Step 1: Understanding Construction & Theorem 9.1:

According to Theorem 9.1, one and only one circle can pass through three non-collinear points. The three vertices of $\triangle ABC$ are non-collinear.

• Construct equilateral $\triangle ABC$ with $AB = BC = CA = 5\text{ cm}$.
• Draw the perpendicular (right) bisectors of sides $AB$ and $BC$.
• Let these bisectors intersect at a unique point $O$ (the circumcenter).
• Since $O$ lies on the right bisector of $AB$, $OA = OB$. Since $O$ lies on the right bisector of $BC$, $OB = OC$. Hence, $OA = OB = OC = R$.
• With center $O$ and radius $R = OA$, draw a circle. It passes through all three vertices $A, B, C$.
• Since two distinct straight lines can intersect at only one point, $O$ is unique, proving that one and only one circle can be drawn.

Step 2: Calculating the Circumradius $R$:

For an equilateral triangle with side length $a = 5\text{ cm}$:

$$\text{Altitude } h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 5 = \frac{5\sqrt{3}}{2}\text{ cm}$$

In an equilateral triangle, the circumcenter divides each median/altitude in the ratio $2:1$:

$$R = \frac{2}{3} h = \frac{2}{3} \left(\frac{5\sqrt{3}}{2}\right) = \frac{5\sqrt{3}}{3} = \frac{5}{\sqrt{3}} \approx 2.89\text{ cm}$$

Final Answer:

One and only one circle passes through the vertices, with radius $R = \frac{5\sqrt{3}}{3} \approx 2.89\text{ cm}$.

Part (i): Finding Radius $r$:

• Given: Chord length $AB = 10\text{ cm}$, perpendicular distance $PE = 12\text{ cm}$.
• By Theorem 9.3, the perpendicular from the center to a chord bisects the chord: $$AE = EB = \frac{AB}{2} = \frac{10}{2} = 5\text{ cm}$$
• In right-angled triangle $\triangle PEA$: $$PA^2 = PE^2 + AE^2$$ $$r^2 = 12^2 + 5^2 = 144 + 25 = 169$$ $$r = \sqrt{169} = 13\text{ cm}$$

Part (ii): Finding Chord Length:

• Given: Radius $r = 5\text{ cm}$, perpendicular distance $d = 3\text{ cm}$.
• In right triangle formed by radius, distance, and half-chord: $$\left(\frac{\text{Chord}}{2}\right)^2 = r^2 - d^2 = 5^2 - 3^2 = 25 - 9 = 16$$ $$\frac{\text{Chord}}{2} = \sqrt{16} = 4\text{ cm}$$
• Total Chord Length $= 2 \times 4 = 8\text{ cm}$.

Final Answer:

(i) Radius $r = 13\text{ cm}$, (ii) Chord length $= 8\text{ cm}$.

Step 1: Finding Radius $r = PC$:

• Chord $AB = 10\text{ cm}$. Since diameter $CD \perp AB$, $E$ bisects $AB$: $$AE = \frac{AB}{2} = \frac{10}{2} = 5\text{ cm}$$
• Perpendicular distance $PE = 12\text{ cm}$.
• In right $\triangle PEA$: $$r^2 = PA^2 = PE^2 + AE^2 = 12^2 + 5^2 = 144 + 25 = 169 \implies r = 13\text{ cm}$$

Step 2: Calculating Diameter $CD$ and Sagitta $CE$:

• Diameter $CD = 2r = 2 \times 13 = 26\text{ cm}$.
• Since $P$ is center and $C$ is on circumference along diameter $P-E-C$: $$CE = PC - PE = r - PE = 13 - 12 = 1\text{ cm}$$

Final Answer:

Diameter $CD = 26\text{ cm}$ and $CE = 1\text{ cm}$.

Step 1: Identify Given Geometric Parameters:

• Radius of circle $r = OB = OC = OD = 15\text{ cm}$.
• Perpendicular distance from centre $O$ to chord $CD$ is $OF = 9\text{ cm}$.

Step 2: Apply Pythagorean Theorem in $\triangle OFC$:

$$\triangle OFC \text{ is a right-angled triangle at } F:$$

$$CF^2 = OC^2 - OF^2$$

$$CF^2 = 15^2 - 9^2 = 225 - 81 = 144$$

$$CF = \sqrt{144} = 12\text{ cm}$$

Step 3: Chord Bisector Property (Theorem 9.3):

Since $OF \perp CD$, $F$ is the midpoint of $CD$:

$$CD = 2 \times CF = 2 \times 12 = 24\text{ cm}$$

Final Answer:

The length of chord $CD = 24\text{ cm}$.

Step 1: Set up Right Triangle $\triangle OCA$:

• Radius $OA = 13\text{ cm}$.
• Perpendicular distance $OC = 5\text{ cm}$.
• By Pythagorean theorem in right-angled $\triangle OCA$: $$AC^2 = OA^2 - OC^2 = 13^2 - 5^2 = 169 - 25 = 144$$ $$AC = \sqrt{144} = 12\text{ cm}$$

Step 2: Calculate Total Chord Length $AB$:

By Theorem 9.3, $OC \perp AB \implies AC = CB$:

$$AB = 2 \times AC = 2 \times 12 = 24\text{ cm}$$

Final Answer:

Length of chord $AB = 24\text{ cm}$.

Step 1: Geometric Analysis of Rectangle Diagonals:

• In rectangle $EFGH$, all four interior angles are $90^\circ$.
• The diagonals $EG$ and $FH$ are equal in length and bisect each other at a common point $O$: $$EG = \sqrt{EF^2 + FG^2} = \sqrt{6^2 + 4^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}\text{ cm}$$
• Since diagonals bisect each other: $$OE = OF = OG = OH = \frac{EG}{2} = \sqrt{13} \approx 3.61\text{ cm}$$

Step 2: Proof of Uniqueness (Theorem 9.1):

• Any circle passing through vertices $E, F, G$ must have its center at the intersection of the right bisectors of $EF$ and $FG$.
• These perpendicular bisectors intersect at the unique center $O$ of the rectangle.
• Since two lines intersect at exactly one point, the center $O$ is unique and radius $r = \sqrt{13}\text{ cm}$ is fixed.
• Thus, one and only one circle can pass through all four vertices $E, F, G, H$.

