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source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch8
exercise: ex8.1
difficulty: intermediate
---
Class 9 | Ch8: Quadrilaterals | Exercise 8.1
Key Theorems:
- Angle sum property: Sum of angles of quadrilateral = 360°.
- Parallelogram: opposite sides equal, opposite angles equal, diagonals bisect each other.
- Mid-point theorem: Line joining midpoints of two sides of triangle is parallel to third side and half its length.
Q1. The angles of a quadrilateral are in ratio 3:5:9:13. Find all angles.
Answer: 3x+5x+9x+13x=360° → 30x=360° → x=12°. Angles: 36°, 60°, 108°, 156°. ✅
Q2. If diagonals of a parallelogram are equal, show it is a rectangle.
Answer: In parallelogram ABCD, AC=BD. In △ABC and △ABD: AB=AB (common), BC=AD (opp sides of ∥gram), AC=BD (given). △ABC≅△ABD by SSS. ∠ABC=∠BAD by CPCT. But ∠ABC+∠BAD=180° (co-interior). 2∠ABC=180° → ∠ABC=90°. All angles=90°. ∴ ABCD is rectangle. ✅
Q3. Show diagonals of a rhombus are perpendicular to each other.
Answer: In rhombus ABCD, AB=BC=CD=DA. Diagonals AC and BD bisect each other at O. In △AOB and △COB: AO=CO (diagonals bisect), OB=OB, AB=CB (sides of rhombus). △AOB≅△COB by SSS. ∠AOB=∠COB. But ∠AOB+∠COB=180° → ∠AOB=90°. ∴ Diagonals ⊥. ✅
Q4. Show that diagonals of a square are equal and bisect each other at right angles.
Answer: Square = rectangle + rhombus. From Q2: equal diagonals (rectangle property). From Q3: perpendicular diagonals (rhombus property). Also diagonals bisect each other (parallelogram property). ✅
Q5. Show that if diagonals of a quadrilateral bisect each other at right angles, it is a rhombus.
Answer: Let diagonals bisect at O at 90°. In △AOB, △BOC, △COD, △DOA: AO=CO, BO=DO, ∠AOB=∠BOC=∠COD=∠DOA=90°. By SAS: all four triangles congruent. ∴ AB=BC=CD=DA → rhombus. ✅
Q6. Diagonal AC of parallelogram ABCD bisects ∠A. Show it bisects ∠C and ABCD is a rhombus.
Answer: ∠DAC=∠BAC (AC bisects ∠A). ∠DAC=∠DCA (AD∥BC, alternate interior). So ∠BAC=∠DCA. Also ∠BAC=∠BCA (AB∥DC, alternate interior). So ∠BCA=∠DCA → AC bisects ∠C. In △ABC: ∠BAC=∠BCA → AB=BC. ∴ ABCD is rhombus. ✅
Q7. ABCD is a rhombus. Show that diagonal AC bisects ∠A and ∠C, and BD bisects ∠B and ∠D.
Answer: In △ABC: AB=BC (rhombus) → ∠BAC=∠BCA. AB∥DC → ∠BAC=∠DCA (alternate). So ∠BCA=∠DCA → AC bisects ∠C. Similarly for ∠A and BD. ✅
Q8. ABCD is rectangle with ∠ABD=25°. Find ∠DBC.
Answer: In rectangle, ∠ABC=90°. ∠DBC=∠ABC-∠ABD=90°-25°=65°. ✅
Q9. Diagonals of a parallelogram bisect each other. Given OA=OC and OB=OD. Show ABCD is a parallelogram.
Answer: In △AOB and △COD: OA=OC, OB=OD, ∠AOB=∠COD (vertically opposite). △AOB≅△COD by SAS. AB=CD and ∠OAB=∠OCD (alternate interior) → AB∥CD. Similarly AD∥BC. ∴ ABCD is parallelogram. ✅
Q10. ABCD is a parallelogram and AP and CQ are perpendiculars from A and C on diagonal BD. Show △APB≅△CQD.
Answer: In △APB and △CQD: AB=CD (opposite sides), ∠ABP=∠CDQ (alternate interior, AB∥CD), ∠APB=∠CQD=90°. ∴ △APB≅△CQD by AAS. ✅
Q11. In △ABC and △DEF, AB∥DE, AB=DE, BC∥EF, BC=EF. Show AC∥DF and AC=DF.
Answer: ABDE is a parallelogram (AB∥DE, AB=DE) → AD∥BE and AD=BE. BCEF is a parallelogram (BC∥EF, BC=EF) → BE∥CF and BE=CF. So AD∥CF and AD=CF → ADFC is parallelogram → AC∥DF and AC=DF. ✅
Q12. ABCD is a trapezium with AB∥DC. BD is diagonal and E is midpoint of AD. Through E, line parallel to DC meets BC at F. Show F is midpoint of BC.
Answer: In △ABD, EG∥AB (E midpoint of AD, G on BD). By midpoint theorem: G is midpoint of BD. In △BDC, GF∥DC (given, and G midpoint of BD). By converse of midpoint theorem: F is midpoint of BC. ✅ |