1Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.
Answer & Reference:
Consider two congruent circles with centres O and O′ and equal chords AB and CD.
In △AOB and △CO′D:
OA = O′C (radii of congruent circles)
OB = O′D (radii of congruent circles)
AB = CD (given)
∴ △AOB ≅ △CO′D (SSS congruence)
⇒ ∠AOB = ∠CO′D (CPCT). (Based on Theorem 9.1 logic, Page not numbered but follows Section 9.1)
2Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.
Answer & Reference:
Let ∠AOB = ∠CO′D in congruent circles with centres O and O′.
OA = O′C, OB = O′D (radii of congruent circles)
∠AOB = ∠CO′D (given)
∴ △AOB ≅ △CO′D (SAS)
⇒ AB = CD (CPCT). (Converse of Theorem 9.1, Section 9.1)
3Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.
Answer & Reference:
Let centres be O and O′, with OO′ = 4 cm. Let common chord be AB, and M its midpoint.
Then OM ⊥ AB and O′M ⊥ AB.
Let OM = x ⇒ O′M = 4 − x.
In △OAM: OA² = OM² + AM² ⇒ 25 = x² + AM²
In △O′AM: O′A² = O′M² + AM² ⇒ 9 = (4 − x)² + AM²
Subtract: 25 − 9 = x² − (16 − 8x + x²) ⇒ 16 = 8x − 16 ⇒ 8x = 32 ⇒ x = 4
But x = 4 ⇒ O′M = 0 ⇒ M coincides with O′ ⇒ AM² = 25 − 16 = 9 ⇒ AM = 3 cm
∴ AB = 2 × 3 = 6 cm. (Uses Theorem 9.3: perpendicular from centre bisects chord)
4If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.
Answer & Reference:
Let chords AB and CD intersect at E, with AB = CD.
Draw OM ⊥ AB, ON ⊥ CD. Since AB = CD, OM = ON (Theorem 9.5).
In right △OME and △ONE:
OM = ON, OE = OE ⇒ △OME ≅ △ONE (RHS)
⇒ ME = NE.
Since M, N are midpoints: AE = AM − ME, CE = CN − NE ⇒ AE = CE and BE = DE. (Based on Theorems 9.3 and 9.5, Section 9.3)
5If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.
Answer & Reference:
Using same setup as Q4: △OME ≅ △ONE ⇒ ∠OEM = ∠OEN.
Thus, OE makes equal angles with AB and CD. (Follows from congruence in Q4 proof)
6If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.
Answer & Reference:
Draw OM ⊥ AD. Since OM ⊥ chord AD of outer circle, AM = MD (Theorem 9.3).
Similarly, OM ⊥ chord BC of inner circle ⇒ BM = MC.
Subtract: AM − BM = MD − MC ⇒ AB = CD. (Uses Theorem 9.3, Section 9.2)
Fig. 9.12: Line intersecting concentric circles
7Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5 m. Distance between Reshma–Salma and Salma–Mandip is 6 m each. Find distance between Reshma and Mandip.
Answer & Reference:
Let R, S, M be positions. RS = SM = 6 m, OR = OS = OM = 5 m.
Draw OK ⊥ RM. In △ORS, OR = OS = 5, RS = 6.
Altitude from O to RS: h = √(5² − 3²) = 4 m.
Area of △ORS = ½ × 6 × 4 = 12 m².
Also, area = ½ × RM × OK. But by symmetry, S lies on perpendicular bisector of RM.
Using coordinates: Place S at (0,0), R at (−3, y), M at (3, y), with distance to O(0, k) = 5.
From RS = 6: distance between (−3,y) and (0,0) = 6 ⇒ 9 + y² = 36 ⇒ y = √27 = 3√3.
Distance from O(0,k) to S(0,0): |k| = d ⇒ but OS = 5 ⇒ k = ±5.
Distance from O to R: √(9 + (y − k)²) = 5 ⇒ 9 + (3√3 − k)² = 25 ⇒ (3√3 − k)² = 16 ⇒ 3√3 − k = ±4.
Take k = 3√3 − 4 ≈ 5.196 − 4 = 1.196 ⇒ then RM = distance between (−3,3√3) and (3,3√3) = 6 m? But check circle radius.
Better: Use triangle ORM isosceles with OR=OM=5, and S midpoint of arc? Actually, △RSM is isosceles with RS=SM=6.
Let RM = 2x. Midpoint N of RM ⇒ SN ⊥ RM.