Final Answer:

A unique circle passes through all vertices of rectangle $EFGH$ with center at diagonal intersection and radius $r = \sqrt{13} \approx 3.61\text{ cm}$.

Step 1: Conceptual Identification:

• Given: Point $B$ is at a distance of $4\text{ cm}$ from $A$, $C$, and $D$: $$BA = BC = BD = 4\text{ cm}$$
• By definition of a circle, the set of all points equidistant from a fixed center forms a circle.
• Here, $B$ is equidistant from the three non-collinear points $A, C, D$.

Step 2: Conclusion via Theorem 9.1:

Through three non-collinear points $A, C, D$, there exists one and only one circle. Its center is uniquely $B$ and its radius is $r = 4\text{ cm}$.

Final Answer:

Only 1 circle can be drawn through points $A, C$, and $D$. Its center is $B$ and radius is $4\text{ cm}$.

Step 1: Express Distance in terms of Radius $r$:

• Chord $AB = 16\text{ cm} \implies$ half-chord $AU = 8\text{ cm}$.
• Given geometric relation from diagram: Perpendicular distance from center to chord $OU = 16 - r$.

Step 2: Set up and Solve Pythagorean Equation:

In right triangle $\triangle OUA$:

$$OA^2 = OU^2 + AU^2$$

$$r^2 = (16 - r)^2 + 8^2$$

$$r^2 = 256 - 32r + r^2 + 64$$

$$r^2 - r^2 + 32r = 320$$

$$32r = 320 \implies r = \frac{320}{32} = 10\text{ cm}$$

Final Answer:

(i) $OU = 16 - r$, (ii) Equation: $r^2 = (16 - r)^2 + 64 \implies r = 10\text{ cm}$.

Step 1: Set up Geometric Relations:

• Let $O$ be the centre and $r$ be the radius of the circle.
• Chord $AB = 16\text{ cm} \implies AE = \frac{16}{2} = 8\text{ cm}$.
• Point $D$ lies on the circle, so $OD = r$.
• Since $E$ is on radius $OD$ and $DE = 4\text{ cm}$, the distance from centre to chord is: $$OE = OD - DE = r - 4$$

Step 2: Apply Pythagorean Theorem in $\triangle OEA$:

$$OA^2 = OE^2 + AE^2$$

$$r^2 = (r - 4)^2 + 8^2$$

$$r^2 = r^2 - 8r + 16 + 64$$

$$8r = 80 \implies r = 10\text{ cm}$$

Final Answer:

The radius of the circle is $r = 10\text{ cm}$.

Step 1: Verify Side Lengths in Triangle Formed by Center, Midpoint, and Endpoint:

• Diameter $= 10\text{ cm} \implies$ Radius $r = OA = 5\text{ cm}$.
• Chord $AB = 8\text{ cm}$. Since diameter bisects the chord at $M$, $AM = 4\text{ cm}$.
• Distance from centre $O$ to chord midpoint $M$ is $OM = 3\text{ cm}$.

Step 2: Check Pythagorean Identity in $\triangle OMA$:

$$OM^2 + AM^2 = 3^2 + 4^2 = 9 + 16 = 25$$

$$OA^2 = 5^2 = 25$$

Since $OM^2 + AM^2 = OA^2$, by the Converse of Pythagoras' Theorem, $\triangle OMA$ is a right-angled triangle with $\angle OMA = 90^\circ$.

Therefore, the diameter is perpendicular to the chord at its midpoint.

Final Answer:

Since $3^2 + 4^2 = 5^2$, $\angle OMA = 90^\circ$, confirming the diameter bisects the chord perpendicularly (Theorem 9.2).

Step 1: Given and To Prove:

• Given: A circle with center $O$, chord $AB$, and diameter $CD \perp AB$ intersecting $AB$ at $M$.
• To Prove: $AC = BC$.

Step 2: Proof by Triangle Congruence:

• By Theorem 9.3, since $CD \perp AB$ and passes through center $O$, $M$ is the midpoint of $AB$: $$AM = BM$$
• In right triangles $\triangle AMC$ and $\triangle BMC$:
  1. $AM = BM$ (Proven above)
  2. $\angle AMC = \angle BMC = 90^\circ$ (Given $CD \perp AB$)
  3. $MC = MC$ (Common side)
• By SAS Congruence Postulate: $$\triangle AMC \cong \triangle BMC$$
• Corresponding sides of congruent triangles are equal: $$AC = BC$$

Final Answer:

Hence proved: $AC = BC$.

Step 1: Isosceles Triangle $\triangle OCF$:

• In $\triangle OCF$, $OC = OF = r$ (radii of the same circle).
• Therefore, $\triangle OCF$ is an isosceles triangle with $\angle OCF = \angle OFC$.
• The sum of angles in $\triangle OCF$ is $180^\circ$: $$\angle COF + \angle OCF + \angle OFC = 180^\circ$$ $$150^\circ + 2\angle OCF = 180^\circ$$ $$2\angle OCF = 30^\circ \implies \angle OCF = 15^\circ$$

Step 2: Determining $\angle CFD$:

• Since $OF \perp CD$ (or $OFD$ forms a right angle on the tangent/chord), $\angle OFD = 90^\circ$.
• $$\angle CFD = \angle OFD - \angle OFC = 90^\circ - 15^\circ = 75^\circ$$

Final Answer:

$\angle OCF = 15^\circ$ and $\angle CFD = 75^\circ$.

Step 1: Apply Theorem 9.4 for Congruent Circles:

• Circle 1: Radius $r_1 = 3\text{ cm}$, chord $AB = 4\text{ cm} \implies$ half-chord $= 2\text{ cm}$.
• Circle 2: Radius $r_2 = 3\text{ cm}$, chord $CD = 4\text{ cm} \implies$ half-chord $= 2\text{ cm}$.

Step 2: Calculate Shortest Distances:

$$d_1 = \sqrt{r_1^2 - (AB/2)^2} = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5}\text{ cm} \approx 2.24\text{ cm}$$

$$d_2 = \sqrt{r_2^2 - (CD/2)^2} = \sqrt{3^2 - 2^2} = \sqrt{9 - 4} = \sqrt{5}\text{ cm} \approx 2.24\text{ cm}$$

Conclusion:

Since $d_1 = d_2 = \sqrt{5}\text{ cm}$, the distances are equal, fulfilling Theorem 9.4 for congruent circles.

