reflected along CD. Prove AB∥CD. |
| Answer: Draw BM⊥PQ at B and CN⊥RS at C. BM∥CN (both ⊥ to parallel mirrors). ∠ABM=∠MBC (angle of incidence=angle of reflection at B). ∠BCN=∠NCD (at C). ∠MBC=∠BCN (alternate interior, BM∥CN). So ∠ABM=∠NCD. AB∥CD (alternate interior angles equal). ✅ Proved. |
