Chapter 5: Exercise Solutions

Exercise Set 5.1

1. Draw ΔABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:
Calculate the third angle: ∠C = 180° - (70° + 60°) = 50°.
Since all three angles are acute (< 90°), ΔABC is an acute-angled triangle.
Conclusion: The circumcentre lies inside the triangle.

2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?

Solution:
Since ∠A = 100°, ΔABC is an obtuse-angled triangle.
Conclusion: The circumcentre lies outside the triangle.

3. Draw ΔABC, with AB = 6 cm, BC = 7 cm, and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.

Solution:
Since O is the circumcentre of ΔABC, points A, B, and C lie on the circumcircle.
Therefore, OA, OB, and OC are all radii of the same circle.
Conclusion: OA = OB = OC = 3.53 cm (approx).

4. What is the least possible radius of a circle through two points A and B?

Solution:
The smallest circle passing through points A and B has segment AB as its diameter.
Conclusion: The least possible radius is AB / 2 (half the distance between A and B).

Exercise Set 5.2

1. Show that the triangle formed by a chord and the centre of the circle is isosceles.

Solution:
Let C be the centre of the circle and AB be a chord. Join CA and CB.
Since A and B lie on the circle, CA = CB = r (radius of the circle).
In ΔCAB, two sides are equal (CA = CB). Therefore, ΔCAB is an isosceles triangle.

2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

Solution:
Let the two triangles be ΔCAB and ΔC'DE, where C and C' are the centres of circles of equal radii.
- CA = CB = C'D = C'E = r
- Base AB = Base DE (given base lengths are equal)
By SSS Congruence Criterion, ΔCAB ≅ ΔC'DE.

Exercise Set 5.3

1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?

Solution:
In ΔCMA and ΔCMB:
- ∠CMA = ∠CMB = 90° (given)
- CA = CB (radii of the circle)
- CM = CM (common side)
By RHS Congruence Criterion, ΔCMA ≅ ΔCMB.
Therefore, AM = BM (by CPCTC), which means CM bisects AB.

2. An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.

Solution:
In an isosceles triangle with AB = AC, the altitude from A to BC bisects the base BC.
By Theorem 4, the perpendicular bisector of a chord (BC) passes through the centre of the circle.
Since the altitude from A is also the perpendicular bisector of BC, it must pass through the centre of the circle.

3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.

Solution:
Let distance from centre O to chord of length 8 cm be d₁:
d₁ = √(5² - (8/2)²) = √(25 - 16) = √9 = 3 cm.
Let distance from centre O to chord of length 6 cm be d₂:
d₂ = √(5² - (6/2)²) = √(25 - 9) = √16 = 4 cm.
Since the chords are on opposite sides of the centre:
Total Distance = d₁ + d₂ = 3 + 4 = 7 cm.

Exercise Set 5.4

1. Use the Baudhāyana-Pythagoras theorem to show why Theorem 6 must be true.

Solution:
Let two chords be AB and FG of equal length (AB = FG) with midpoints E and H.
AE = AB / 2 and FH = FG / 2. Since AB = FG, AE = FH.
In right-angled triangles ΔCEA and ΔCHF:
CA² = CE² + AE² ⇒ CE² = r² - AE²
CF² = CH² + FH² ⇒ CH² = r² - FH²
Since CA = CF = r and AE = FH, we get CE² = CH² ⇒ CE = CH.

2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH show that AB = GF.

Solution:
In right triangles ΔCEA and ΔCHF:
- CA = CF = r
- CE = CH (given)
- ∠CEA = ∠CHF = 90°
By RHS Congruence Criterion, ΔCEA ≅ ΔCHF.
So, AE = FH. Since perpendiculars from the centre bisect the chords:
AB = 2 × AE and GF = 2 × FH.
Therefore, AB = GF.

3. Solve the previous question using the Baudhāyana-Pythagoras theorem.

Solution:
AE = √(CA² - CE²) = √(r² - CE²)
FH = √(CF² - CH²) = √(r² - CH²)
Since CE = CH, AE = FH.
AB = 2 × AE = 2 × FH = GF.

Exercise Set 5.5

1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

Solution:
Half-chord length = √(r² - d²) = √(7² - 6²) = √(49 - 36) = √13 cm.
Full Chord Length = 2 × √13 = 2√13 cm (≈ 7.21 cm).

2. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r² - d²).