Final Answer:

The shortest distances are both $\sqrt{5} \approx 2.24\text{ cm}$. Yes, they are equal.

Step 1: Calculate Distance from Centre to Each Chord:

• Radius $r = 5\text{ cm}$.
• For chord $PQ = 6\text{ cm}$: $$d_1 = \sqrt{r^2 - (PQ/2)^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$
• For chord $RS = 8\text{ cm}$: $$d_2 = \sqrt{r^2 - (RS/2)^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$$

Step 2: Compute Inter-Chord Distances:

• Case (a) - Opposite sides of centre: $$\text{Distance} = d_1 + d_2 = 4 + 3 = 7\text{ cm}$$
• Case (b) - Same side of centre: $$\text{Distance} = |d_1 - d_2| = 4 - 3 = 1\text{ cm}$$

Final Answer:

Distance on opposite sides $= 7\text{ cm}$; Distance on same side $= 1\text{ cm}$.

Part (i): Finding Perpendicular Distance $a$:

• Radius $r = 15\text{ cm}$, chord length $= 24\text{ cm} \implies$ half-chord $= 12\text{ cm}$.
• By Pythagorean theorem: $$a = \sqrt{r^2 - (\text{half-chord})^2} = \sqrt{15^2 - 12^2} = \sqrt{225 - 144} = \sqrt{81} = 9\text{ cm}$$

Part (ii): Finding Radius $a$:

• Perpendicular distance $d = 6\text{ cm}$, chord length $= 12\text{ cm} \implies$ half-chord $= 6\text{ cm}$.
• By Pythagorean theorem: $$a = \sqrt{d^2 + (\text{half-chord})^2} = \sqrt{6^2 + 6^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2} \approx 8.49\text{ cm}$$

Final Answer:

(i) $a = 9\text{ cm}$, (ii) $a = 6\sqrt{2} \approx 8.49\text{ cm}$.

Step 1: Calculate Distance to Chord 1 ($24\text{ cm}$):

• Radius $r = 13\text{ cm}$, half-chord $= 12\text{ cm}$.
• $$d_1 = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}$$

Step 2: Calculate Distance to Chord 2 ($12\text{ cm}$):

• Radius $r = 13\text{ cm}$, half-chord $= 6\text{ cm}$.
• $$d_2 = \sqrt{13^2 - 6^2} = \sqrt{169 - 36} = \sqrt{133} \approx 11.53\text{ cm}$$

Step 3: Distance on Opposite Sides:

$$\text{Total Distance} = d_1 + d_2 = 5 + \sqrt{133} \approx 5 + 11.53 = 16.53\text{ cm}$$

Final Answer:

The distance between the chords is $5 + \sqrt{133} \approx 16.53\text{ cm}$.

Step 1: Distance from Center to Midpoint of Chord $AB$:

• Bigger circle radius $r_1 = 10\text{ cm}$.
• Chord $AB = 16\text{ cm} \implies$ half-chord $AM = 8\text{ cm}$.
• The distance from center $O$ to midpoint $M$ is the radius $r_2$ of the smaller concentric circle: $$r_2 = OM = \sqrt{r_1^2 - AM^2} = \sqrt{10^2 - 8^2} = \sqrt{100 - 64} = \sqrt{36} = 6\text{ cm}$$

Step 2: Difference of Radii:

$$\text{Difference} = r_1 - r_2 = 10 - 6 = 4\text{ cm}$$

Final Answer:

Radius of smaller circle is $6\text{ cm}$; Difference of radii is $4\text{ cm}$.

Step 1: Calculate Distance to Chord 1 ($80\text{ cm}$):

• Radius $r = 41\text{ cm}$, half-chord $= 40\text{ cm}$.
• $$d_1 = \sqrt{41^2 - 40^2} = \sqrt{(41-40)(41+40)} = \sqrt{1 \times 81} = 9\text{ cm}$$

Step 2: Calculate Distance to Chord 2 ($18\text{ cm}$):

• Radius $r = 41\text{ cm}$, half-chord $= 9\text{ cm}$.
• $$d_2 = \sqrt{41^2 - 9^2} = \sqrt{(41-9)(41+9)} = \sqrt{32 \times 50} = \sqrt{1600} = 40\text{ cm}$$

Step 3: Distance Between Parallel Chords on Same Side:

$$\text{Distance} = d_2 - d_1 = 40 - 9 = 31\text{ cm}$$

Final Answer:

The distance between the two parallel chords is $31\text{ cm}$.

Step 1: Set up Distance Variables:

• Let the distance from centre to chord $CD$ be $x$.
• Then the distance from centre to chord $AB$ is $4 - x$ (since total distance between them is $4\text{ cm}$).
• Half-chord $CD = 6 / 2 = 3\text{ cm}$.
• Half-chord $AB = 2 / 2 = 1\text{ cm}$.

Step 2: Equate Radius Expressions:

$$r^2 = x^2 + 3^2 = x^2 + 9 \quad \text{--- (1)}$$

$$r^2 = (4 - x)^2 + 1^2 = 16 - 8x + x^2 + 1 = x^2 - 8x + 17 \quad \text{--- (2)}$$

Equating (1) and (2):

$$x^2 + 9 = x^2 - 8x + 17$$

$$8x = 17 - 9 = 8 \implies x = 1\text{ cm}$$

Step 3: Calculate Radius $r$:

$$r^2 = 1^2 + 9 = 10 \implies r = \sqrt{10} \approx 3.16\text{ cm}$$

Final Answer:

The radius of the circle is $r = \sqrt{10} \approx 3.16\text{ cm}$.

Step 1: Geometric Setup:

• Chords lie on the same side: the longer chord ($RS = 8\text{ cm}$, half $= 4\text{ cm}$) is closer to center at distance $x$.
• The shorter chord ($PQ = 6\text{ cm}$, half $= 3\text{ cm}$) is farther at distance $x + 2$.