In △RSN: RN = x, RS = 6 ⇒ SN = √(36 − x²)
O lies on SN. Let ON = d ⇒ OS = |SN − d| = 5, OR² = x² + d² = 25
So: (√(36 − x²) − d)² = 25 and x² + d² = 25
Expand: 36 − x² − 2d√(36 − x²) + d² = 25 ⇒ (36 − x² + d²) − 2d√(36 − x²) = 25
But x² + d² = 25 ⇒ d² = 25 − x² ⇒ 36 − x² + 25 − x² = 61 − 2x²
So: 61 − 2x² − 2d√(36 − x²) = 25 ⇒ 36 − 2x² = 2d√(36 − x²)
Divide 2: 18 − x² = d√(36 − x²)
Square both sides: (18 − x²)² = d²(36 − x²) = (25 − x²)(36 − x²)
Expand: 324 − 36x² + x⁴ = 900 − 25x² − 36x² + x⁴ ⇒ 324 − 36x² = 900 − 61x² ⇒ 25x² = 576 ⇒ x² = 23.04 ⇒ x = 4.8
∴ RM = 2x = 9.6 m. (Uses chord properties and Pythagoras; no direct theorem, but based on Section 9.2–9.3)
8A circular park of radius 20 m. Three boys sit at equal distance on boundary. Find length of string of each toy telephone.
Answer & Reference:
They form an equilateral triangle inscribed in circle of radius 20 m.
For equilateral triangle, R = a / √3 ⇒ a = R√3 = 20√3 m.
But correct formula: R = a / (√3) is wrong.
Correct: In equilateral △, centroid = circumcentre. Height h = (√3/2)a.
Circumradius R = (2/3)h = (2/3)(√3/2)a = a/√3 ⇒ a = R√3 = 20√3 ≈ 34.64 m.
However, standard formula: R = a / (√3) ⇒ a = R√3.
∴ Length of each side = 20√3 m. (Based on properties of equilateral triangle and circle)
9In Fig. 9.23, A, B, C on circle, ∠BOC = 30°, ∠AOB = 60°. D on circle not on arc ABC. Find ∠ADC.
Answer & Reference:
∠AOC = ∠AOB + ∠BOC = 60° + 30° = 90°.
Angle subtended by arc ABC at centre = 90°.
∠ADC is angle subtended by same arc at circumference.
∴ ∠ADC = ½ × ∠AOC = ½ × 90° = 45°. (Theorem 9.7: angle at centre is double angle at circumference, Section 9.4)
Fig. 9.23
10A chord equals radius. Find angle subtended at point on minor arc and major arc.
Answer & Reference:
Chord AB = radius ⇒ △OAB equilateral ⇒ ∠AOB = 60°.
Angle on minor arc (reflex side): ½ × reflex ∠AOB = ½ × (360° − 60°) = 150°.
Angle on major arc (minor side): ½ × 60° = 30°.
∴ Angles are 150° (minor arc point) and 30° (major arc point). (Theorem 9.7, Section 9.4)
11In Fig. 9.24, ∠PQR = 100°, P,Q,R on circle, centre O. Find ∠OPR.
Answer & Reference:
∠PQR = 100° is angle in segment ⇒ subtended by arc PR.
∴ ∠POR = 2 × ∠PQR = 200° (reflex angle at centre).
So minor ∠POR = 360° − 200° = 160°.
In △OPR, OP = OR ⇒ ∠OPR = ∠ORP.
∠OPR = (180° − 160°)/2 = 10°. (Theorem 9.7, Section 9.4)
Answer & Reference:
In △ABC: ∠BAC = 180° − 69° − 31° = 80°.
∠BDC and ∠BAC subtend same arc BC ⇒ ∠BDC = ∠BAC = 80°. (Theorem 9.8: angles in same segment are equal, Section 9.4)
13In Fig. 9.26, A,B,C,D on circle. AC, BD intersect at E. ∠BEC = 130°, ∠ECD = 20°. Find ∠BAC.
Answer & Reference:
∠BEC = 130° ⇒ ∠CED = 50° (linear pair).
In △CED: ∠CDE = 180° − 50° − 20° = 110°.
But ∠BAC = ∠BDC (same segment, arc BC) = ∠CDE = 110°? Wait.
Actually, ∠BAC = ∠BDC (angles in same segment).
∠BDC = ∠CDE = ? From △CED: ∠CDE = 110°, but ∠BDC is same as ∠CDE.
However, ∠ECD = 20° = ∠ACD.
Note: ∠BAC = ∠BDC.
In △BEC: ∠EBC = 180° − 130° − ∠ECB.
