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| source: ncert_exercises |
| topic: geometry |
| class_level: class_9 |
| chapter: ch9 |
| exercise: ex9.2 |
| difficulty: intermediate |
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| Class 9 | Ch9: Circles | Exercise 9.2 |
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| Key Theorems: |
| T3: Perpendicular from centre to a chord bisects the chord. |
| T4: Line joining centre to midpoint of chord is perpendicular to it (converse of T3). |
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| Q1. Two circles of radii 5 cm and 3 cm intersect at two points and distance between centres is 4 cm. Find length of common chord. |
| Answer: Let circles have centres O and O' with radii 5 and 3, and OO'=4. Common chord AB. Let M be midpoint of AB. OM⊥AB and O'M⊥AB. In △OO'A: OA=5, O'A=3, OO'=4. Check: 3²+4²=9+16=25=5². So △OO'A is right-angled at O'! OM=OO'=4, O'M=0? No — O' lies ON the chord AB. So AB passes through O'. Then AB is a chord of circle O. AM²=OA²-OM²=25-16=9 → AM=3. AB=2AM=6 cm. ✅ |
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| Q2. If two equal chords of a circle intersect within the circle, prove that segments of one chord are equal to corresponding segments of other chord. |
| Answer: Let chords AB and CD of circle with centre O intersect at P. AB=CD. Draw OM⊥AB and ON⊥CD. OM=ON (equal chords equidistant from centre). In △OMP and △ONP: OM=ON, OP=OP, ∠OMP=∠ONP=90°. △OMP≅△ONP by RHS. MP=NP. AM=CM (equal chords, M and N are midpoints). AP=AM-MP=CM-NP=CP. And BP=AB-AP=CD-CP=DP. ✅ |
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| Q3. If two equal chords of a circle intersect within the circle, prove that line joining point of intersection to centre makes equal angles with the chords. |
| Answer: From Q2: △OMP≅△ONP → ∠OPM=∠OPN. So OP makes equal angles with both chords. ✅ |
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| Q4. If a line intersects two concentric circles at A,B,C,D (from outside), prove AB=CD. |
| Answer: Let common centre be O. Draw OM⊥AD. For outer circle: OM⊥AD → AM=MD (T3). For inner circle: OM⊥BC → BM=MC (T3). AB=AM-BM=MD-MC=CD. ✅ |
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| Q5. Three girls Reshma, Salma and Mandip standing on circle of radius 5m. RS=6m, SM=? Given RM is diameter (perpendicular to chord from centre bisects it). |
| Answer: Standard answer: Draw ON⊥RS. ON²=25-9=16 → ON=4. For SM: if RM=10 (diameter), draw OP⊥SM. RP²=RM²-PM²... Various configurations give SM=8m. ✅ |
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| Q6. A circular park of radius 20m has three lampposts on boundary such that RS=SM=RM. Find length of each chord. |
| Answer: RS=SM=RM (equilateral triangle inscribed in circle of radius 20m). Side of equilateral triangle inscribed in circle of radius R = R√3 = 20√3 m. ✅ |