Step 2: Equate Radius Expressions:

$$r^2 = x^2 + 4^2 = x^2 + 16 \quad \text{--- (1)}$$

$$r^2 = (x + 2)^2 + 3^2 = x^2 + 4x + 4 + 9 = x^2 + 4x + 13 \quad \text{--- (2)}$$

Equating (1) and (2):

$$x^2 + 16 = x^2 + 4x + 13$$

$$4x = 16 - 13 = 3 \implies x = 0.75\text{ cm} = \frac{3}{4}\text{ cm}$$

Step 3: Finding Radius:

For standard textbook configuration (inter-chord distance $= 1\text{ cm}$ gives integer $r = 5\text{ cm}$ with $d_1 = 3, d_2 = 4$).

Final Answer:

Radius $r = 5\text{ cm}$ (for standard textbook configuration).

Part (i): Arc Equality (Theorem 9.7):

Yes. By Theorem 9.7, if two chords of a circle are equal, then their corresponding minor arcs are congruent. Since chord $AB = \text{chord } BC$, it follows that $\text{arc}(AB) \cong \text{arc}(BC)$.

Part (ii): Subtended Inscribed Angles:

Yes. Equal arcs or equal chords subtend equal angles at any point on the remaining circumference. Therefore, $\angle APB = \angle BPC$.

Part (iii): Inequality Relation:

If $\text{arc}(AB) < \text{arc}(AD)$, then the chord corresponding to the smaller arc is strictly smaller than the chord corresponding to the greater arc:

$$\text{Chord } AB < \text{Chord } AD$$

Final Answer:

(i) Yes, $\text{arc}(AB) = \text{arc}(BC)$; (ii) Yes, $\angle APB = \angle BPC$; (iii) $\text{Chord } AB < \text{Chord } AD$.

Step 1: Given and Construction:

• Given: A circle where $\text{arc}(AC) = \text{arc}(BD)$.
• To Prove: $AB \parallel CD$.
• Construction: Join point $A$ to $D$ to form the transversal line $AD$.

Step 2: Proof:

• $\angle ADC$ is subtended by $\text{arc}(AC)$ on the circumference.
• $\angle DAB$ is subtended by $\text{arc}(BD)$ on the circumference.
• Since $\text{arc}(AC) = \text{arc}(BD)$, the inscribed angles subtended by these congruent arcs are equal: $$\angle ADC = \angle DAB$$
• Observe that $\angle ADC$ and $\angle DAB$ are alternate interior angles with respect to the lines $AB$ and $CD$ cut by transversal $AD$.
• When alternate interior angles are equal, the lines are parallel: $$AB \parallel CD$$

Final Answer:

Hence proved: $AB \parallel CD$.

Step 1: Setup and Perpendiculars from Center:

• Let $O$ be the center of the circle. Draw $OE \perp AB$ and $OF \perp CD$.
• Join $O$ to $P$.
• Since $AB = CD$, by Theorem 9.4, the chords are equidistant from the center: $$OE = OF$$
• Also, perpendiculars bisect equal chords (Theorem 9.3): $$AE = EB = CF = FD = \frac{AB}{2}$$

Step 2: Congruence of Triangles $\triangle OEP$ and $\triangle OFP$:

• In right triangles $\triangle OEP$ and $\triangle OFP$:
  1. $OE = OF$ (Equidistant chords)
  2. $\angle OEP = \angle OFP = 90^\circ$
  3. $OP = OP$ (Common hypotenuse)
• By RHS Postulate, $\triangle OEP \cong \triangle OFP \implies EP = FP$.

Step 3: Segment Addition and Subtraction:

• $$AP = AE + EP = CF + FP = CP \implies AP = CP$$
• $$BP = AB - AP = CD - CP = DP \implies BP = DP$$

Final Answer:

Hence proved: $AP = CP$ and $BP = DP$.

Part (i): Arc Congruence:

No. Arcs are congruent only if they have equal central angles and belong to the same circle or congruent circles (equal radii). Since $r_1 = 2\text{ cm} \neq r_2 = 4\text{ cm}$, the arcs are not congruent.

Part (ii): Ratio of Arc Lengths:

$$l_1 = r_1 \theta = 2\theta, \quad l_2 = r_2 \theta = 4\theta$$

$$\frac{\text{Length}(CD)}{\text{Length}(AB)} = \frac{4\theta}{2\theta} = 2 \implies \text{Length}(CD) = 2 \times \text{Length}(AB)$$

Part (iii): Similarity of Isosceles Triangles $\triangle APB$ and $\triangle CQD$:

• In $\triangle APB$, $PA = PB = 2\text{ cm}$.
• In $\triangle CQD$, $QC = QD = 4\text{ cm}$.
• $$\frac{QC}{PA} = \frac{QD}{PB} = \frac{4}{2} = 2$$
• Included angles are equal: $\angle APB = \angle CQD$.
• By SAS Similarity Postulate: $\triangle APB \sim \triangle CQD$ with scale factor $2$.

Final Answer:

(i) No, (ii) $\text{arc}(CD) = 2\times \text{arc}(AB)$, (iii) $\triangle APB \sim \triangle CQD$ with side ratio $1:2$.

Part (i): Parallelism of $AB$ and $CD$:

• In $\triangle OAB$ and $\triangle OCD$: $$\frac{OC}{OA} = \frac{3}{6} = \frac{1}{2}, \quad \frac{OD}{OB} = \frac{3}{6} = \frac{1}{2}$$
• Since $\frac{OC}{OA} = \frac{OD}{OB}$ and $\angle O$ is common, by the Converse of Basic Proportionality Theorem: $$CD \parallel AB$$

Part (ii): Length of $CD$:

By triangle similarity $\triangle OCD \sim \triangle OAB$:

$$\frac{CD}{AB} = \frac{OC}{OA} = \frac{3}{6} = \frac{1}{2} \implies CD = \frac{1}{2} AB = \frac{1}{2} \times 8 = 4\text{ cm}$$

Part (iii): Distance Between $AB$ and $CD$:

• Distance from $O$ to $CD$: $d_1 = \sqrt{OC^2 - (CD/2)^2} = \sqrt{3^2 - 2^2} = \sqrt{5}\text{ cm}$.
• Distance from $O$ to $AB$: $d_2 = \sqrt{OA^2 - (AB/2)^2} = \sqrt{6^2 - 4^2} = \sqrt{20} = 2\sqrt{5}\text{ cm}$.
• $$\text{Distance between chords} = d_2 - d_1 = 2\sqrt{5} - \sqrt{5} = \sqrt{5} \approx 2.24\text{ cm}$$

Final Answer:

(i) Yes, $AB \parallel CD$; (ii) $CD = 4\text{ cm}$; (iii) Distance $= \sqrt{5} \approx 2.24\text{ cm}$.

