Chapter-5 I’m Up and Down, and Round and Round

Exercise Set 5.1

Question 1

Triangle Construction: Draw AB = 5 cm. At A, draw an angle of 70°, and at B, draw an angle of 60°. Let the rays intersect at C.
Third Angle Calculation: Angle C = 180° – (70° + 60°) = 50°.
Circumcentre Location: Since all three interior angles (70°, 60°, 50°) are acute (less than 90°), ΔABC is an acute-angled triangle. Therefore, the circumcentre lies inside the triangle.

Question 2

Triangle Construction: Draw AB = 5 cm. At A, draw an angle of 100°. Mark point C on this ray such that AC = 4 cm. Join BC.
Circumcentre Location: Since Angle A = 100° (greater than 90°), ΔABC is an obtuse-angled triangle. Therefore, the circumcentre lies outside the triangle.

Question 3

Measurement: The distance from the circumcentre O to any vertex of the triangle represents the circumradius R.
Conclusion: OA = OB = OC.

Question 4

Least Possible Radius: The smallest circle passing through two points A and B is the circle where segment AB serves as the diameter.
Value: The least possible radius is half of AB (1/2 AB).

Think, Draw and Infer

Question 1

Existence of Point P: No, such a point P cannot exist for three distinct collinear points.
Perpendicular Bisectors: The perpendicular bisectors of AB and BC are parallel to each other because both are perpendicular to the same line segment AC.
Circle Through Collinear Points: No, a circle cannot pass through three collinear points.
Line Cutting a Circle: No, a straight line can intersect a circle in at most two distinct points.

Question 2

Congruent Triangles Sharing a Circumcircle: Yes, infinitely many congruent triangles can share the same circumcircle (e.g., by rotating or reflecting ΔABC within the circle).

Exercise Set 5.2

Question 1

Let O be the centre of the circle and AB be a chord.
Consider ΔOAB. The sides OA and OB are radii of the circle, so OA = OB = r.
Since two sides of ΔOAB are equal in length, ΔOAB is an isosceles triangle.

Question 2

Let two chords be AB and CD in circles with centre O (or radius r), such that AB = CD.
In ΔOAB and ΔOCD:

OA = OC = r (radii)
OB = OD = r (radii)
AB = CD (given equal base length)
By SSS Congruence Criterion, ΔOAB is congruent to ΔOCD.

Exercise Set 5.3

Question 1

Given: In ΔOAM and ΔOBM, OM is perpendicular to AB, so Angle OMA = Angle OMB = 90°.
Proof:

Angle OMA = Angle OMB = 90° (Given)
OA = OB = r (Hypotenuse, radii of the circle)
OM = OM (Common side)

By RHS Congruence Criterion, ΔOAM is congruent to ΔOBM.
Therefore, AM = BM (by CPCTC), which proves the perpendicular from the centre bisects the chord.

Question 2

In isosceles ΔABC with AB = AC, the altitude from A to BC is also the perpendicular bisector of BC.
The centre of the circumcircle of ΔABC must lie on the perpendicular bisector of every chord/side of the triangle.
Since the altitude from A to BC is the perpendicular bisector of BC, it must pass through the centre of the circle.

Question 3

Let O be the centre, AB = 6 cm, and CD = 8 cm.
Perpendicular distance from O to AB: d1 = √(5² – 3²) = √(25 – 9) = 4 cm.
Perpendicular distance from O to CD: d2 = √(5² – 4²) = √(25 – 16) = 3 cm.
Since the chords lie on opposite sides of the centre, the distance between their midpoints is d1 + d2 = 4 + 3 = 7 cm.

Exercise Set 5.4

Question 1

In right-angled triangle OMA (where M is the midpoint of chord AB, OM = d, AM = L/2, OA = r): OM² + AM² = OA² ⇒ d² + (L/2)² = r².
Rearranging gives (L/2)² = r² – d² ⇒ L = 2√(r² – d²).
If two chords have equal lengths (L1 = L2), then r² – d1² = r² – d2² ⇒ d1 = d2. Hence, equal chords are equidistant from the centre.

Question 2

In right triangles CEB and CHG:

Angle CEB = Angle CHG = 90° (Given)
CB = CG = r (Hypotenuse, radii of the circle)
CE = CH (Given)

By RHS Congruence Criterion, ΔCEB is congruent to ΔCHG ⇒ EB = HG.
Similarly, ΔCEA is congruent to ΔCHF ⇒ EA = HF.
Adding both: AB = EB + EA = HG + HF = GF. Thus, AB = GF.

Question 3

Using Baudh&amacron;yana–Pythagoras theorem in ΔCEA: EA² = CA² – CE² = r² – CE².
In ΔCHF: HF² = CF² – CH² = r² – CH².
Since CE = CH, EA² = HF² ⇒ EA = HF.
Since perpendiculars bisect the chords, AB = 2 × EA and GF = 2 × HF ⇒ AB = GF.

Exercise Set 5.5

Question 1

Given r = 7 cm and d = 6 cm.
Chord length = 2√(r² – d²) = 2√(7² – 6²) = 2√(49 – 36) = 2√13 cm (approx. 7.21 cm).

Question 2

Let M be the midpoint of chord AB, O be the centre, OA = r, OM = d.
In right-angled triangle OMA: AM = √(OA² – OM²) = √(r² – d²).
Since OM perpendicular to AB bisects the chord, AB = 2 × AM = 2√(r² – d²).

Question 3

No, we cannot conclude that CD = 2 × AB.
Reason: Let d1 and d2 be the distances of chords AB and CD from the centre, with d1 = 2 × d2.

