Circles
Practice Questions

Equal chords subtend equal angles at the centre.

Perpendicular from centre to a chord bisects the chord.

Theorem: Angle at centre is double the angle at circumference.

Angles in the same segment of a circle are equal.

Both A and R are true and R explains A.

Angle at circumference = 1/2 (180°) = 90°. Both A and R are true.

Perpendicular bisects chord: half chord = 8 cm. Distance d = √(10² - 8²) = √36 = 6 cm.

Half chord = √(5² - 3²) = 4 cm. Full chord = 8 cm.

Reflex ∠AOC = 360° - (90+120) = 150°. Angle at B = 1/2 (150°) = 75°.

Opposite angles sum to 180°: ∠C = 180° - 70° = 110°.

Angles in semicircle = 90° for all 4 corners ⇒ Rectangle.

Join PO and extend to Q. Exterior angle theorem in △AOP and △BOP gives ∠AOB = 2 ∠APB.

Draw central angles. Sum of central angles = 360° ⇒ Sum of opposite angles = 1/2 (360°) = 180°.

Draw perpendiculars from top vertices to base. Right triangles congruent ⇒ Base angles equal ⇒ Opposite angles supplementary ⇒ Cyclic.

Angles in same segment: ∠ACP = ∠ABP and ∠QCD = ∠QBD. Vertically opposite angles ∠ABP = ∠QBD ⇒ ∠ACP = ∠QCD.

Isosceles triangle with sides 6, 6 and base chord. Perpendicular from center = √(5² - 3²) = 4m. Half chord = (6 × 4) / 5 = 4.8m. Full distance = 9.6 m.

Radius r = 2/3 h = 20 ⇒ h = 30m. Side a = 2h / √3 = 60 / √3 = 20√3 m.

In △ABC: ∠BAC = 180° - (69+31) = 80°. Angles in same segment: ∠BDC = ∠BAC = 80°.

∠DEC = 180° - 130° = 50°. In △CDE: ∠EDC = 180° - (50+20) = 110°. ∠BAC = ∠BDC = 110°.

Let circles intersect at D. ∠ADB = 90° (semicircle) and ∠ADC = 90° (semicircle). ∠ADB + ∠ADC = 180° ⇒ BDC is straight line.