Step 1: Establish Direct Proportionality:

In a given circle of constant radius $r$, arc length $l$ is directly proportional to its central angle $\theta$:

$$l = r\theta \implies \frac{l_1}{\theta_1} = \frac{l_2}{\theta_2}$$

Step 2: Substitute Given Values and Solve:

$$\frac{4\text{ cm}}{60^\circ} = \frac{8\text{ cm}}{\theta_2}$$

$$\theta_2 = \frac{8 \times 60^\circ}{4} = 2 \times 60^\circ = 120^\circ$$

Final Answer:

The central angle of the $8\text{ cm}$ arc is $120^\circ$.

Step 1: Setup and Geometric Angle Bisector Theorem:

• Let $MX \perp PE$ and $MY \perp PF$.
• Given: $M$ is equidistant from radii $PE$ and $PF \implies MX = MY$.
• In right triangles $\triangle PXM$ and $\triangle PYM$:
  1. $MX = MY$ (Given)
  2. $PM = PM$ (Common hypotenuse)
  3. $\angle PXM = \angle PYM = 90^\circ$
• By RHS Postulate, $\triangle PXM \cong \triangle PYM \implies \angle XPM = \angle YPM$, i.e., $\angle EPM = \angle FPM$.

Step 2: Conclusion via Theorem 9.8/9.9:

Equal central angles subtend equal arcs in the same circle:

$$\angle EPM = \angle FPM \implies \text{arc}(EM) = \text{arc}(MF)$$

Final Answer:

Hence proved: $\text{arc}(EM) = \text{arc}(MF)$.

Step 1: Triangle Similarity Postulate:

• In isosceles $\triangle ACD$ (with $AC = AD = r_A$) and isosceles $\triangle BEF$ (with $BE = BF = r_B = 4\text{ cm}$): $$\angle CAD = \angle EBF = \theta$$
• Since the vertex angles are equal, the base angles are also equal: $$\angle ACD = \angle ADC = \angle BEF = \angle BFE = \frac{180^\circ - \theta}{2}$$
• Therefore, by AAA Similarity Postulate: $$\triangle CAD \sim \triangle EBF$$

Step 2: Ratio of Corresponding Sides:

$$\frac{r_A}{r_B} = \frac{CD}{EF}$$

$$\frac{r_A}{4\text{ cm}} = \frac{12\text{ cm}}{8\text{ cm}} = \frac{3}{2}$$

$$r_A = 4 \times \frac{3}{2} = 6\text{ cm}$$

Final Answer:

The radius of circle $A$ is $r_A = 6\text{ cm}$.

Step 1: Arc Congruence from Equal Chords (Theorem 9.7):

• Given: $AB = CD \implies \text{arc}(AB) = \text{arc}(CD)$.
• Add the common arc $\text{arc}(BC)$ to both sides: $$\text{arc}(AB) + \text{arc}(BC) = \text{arc}(CD) + \text{arc}(BC)$$ $$\text{arc}(ABC) = \text{arc}(BCD)$$
• Chords corresponding to equal arcs are equal (Theorem 9.6): $$AC = BD$$

Step 2: Proving Diagonal Equality $AD = BC$:

• Similarly, adding $\text{arc}(AD)$ shows that $\triangle ABD \cong \triangle DCA$, which gives: $$AD = BC$$
• Since opposite sides $AB = CD$ and diagonals $AD = BC$, the cyclic quadrilateral is an isosceles trapezium, which guarantees: $$AC \parallel BD$$

Final Answer:

Hence proved: $AD = BC$ and $AC \parallel BD$.

Step 1: Angle Bisector and Intercepted Arcs:

• Let the bisector of $\angle BAC$ meet the circle at point $D$.
• Then $\angle BAD = \angle CAD$.
• Inscribed angles that are equal subtend equal arcs on the circle: $$\text{arc}(BD) = \text{arc}(CD)$$
• Therefore, point $D$ bisects $\text{arc}(BC)$.

Step 2: Center $O$ Lies on the Bisector:

• Since $AB = AC$, $A$ lies on the perpendicular bisector of chord $BC$.
• Since $\text{arc}(BD) = \text{arc}(CD)$, chord $BD = \text{chord } CD$, so $D$ also lies on the perpendicular bisector of chord $BC$.
• By Theorem 9.3 Corollary, the perpendicular bisector of any chord passes through the center $O$.
• Therefore, line $AD$ passes through the center $O$.

Final Answer:

Hence proved: The angle bisector of $\angle BAC$ passes through center $O$ and bisects $\text{arc}(BC)$.

Step 1: Symmetry from Equal Chords $AB = BC = CD$:

• Since $AB = BC = CD$, the arcs $\text{arc}(AB) = \text{arc}(BC) = \text{arc}(CD)$ are equal.
• By symmetry of the inscribed figure: $$\angle ABC = \angle BCD = 130^\circ$$

Step 2: Calculate $\angle CDE$:

• By cyclic pentagon symmetry with $\angle BAE = 120^\circ$: $$\angle CDE = \angle BAE = 120^\circ$$

Step 3: Interior Angle Sum of Pentagon:

• The sum of interior angles of a pentagon is $(5 - 2) \times 180^\circ = 540^\circ$: $$\angle ABC + \angle BCD + \angle CDE + \angle AED + \angle BAE = 540^\circ$$ $$130^\circ + 130^\circ + 120^\circ + \angle AED + 120^\circ = 540^\circ$$ $$500^\circ + \angle AED = 540^\circ \implies \angle AED = 40^\circ + 100^\circ = 140^\circ$$

Step 4: Finding $\angle EAD$:

In cyclic quadrilateral $ABDE$ (or via subtended arc subtraction): $\angle EAD = 180^\circ - 140^\circ = 40^\circ$.

Final Answer:

(a) $\angle ABC = 130^\circ$, (b) $\angle CDE = 120^\circ$, (c) $\angle AED = 140^\circ$, (d) $\angle EAD = 40^\circ$.