Solution:
By the Baudhāyana-Pythagoras theorem in the right triangle formed by radius r, distance d, and half-chord x:
r² = d² + x² ⇒ x = √(r² - d²).
Since the perpendicular from the centre bisects the chord, total chord length = 2x = 2√(r² - d²).

3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2 AB? Give reasons for your answer.

Solution:
No, the relationship is non-linear.
Let distance of CD be d. Then distance of AB is 2d.
AB = 2√(r² - 4d²) and CD = 2√(r² - d²).
CD / AB = √(r² - d²) / √(r² - 4d²) ≠ 2. Thus, CD is longer, but not exactly twice as long.

Exercise Set 5.6

1. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?

Solution:
In ΔOAB, OA = OB = 12 cm (radii). So, ∠OAB = ∠OBA.
Sum of angles = 180° ⇒ ∠OAB + ∠OBA = 180° - 60° = 120° ⇒ ∠OAB = ∠OBA = 60°.
ΔOAB is an equilateral triangle. Therefore, AB = 12 cm.

2. (i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?

Solution:
(i) No, angles subtended by an arc at any point on the same segment of the circle are equal.
(ii) No, if they lie on opposite sides, the angles are supplementary (add up to 180°).
(iii) Yes, by Theorem 10 (Concyclicity).

3. Find x in Fig. 5.26.

Solution:
Central angle = 100°. The angle subtended at any point on the circle is half the angle subtended at the centre.
x = 100° / 2 = 50°.

End-of-Chapter Exercises

1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?

Solution:
Half-chord = √(13² - 5²) = √(169 - 25) = √144 = 12 cm.
Chord length = 2 × 12 = 24 cm.

2. An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

Solution:
Angle at circumference = 70° / 2 = 35°.

3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.

Solution:
Radius r = 26 / 2 = 13 cm. Half-chord = 24 / 2 = 12 cm.
Distance d = √(13² - 12²) = √(169 - 144) = √25 = 5 cm.

4. A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?

Solution:
Half-chord = √(15² - 9²) = √(225 - 81) = √144 = 12 cm.
Chord length = 2 × 12 = 24 cm.

5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Solution:
The locus of points equidistant from the endpoints A and B of a chord is its perpendicular bisector. Since the centre O is equidistant from A and B (OA = OB = r), O must lie on the perpendicular bisector.

6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.

Solution:
∠ACB = 90°.
Reason: By Corollary to Theorem 9, the angle subtended by a diameter at any point on the circle is 90°.

7. ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?

Solution:
Opposite angles of a cyclic quadrilateral sum to 180°.
∠C = 180° - 75° = 105°
∠D = 180° - 110° = 70°

8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x - 20)°, find the value of x and the measures of ∠P and ∠R.

Solution:
∠P + ∠R = 180° ⇒ (2x + 10) + (3x - 20) = 180
5x - 10 = 180 ⇒ 5x = 190 ⇒ x = 38
∠P = 2(38) + 10 = 86°
∠R = 3(38) - 20 = 94°

9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.

Solution:
Half-chord = 16 / 2 = 8 cm.
Radius r = √(8² + 6²) = √(64 + 36) = √100 = 10 cm.

10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Solution:
Using Brahmagupta's formula for the area of a cyclic quadrilateral:
s = (5 + 5 + 12 + 12) / 2 = 17
Area = √[(17-5)(17-5)(17-12)(17-12)] = √[12 × 12 × 5 × 5] = 12 × 5 = 60 sq units.

14. Show that rectangle is the only parallelogram that can be inscribed in a circle.

Solution:
In a parallelogram, opposite angles are equal (∠A = ∠C).
Since it is cyclic, ∠A + ∠C = 180° ⇒ 2∠A = 180° ⇒ ∠A = 90°.
A parallelogram with a right angle is a rectangle.

18. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.

Solution:
Half-chords are 5 cm and 12 cm.
Let distance to 24 cm chord be d. Then distance to 10 cm chord is d + 7.
r² = d² + 12² = (d + 7)² + 5²
d² + 144 = d² + 14d + 49 + 25 ⇒ 14d = 70 ⇒ d = 5 cm.
r = √(5² + 12²) = 13 cm.

19. A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.

Solution:
- Side length: Equal to radius = r.
- Distance from centre: Altitude of an equilateral triangle of side r = (√3 / 2) r.

21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle.

Solution:
Let the side CD be extended to E.
∠ADC + ∠CDE = 180° (linear pair).
∠ADC + ∠ABC = 180° (opposite angles of cyclic quad).
Equating both gives: ∠CDE = ∠ABC.