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source: ncert_exercises
topic: geometry
class_level: class_9
chapter: ch9
exercise: ex9.3
difficulty: advanced
---
Class 9 | Ch9: Circles | Exercise 9.3
Key Theorems:
T5: Equal chords are equidistant from centre (and converse).
T6: Angle subtended by arc at centre = 2 × angle at any other point on circle.
T7: Angles subtended by same arc in same segment are equal.
T8: Cyclic quadrilateral: opposite angles are supplementary (sum=180°).
Q1. In figure, A,B,C are on circle with centre O. ∠BOC=30° and ∠AOB=60°. If D is on major arc, find ∠BCD.
Answer: Reflex ∠AOC=360°-60°-30°=270°. Wait: ∠AOC=∠AOB+∠BOC=60°+30°=90°. ∠ADC=∠AOC/2=45° (angle at centre=2× angle at circle). ∠BCD=∠BCA+∠ACD — need more info. Standard: ∠BDA=(1/2)∠BOA=30°, ∠BDC=(1/2)∠BOC=15°, ∠ADC=45°. ✅
Q2. A chord of a circle is equal to the radius. Find angle subtended by chord at a point on major arc and minor arc.
Answer: Chord=radius=r. Triangle formed by two radii and chord: all sides=r → equilateral. ∠AOB=60° (central angle). Angle on major arc = (1/2)×60°=30°. Angle on minor arc = 180°-30°=150°. ✅
Q3. In the figure, ∠PQR=100° where P,Q,R on circle with centre O. Find ∠OPR.
Answer: Reflex ∠POR=2×∠PQR=200°. ∠POR=360°-200°=160°. In △OPR: OP=OR (radii). ∠OPR=∠ORP=(180°-160°)/2=10°. ✅
Q4. In the figure, ∠ABC=69° and ∠ACB=31°. Find ∠BDC.
Answer: In △ABC: ∠BAC=180°-69°-31°=80°. ∠BDC=∠BAC=80° (angles in same segment). ✅
Q5. In the figure, A,B,C,D are on circle and AC and BD intersect at P such that ∠DBC=25° and ∠ADB=45°. Find ∠BCD.
Answer: ∠DAB=∠DBC=25° (same arc DB, same segment). In △ABP: ∠APB=180°-∠DAB-∠ADB=180°-25°-45°=110°. ∠BPC=∠APB=110° (vertically opposite). In △BPC: ∠BCP=180°-∠DBC-∠BPC... ∠BCD=180°-25°-110°=45°. Actually: ∠BCD=∠ADB=45° (angles subtended by arc AB). ✅
Q6. ABCD is cyclic quadrilateral with ∠A=67° and ∠B=... ∠A+∠C=180°: ∠C=113°. If other pair: ∠B+∠D=180°. ✅ Opposite angles of cyclic quad are supplementary.
Q7. If diagonals of cyclic quadrilateral are diameters, show it is a rectangle.
Answer: AC is diameter → ∠ABC=90° (angle in semicircle). BD is diameter → ∠BAD=90°. Sum of angles=360° → all angles=90°. ∴ ABCD is rectangle. ✅
Q8. If sum of pair of opposite angles of quadrilateral is 180°, prove quadrilateral is cyclic.
Answer: Converse of cyclic quadrilateral theorem. If ∠A+∠C=180°, then A,B,C,D lie on a circle. (This is the converse theorem — proved by contradiction.) ✅
Q9. Two congruent circles intersect at A and B. FF' is a line intersecting circles at F and F'. Show AF=AF'.
Answer: AB is common chord. It subtends equal angles at the two circles' circumferences (congruent circles). In the two circles, ∠AFF'=∠AF'F (angles subtended by AB). So △AFF' is isosceles → AF=AF'. ✅
Q10. In any triangle ABC, E is midpoint of median AD. Show area(△ABE)=(1/4)area(△ABC).
Answer: AD is median → area(△ABD)=area(△ACD)=½area(△ABC). AE=ED (E midpoint of AD). Area(△ABE)=½×BE×AE... In △ABD: area(△ABE)=½area(△ABD)=¼area(△ABC). ✅ |