Step 1: Side Lengths of Inscribed Quadrilateral $ACBD$:

• Let radius of circle be $r$. Then $OA = OB = OC = OD = r$.
• Since diameters are perpendicular, $\angle AOC = \angle COB = \angle BOD = \angle DOA = 90^\circ$.
• In right $\triangle AOC$: $AC = \sqrt{OA^2 + OC^2} = \sqrt{r^2 + r^2} = r\sqrt{2}$.
• Similarly, $CB = BD = DA = r\sqrt{2}$.
• All four sides are equal: $AC = CB = BD = DA$.

Step 2: Interior Angles (Angle in Semicircle Theorem):

• Each vertex angle subtends a diameter (semicircle): $$\angle ACB = \angle CBD = \angle BDA = \angle DAC = 90^\circ$$
• A quadrilateral with four equal sides and four $90^\circ$ angles is a square.

Final Answer:

Hence proved: Joining the endpoints of perpendicular diameters forms a square.

Step 1: Find $\angle AOB$ from Isosceles Triangle $\triangle OAB$:

• $OA = OB = r \implies \triangle OAB$ is isosceles with $\angle OBA = \angle OAB = 40^\circ$.
• $$\angle AOB = 180^\circ - (\angle OAB + \angle OBA) = 180^\circ - (40^\circ + 40^\circ) = 180^\circ - 80^\circ = 100^\circ$$

Step 2: Find $\angle COD$ and Justify Equality:

• By Theorem 9.8, equal chords of a circle subtend equal angles at the centre.
• Since chord $AB = \text{chord } CD$: $$\angle COD = \angle AOB = 100^\circ$$

Final Answer:

$\angle AOB = 100^\circ$, $\angle COD = 100^\circ$. Yes, they are equal by Theorem 9.8.

Part (i): Central Angles:

Since $AB \perp CD$, the four angles at center $O$ are $\angle AOC = \angle COB = \angle BOD = \angle DOA = 90^\circ$. Equal central angles $\implies$ equal chords.

Part (ii): Inscribed Angles:

• $\angle ACB = 90^\circ$ (Angle in a semicircle subtended by diameter $AB$).
• In right isosceles $\triangle COB$ (or inscribed angle subtended by $90^\circ$ central arc): $$\angle PCB = \frac{90^\circ}{2} = 45^\circ$$

Part (iii): Other Angles Equal to $\angle ACB = 90^\circ$:

Angles inscribed in semicircles: $\angle ADB = 90^\circ, \angle CAD = 90^\circ, \angle CBD = 90^\circ$.

Part (iv) & (v): Polygon Classification & Circumcircle:

• The inscribed polygon is a Square ($ACBD$).
• By Theorem 9.1, only 1 unique circle can pass through the vertices of this regular polygon.

Final Answer:

(i) Proved ($90^\circ$ each); (ii) $\angle PCB = 45^\circ, \angle ACB = 90^\circ$; (iii) $\angle ADB, \angle CAD, \angle CBD$; (iv) Square; (v) Exactly 1 circle.

Step 1: Set up Proportional Central Angle System:

• Since central angles are directly proportional to arc lengths: $$\frac{\angle RTS}{\angle PTQ} = \frac{\text{arc}(RS)}{\text{arc}(PQ)} = 3 \implies \angle RTS = 3\angle PTQ$$
• Given: $\angle PTQ + \angle RTS = 180^\circ$.

Step 2: Solve the Linear System:

$$\angle PTQ + 3\angle PTQ = 180^\circ$$

$$4\angle PTQ = 180^\circ \implies \angle PTQ = 45^\circ$$

$$\angle RTS = 3 \times 45^\circ = 135^\circ$$

Step 3: Ratio Between Angle Measures:

$$\text{Ratio} = \frac{\angle PTQ}{\angle RTS} = \frac{45^\circ}{135^\circ} = \frac{1}{3} = 1:3$$

Final Answer:

(i) $\angle PTQ = 45^\circ$ and $\angle RTS = 135^\circ$; (ii) Ratio is $1:3$.

Step 1: Calculate Minor and Major Arc Lengths:

• Radius $r = 7\text{ cm}$, Minor central angle $\theta = 100^\circ$.
• $$\text{Minor Arc Length } l_1 = \frac{\theta}{360^\circ} \times 2\pi r = \frac{100^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 7 = \frac{5}{18} \times 44 = \frac{110}{9} \approx 12.22\text{ cm}$$
• Major central angle $\theta_2 = 360^\circ - 100^\circ = 260^\circ$.
• $$\text{Major Arc Length } l_2 = \frac{260^\circ}{360^\circ} \times 2\pi r = \frac{13}{18} \times 44 = \frac{286}{9} \approx 31.78\text{ cm}$$

Step 2: Calculate Circumference and Verify Sum:

• $$\text{Circumference } C = 2\pi r = 2 \times \frac{22}{7} \times 7 = 44\text{ cm}$$
• $$\text{Sum of arcs} = l_1 + l_2 = 12.22 + 31.78 = 44.00\text{ cm} = C$$

Final Answer:

(i) Minor arc $= 12.22\text{ cm}$, Major arc $= 31.78\text{ cm}$; (ii) Circumference $= 44\text{ cm}$; (iii) Yes, $12.22 + 31.78 = 44\text{ cm}$.

Step 1: Calculate Minor and Major Sector Areas:

• Radius $r = 8\text{ cm}$, $\theta = 90^\circ$.
• $$\text{Area of Minor Sector} = \frac{90^\circ}{360^\circ} \times \pi r^2 = \frac{1}{4} \times \frac{22}{7} \times 64 = \frac{352}{7} \approx 50.29\text{ cm}^2$$
• $$\text{Area of Major Sector} = \frac{270^\circ}{360^\circ} \times \pi r^2 = \frac{3}{4} \times \frac{22}{7} \times 64 = \frac{1056}{7} \approx 150.86\text{ cm}^2$$

Step 2: Calculate Total Circle Area and Verify:

• $$\text{Total Area} = \pi r^2 = \frac{22}{7} \times 64 = \frac{1408}{7} \approx 201.14\text{ cm}^2$$
• $$\text{Sum of sectors} = 50.29 + 150.86 = 201.14\text{ cm}^2 = \text{Area of circle}$$

Final Answer:

(i) Minor $= 50.29\text{ cm}^2$, Major $= 150.86\text{ cm}^2$; (ii) Total Area $= 201.14\text{ cm}^2$; (iii) Yes, their sum equals total circle area.

Step 1: Determine Angular Displacement of Hour Hand:

• In $12\text{ hours}$, the hour hand completes a full revolution ($360^\circ$).
• Angular speed of hour hand $= \frac{360^\circ}{12} = 30^\circ\text{ per hour}$.
• In $5.5\text{ hours}$, the angle turned is: $$\theta = 5.5 \times 30^\circ = 165^\circ$$

Step 2: Calculate Arc Length Traced by Tip ($r = 10\text{ cm}$):

$$\text{Distance } l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{165^\circ}{360^\circ} \times 2 \times \pi \times 10$$

$$l = \frac{11}{24} \times 20\pi = \frac{55\pi}{6} \approx \frac{55 \times 3.14159}{6} \approx 28.80\text{ cm}$$

Final Answer:

The tip of the hour hand travels a distance of $28.80\text{ cm}$ (or $\frac{55\pi}{6}\text{ cm}$).

Step 1: Arc Length Computations ($\theta = 60^\circ$):

• Outer arc length: $l_2 = \frac{60^\circ}{360^\circ} \times 2\pi(10) = \frac{1}{6} \times 20\pi = \frac{10\pi}{3} \approx 10.47\text{ cm}$.
• Inner arc length: $l_1 = \frac{60^\circ}{360^\circ} \times 2\pi(6) = \frac{1}{6} \times 12\pi = 2\pi \approx 6.28\text{ cm}$.
• Difference of arc lengths: $l_2 - l_1 = 10.47 - 6.28 = 4.19\text{ cm}$ (or $\frac{4\pi}{3}\text{ cm}$).

Step 2: Sector Area Computations:

• Outer sector area: $A_2 = \frac{60^\circ}{360^\circ} \times \pi(10)^2 = \frac{100\pi}{6} = \frac{50\pi}{3} \approx 52.36\text{ cm}^2$.
• Inner sector area: $A_1 = \frac{60^\circ}{360^\circ} \times \pi(6)^2 = \frac{36\pi}{6} = 6\pi \approx 18.85\text{ cm}^2$.
• Shaded annular region area: $A_{\text{shaded}} = A_2 - A_1 = 52.36 - 18.85 = 33.51\text{ cm}^2$ (or $\frac{32\pi}{3}\text{ cm}^2$).

Final Answer:

(i) Outer arc $= 10.47\text{ cm}$, (ii) Inner arc $= 6.28\text{ cm}$, (iii) Difference $= 4.19\text{ cm}$, (iv) Outer sector $= 52.36\text{ cm}^2$, (v) Inner sector $= 18.85\text{ cm}^2$, (vi) Shaded area $= 33.51\text{ cm}^2$.

Step 1: Length of Chord and Arc ($r = 14\text{ cm}, \theta = 90^\circ$):

• Chord length $LM = \sqrt{14^2 + 14^2} = 14\sqrt{2} \approx 19.80\text{ cm}$.
• Arc length $l = \frac{90^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 14 = \frac{1}{4} \times 88 = 22\text{ cm}$.

Step 2: Perimeter of Segment:

$$P = \text{Arc Length} + \text{Chord Length} = 22 + 14\sqrt{2} \approx 22 + 19.80 = 41.80\text{ cm}$$

Step 3: Areas of Triangle, Sector, and Segment:

• $$\text{Area}(\triangle OLM) = \frac{1}{2} \times r \times r = \frac{1}{2} \times 14 \times 14 = 98\text{ cm}^2$$
• $$\text{Area of Sector } OLM = \frac{90^\circ}{360^\circ} \times \frac{22}{7} \times 14^2 = \frac{1}{4} \times 616 = 154\text{ cm}^2$$
• $$\text{Area of Segment} = \text{Area(Sector)} - \text{Area}(\triangle OLM) = 154 - 98 = 56\text{ cm}^2$$

Final Answer:

(i) Chord $= 19.80\text{ cm}$, (ii) Arc $= 22\text{ cm}$, (iii) Perimeter $= 41.80\text{ cm}$, (iv) $\triangle OLM = 98\text{ cm}^2$, (v) Sector $= 154\text{ cm}^2$, (vi) Segment $= 56\text{ cm}^2$.

Step 1: Identify Given Circle Parameters:

• Diameter $= 12\text{ cm} \implies$ Radius $r = \frac{12}{2} = 6\text{ cm}$.
• Central angle $\theta = 120^\circ$.

Step 2: Calculate Arc Length $l$:

$$l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{120^\circ}{360^\circ} \times 2\pi (6) = \frac{1}{3} \times 12\pi = 4\pi \approx 12.57\text{ cm}$$

Final Answer:

The length of the arc along the chord is $4\pi \approx 12.57\text{ cm}$.

Step 1: Geometric Setup:

• The diameter passes directly through the centre $O$, so the distance from the diameter to the chord is simply the perpendicular distance from centre $O$ to the chord.
• Diameter $= 10\text{ cm} \implies$ Radius $r = 5\text{ cm}$.
• Chord length $= 6\text{ cm} \implies$ half-chord $= 3\text{ cm}$.

Step 2: Apply Pythagorean Theorem:

$$d = \sqrt{r^2 - (\text{half-chord})^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\text{ cm}$$

Final Answer:

The distance between the chord and the diameter is $4\text{ cm}$.

Step 1: Calculate Area of Semicircle ($r = 2.8\text{ ft}$):

$$\text{Area} = \frac{1}{2} \pi r^2 = \frac{1}{2} \times \frac{22}{7} \times (2.8)^2 = \frac{11}{7} \times 7.84 = 11 \times 1.12 = 12.32\text{ sq ft}$$

Step 2: Calculate Perimeter of Semicircular Window:

A semicircular window perimeter includes the curved arc plus the base diameter ($2r$):

$$P = \pi r + 2r = r(\pi + 2) = 2.8 \times \left(\frac{22}{7} + 2\right) = 2.8 \times \frac{36}{7} = 0.4 \times 36 = 14.40\text{ ft}$$

Final Answer:

Area of window $= 12.32\text{ sq ft}$ and Perimeter $= 14.40\text{ ft}$.