AB = 2√(r² – 4d2²)
CD = 2√(r² – d2²)
The relationship between the lengths involves non-linear square root operations, so doubling the distance does not halve the chord length or make CD = 2 × AB.

Exercise Set 5.6

Question 1

In ΔAOB, OA = OB = 12 cm (radii) and Angle AOB = 60°.
Since OA = OB, Angle OAB = Angle OBA = (180° – 60°) / 2 = 60°.
Thus, ΔAOB is an equilateral triangle, so AB = OA = OB = 12 cm.

Question 2

(i) No. All angles subtended by a chord on the same major (or minor) arc segment are equal (Angle AXB = Angle AYB).
(ii) Yes, provided X and Y lie on the major arc (or both on the minor arc) on the same side of chord AB.
(iii) No. If X and Y do not lie on the original circle, Y will generally not lie on the circle through A, B, and X.

Question 3

In Fig. 5.26, quadrilateral ADBC is a cyclic quadrilateral inscribed in the circle.
Opposite angles of a cyclic quadrilateral sum to 180°:
Angle D + Angle X = 180° ⇒ 100° + x = 180° ⇒ x = 80°

End-of-Chapter Exercises

#	Problem Summary	Solution / Result
1	Chord length with r = 13 cm, distance d = 5 cm	Half-chord = √(13² – 5²) = 12 cm ⇒ Chord = 2 × 12 = 24 cm
2	Angle on circle subtended by central angle of 70°	Angle on circle = 1/2 × 70° = 35°
3	Distance from centre for chord length 24 cm, diameter 26 cm	r = 13 cm, half-chord = 12 cm ⇒ d = √(13² – 12²) = 5 cm
4	Chord length with r = 15 cm, distance d = 9 cm	Half-chord = √(15² – 9²) = 12 cm ⇒ Chord = 2 × 12 = 24 cm
5	Prove perpendicular bisector of chord passes through centre	Centre O is equidistant from ends A and B (OA = OB = r). Points equidistant from endpoints lie on the perpendicular bisector.
6	Measure of ∠ACB where AB is diameter	Inscribed angle in a semicircle is always 90°.
7	Cyclic quadrilateral ABCD, ∠A = 75°, ∠B = 110°	∠C = 180° – 75° = 105°∠D = 180° – 110° = 70°
8	Cyclic quadrilateral PQRS, ∠P = (2x + 10)°, ∠R = (3x – 20)°	2x + 10 + 3x – 20 = 180 ⇒ 5x = 190 ⇒ x = 38∠P = 86°, ∠R = 94°
9	Radius of circle with chord 16 cm at distance 6 cm	Half-chord = 8 cm ⇒ r = √(8² + 6²) = 10 cm
10	Area of cyclic quadrilateral with sides 5, 5, 12, 12	Split into two right triangles (5, 12, 13) ⇒ Area = 2 × (1/2 × 5 × 12) = 60 sq units
11	Locating circumcentre of cyclic quadrilateral without circle	Construct perpendicular bisectors of non-parallel sides; their intersection is the circumcentre. Check if opposite angles sum to 180°.
12	Intersecting chords of equal length have equal segments	Equal chords are equidistant from centre (OM = ON). RHS congruence gives MP = NP. Since AM = CN, AP = CP and PB = PD.
13	Radius of circle with 6 cm chord at 3 cm distance	Half-chord = 3 cm, distance = 3 cm ⇒ r = √(3² + 3²) = 3√2 cm
14	Show rectangle is the only cyclic parallelogram	Opposite angles in parallelogram are equal (∠A = ∠C), and cyclic gives ∠A + ∠C = 180° ⇒ ∠A = 90° (Rectangle).
15	Diagonals of inscribed rectangle intersect at centre	Diagonals of a rectangle are equal and bisect each other, making the intersection point equidistant from all 4 vertices (circumcentre).
16	Shape formed by midpoints of fixed-length chords	Distance d from centre is constant ⇒ Midpoints form a concentric circle.
17	Center lies on angle bisector of ∠BAC if AB = AC	By SSS, ΔOAB ≅ ΔOAC ⇒ ∠OAB = ∠OAC, so OA bisects ∠BAC.
18	Parallel chords 10 cm and 24 cm on same side, distance 7 cm	√(r² – 25) – √(r² – 144) = 7 ⇒ r = 13 cm
19	Inscribed regular hexagon in circle of radius r	Side = r; Distance from centre = (√3 / 2)r
20	Cyclic quadrilateral MNOP with diameter MN	∠MOP = 180° (central angle) and any inscribed angle subtended by diameter MN is 90°.
21	Exterior angle equal to interior opposite angle	∠ABC + ∠ADC = 180° and ∠CDE + ∠ADC = 180° (linear pair) ⇒ ∠CDE = ∠ABC.
22	No chord is longer than diameter	Chord length = 2√(r² – d²). Maximum occurs when d = 0, giving length = 2r (Diameter).
23	Shortest chord through point A inside circle	Chord length is minimized when distance d from centre is maximized, which occurs when OA ⊥ chord (d = OA).
24	Justify angle in a semicircle is 90° (Fig. 5.30)	OA = OB = OC = r ⇒ Base angles a and b form ΔABC. Angle sum: 2a + 2b = 180° ⇒ a + b = 90°.
25	CC’ and DD’ perpendicular to AB ⇒ MM’ ⊥ AB	Symmetry along diameter AB bisects chords perpendicularly, aligning midpoint segment MM’ along the axis of symmetry perpendicular to AB.
26	Justify sum of opposite angles of cyclic quad is 180° (Fig. 5.31)	Central angle around O sums to 360°. Inscribed angles are half of central angles, so p + q + u + v = 360° / 2 = 180°.