Better: ∠BAC = ∠BDC.
∠BEC = 130° is external to △CED ⇒ ∠BEC = ∠ECD + ∠CDE ⇒ 130° = 20° + ∠CDE ⇒ ∠CDE = 110°.
∴ ∠BAC = ∠BDC = ∠CDE = 110°. (Uses exterior angle theorem and Theorem 9.8)
14ABCD cyclic quadrilateral, diagonals intersect at E. ∠DBC = 70°, ∠BAC = 30°. Find ∠BCD. If AB = BC, find ∠ECD.
Answer & Reference:
∠BAC = ∠BDC = 30° (same segment).
In △DBC: ∠BCD = 180° − ∠DBC − ∠BDC = 180° − 70° − 30° = 80°.
If AB = BC, then ∠BAC = ∠BCA = 30°.
∠BAD = ∠BAC + ∠CAD. But ∠CAD = ∠DBC = 70° (same segment).
So ∠BAD = 30° + 70° = 100°.
In cyclic quad: ∠BCD = 180° − 100° = 80° (consistent).
Now, ∠ECD = ∠BCA = 30°? No.
Since AB = BC, △ABC is isosceles ⇒ ∠BCA = ∠BAC = 30°.
∠ECD is same as ∠ACD.
∠ACB = 30°, and ∠BCD = 80° ⇒ ∠ACD = ∠BCD − ∠ACB = 80° − 30° = 50°.
∴ ∠ECD = 50°. (Uses Theorem 9.8 and cyclic quadrilateral property, Section 9.5)
15If diagonals of a cyclic quadrilateral are diameters, prove it is a rectangle.
Answer & Reference:
Let ABCD be cyclic, with AC and BD diameters.
Then ∠ABC = 90° (angle in semicircle, since AC diameter).
Similarly, ∠BCD = ∠CDA = ∠DAB = 90°.
∴ All angles 90° ⇒ ABCD is a rectangle. (Uses "angle in semicircle is right angle", Section 9.4)
16If non-parallel sides of a trapezium are equal, prove it is cyclic.
Answer & Reference:
Let ABCD be trapezium with AB || CD and AD = BC.
Draw BE || AD. Then ABED is parallelogram ⇒ AD = BE.
But AD = BC ⇒ BE = BC ⇒ ∠BEC = ∠BCE.
Since AB || CD, ∠BAD + ∠ADC = 180°.
Also, ∠ADC = ∠BEC (corresponding).
So ∠BAD + ∠BCD = 180° ⇒ opposite angles supplementary ⇒ cyclic. (Uses converse of Theorem 9.10, Section 9.5)
17Two circles intersect at B and C. Through B, lines ABD and PBQ intersect circles at A,D and P,Q. Prove ∠ACP = ∠QCD.
Answer & Reference:
Join AC, PC, QC, DC.
∠ACP = ∠ABP (angles in same segment of first circle).
∠QCD = ∠QBD (angles in same segment of second circle).
But ∠ABP = ∠QBD (vertically opposite angles).
∴ ∠ACP = ∠QCD. (Uses Theorem 9.8, Section 9.4)
18Circles drawn on two sides of triangle as diameters. Prove intersection point lies on third side.
Answer & Reference:
Let △ABC. Draw circles with AB and AC as diameters.
Let them intersect at D (≠ A).
Then ∠ADB = 90° (angle in semicircle, AB diameter).
∠ADC = 90° (AC diameter).
∴ ∠BDC = ∠ADB + ∠ADC = 180° ⇒ B-D-C collinear.
So D lies on BC. (Example 4 logic, Section 9.5)
19ABC and ADC are right triangles with common hypotenuse AC. Prove ∠CAD = ∠CBD.
Answer & Reference:
Since ∠ABC = ∠ADC = 90°, points A,B,C,D lie on circle with AC as diameter.
∴ ABCD is cyclic.
Then ∠CAD = ∠CBD (angles in same segment CD). (Uses "angle in semicircle" and Theorem 9.8)
20Prove a cyclic parallelogram is a rectangle.
Answer & Reference:
Let ABCD be cyclic parallelogram.
In parallelogram: ∠A = ∠C.
In cyclic quad: ∠A + ∠C = 180°.
⇒ 2∠A = 180° ⇒ ∠A = 90°.
∴ All angles 90° ⇒ rectangle. (Uses Theorem 9.10, Section 9.5)
T1Observation table for cyclic quadrilateral opposite angles.
Answer & Reference:
As per activity in Section 9.5, the table shows: