End-of-Chapter Exercises

Let OM ⟂ AB. Then M bisects AB. In right ΔOMA, AM = √(13²−5²)=12 cm. Hence AB=24 cm.

Angle at the centre = 2 × angle at the circle. Therefore required angle = 70°/2 = 35°.

Radius = 13 cm; half chord = 12 cm. By Pythagoras, d²=13²−12²=25, so d=5 cm.

Half chord = √(15²−9²)=12 cm. Therefore chord = 24 cm.

Let M be midpoint of chord AB. OA=OB, AM=MB and OM is common. Thus ΔOMA≅ΔOMB by SSS. Equal adjacent angles at M form a linear pair, so each is 90°. Hence OM⊥AB; therefore the perpendicular bisector passes through O.

A diameter subtends a right angle at any point on the circle. Hence ∠ACB=90°.

Opposite angles of a cyclic quadrilateral are supplementary. ∠C=180°−75°=105°. ∠D=180°−110°=70°.

(2x+10)+(3x−20)=180 ⇒ 5x−10=180 ⇒ x=38. Then ∠P=86° and ∠R=94°.

Half chord = 8 cm. r²=8²+6²=100, so r=10 cm.

Split it along a suitable diagonal into two 5-12-13 right triangles. Each has area 30 square units, so total area=60 square units.

Construct the perpendicular bisectors of two sides. Their intersection is the circumcentre. Then check whether this point lies inside or outside the quadrilateral.

Equal chords are equidistant from the centre. Join O to their intersection and to the chord midpoints; the resulting right triangles are congruent by RHS. This gives equal corresponding half-segments, hence AP=CP and PB=PD.

Draw AB=6 cm and midpoint M. At M draw a perpendicular and mark O so OM=3 cm. With centre O and radius OA, draw the circle. Since OA²=3²+3², the radius is 3√2 cm.

For a cyclic quadrilateral, opposite angles sum to 180°. In a parallelogram opposite angles are equal. Therefore 2∠A=180°, so ∠A=90°. Similarly all angles are 90°, hence the parallelogram is a rectangle.

Rectangle diagonals are equal and bisect each other. If O is their intersection, OA=OC and OB=OD, and equality of diagonals gives OA=OB. Thus O is equidistant from all vertices and is the centre.

Equal chords are equidistant from the centre. Therefore all midpoints are at the same fixed distance from O, so they form a circle with the same centre.

Equal chords subtend equal central angles, so ∠AOB=∠COA. With OA common and OB=OC, ΔAOB≅ΔAOC by SAS. Hence ∠BAO=∠OAC, so AO bisects ∠BAC.

Half chords are 5 and 12. Let distances from O be d1 (24-cm chord) and d2 (10-cm chord). Then d2−d1=7 and d2²−d1²=12²−5²=119. Hence d2+d1=17, so d2=12,d1=5. Thus r²=12²+5²=169 and r=13 cm.

Each central angle is 60°, so the triangle formed by two radii and one side is equilateral. Thus side=r. If OM is perpendicular to a side, then AM=r/2. Hence OM=√(r²−(r/2)²)=(√3/2)r.

Both angles stand on the same arc MP, with ∠MOP at the centre and ∠MNP at the circle. Therefore the central angle is twice the inscribed angle: ∠MOP=2∠MNP.

∠ABC+∠ADC=180°. Also ∠ADC+∠CDE=180° because they form a linear pair. Comparing the equations gives ∠CDE=∠ABC.

For chord AB with midpoint M, OM⊥AB. In right triangle OMA, AM≤OA. Thus AB=2AM≤2OA, which is the diameter. Equality occurs only when AB passes through O.

For a chord through A, its length decreases as the chord moves farther from O. The maximum distance from O to a line through A occurs when the line is perpendicular to OA. Hence that chord is shortest.

Let AB be the diameter and C a point on the circle. Join OC. Since OA=OB=OC, the two triangles about OC are isosceles. Using the angle-sum property gives ∠ACB=90°; this is the angle-in-a-semicircle theorem.

AB is the axis of reflection. C↔C′ and D↔D′ are symmetric about AB, so the midpoint M of CD maps to the midpoint M′ of C′D′. Therefore MM′ is perpendicular to the symmetry axis AB.

In cyclic ABCD, each inscribed angle is half the central angle of its intercepted arc. The arcs corresponding to opposite angles together make a full 360° at O. Hence ∠A+∠C=½×360°=180°. Similarly ∠B+∠D=180°.