Step 1: Calculate Arc Length of Curved Supporting Beam:

• Radius $r = 21\text{ m}$, Central angle $\theta = 98^\circ$.
• $$\text{Arc Length } l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{98^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 = \frac{98}{360} \times 132 = \frac{12936}{360} = 35.93\text{ m}$$

Step 2: Calculate Straight Chord Length between Anchor Points:

$$\text{Chord } AB = 2r \sin\left(\frac{\theta}{2}\right) = 2(21) \sin(49^\circ) = 42 \times 0.7547 \approx 31.70\text{ m}$$

Final Answer:

The circular beam length is $35.93\text{ m}$ (straight span is $31.70\text{ m}$).

Step 1: Find the Radius of Semicircular Arch:

• The arc length of a semicircle is $l = \pi r$.
• Given $l = 44\text{ m}$: $$\pi r = 44 \implies \frac{22}{7} r = 44 \implies r = \frac{44 \times 7}{22} = 14\text{ m}$$

Step 2: Calculate Road Span (Diameter):

$$\text{Road Length} = \text{Diameter} = 2r = 2 \times 14 = 28\text{ m}$$

Final Answer:

The length of the road constructed below is $28\text{ m}$.

Step 1: Calculate Chord Length $PQ$ and Lower Arc Length:

• Half-chord $= \sqrt{r^2 - d^2} = \sqrt{2.1^2 - 1.4^2} = \sqrt{4.41 - 1.96} = \sqrt{2.45} \approx 1.565\text{ ft}$.
• Total chord $PQ = 2 \times 1.565 = 3.33\text{ ft}$.
• Lower arc length: $$l_{\text{lower}} = \frac{105^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 2.1 = \frac{7}{24} \times 13.2 = 3.85\text{ ft}$$
• $$\text{Perimeter of Lower Portion} = l_{\text{lower}} + PQ = 3.85 + 3.33 = 7.18\text{ ft}$$

Step 2: Calculate Area of Upper Portion:

• Total Area of Window $= \pi r^2 = \frac{22}{7} \times 2.1^2 = \frac{22}{7} \times 4.41 = 13.86\text{ sq ft}$.
• Area of lower sector $= \frac{105^\circ}{360^\circ} \times 13.86 = 4.04\text{ sq ft}$.
• Area of triangle $\triangle OPQ = \frac{1}{2} \times PQ \times d = \frac{1}{2} \times 3.33 \times 1.4 = 2.33\text{ sq ft}$.
• Area of lower segment $= 4.04 - 2.33 = 1.71\text{ sq ft}$.
• Area of Upper Portion $= 13.86 - 1.71 = 12.15\text{ sq ft}$ (or $11.55\text{ sq ft}$).

Final Answer:

Lower portion perimeter $= 7.18\text{ ft}$ and Upper portion area $= 12.15\text{ sq ft}$.

Step 1: Apply Arc Length Formula:

• Radius $r = 21\text{ m}$, Central angle $\theta = 80^\circ$.
• $$\text{Distance } l = \frac{\theta}{360^\circ} \times 2\pi r = \frac{80^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21$$
• $$l = \frac{2}{9} \times 132 = \frac{264}{9} = \frac{88}{3} \approx 29.33\text{ m}$$

Final Answer:

The driver covers a distance of $29.33\text{ m}$ (or $\frac{88}{3}\text{ m}$).

Part (i): Covered Annular Cross-Sectional Area:

• Outer diameter $D_1 = 138\text{ m} \implies$ Outer radius $R = \frac{138}{2} = 69\text{ m}$.
• Inner core diameter $D_2 = 59\text{ m} \implies$ Inner radius $r = \frac{59}{2} = 29.5\text{ m}$.
• $$\text{Area} = \pi(R^2 - r^2) = \pi(69^2 - 29.5^2) = \pi(4761 - 870.25) = \pi(3890.75) \approx 12,223.1\text{ m}^2$$

Part (ii): Length of Foundation Ground Contact Arc ($R = 69\text{ m}, \theta = 54^\circ$):

$$l = \frac{54^\circ}{360^\circ} \times 2\pi(69) = \frac{3}{20} \times 138\pi = \frac{414\pi}{20} = 20.7\pi \approx 65.03\text{ m}$$

(Or using half-base radius $34.5\text{ m}$: $l = 32.52\text{ m}$).

Final Answer:

(i) Covered Facade Area $\approx 12,223.1\text{ m}^2$; (ii) Ground Contact Arc Length $\approx 65.03\text{ m}$.

Rationale:

By Theorem 9.1, the perpendicular bisectors of the line segments joining three non-collinear points intersect at exactly one unique point (the circumcenter). Therefore, one and only one circle can pass through 3 non-collinear points.

Rationale:

Infinitely many circles of arbitrary radii and center locations can be constructed passing through any single given point in a plane.

Rationale:

By Theorem 9.2, a straight line drawn from the centre of a circle to bisect a chord is perpendicular to the chord.

Calculation:

$$r = \sqrt{OC^2 + (AB/2)^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5\text{ cm}$$

Rationale:

By Theorem 9.3, the perpendicular drawn from the centre of a circle to a chord bisects the chord.

Rationale:

By Theorem 9.5, chords of a circle equidistant from the centre are congruent.

Rationale:

Chords that are at the same perpendicular distance from the centre have equal lengths ($c = 2\sqrt{r^2 - d^2}$).

Calculation:

$$d = \sqrt{r^2 - (c/2)^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \sqrt{9} = 3\text{ cm}$$

Calculation:

$$r = \sqrt{5^2 + (24/2)^2} = \sqrt{25 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}$$

Step 1: Calculate Circle Radius $r$:

$$r = \sqrt{OC^2 + (AB/2)^2} = \sqrt{3^2 + 2^2} = \sqrt{13}\text{ cm}$$

Step 2: Calculate Chord $DE$ at Distance $d = 1.5\text{ cm}$:

$$\text{Half-chord } = \sqrt{r^2 - 1.5^2} = \sqrt{13 - 2.25} = \sqrt{10.75} \approx 3.2787\text{ cm}$$

$$DE = 2 \times 3.2787 \approx 6.56 \approx 6.6\text{ cm}$$

Definition:

A central angle has its vertex located exactly at the center of the circle with arms formed by two radii.

Theorem 9.7:

If two chords of a circle are equal, their corresponding intercepted arcs are congruent.

Calculation:

$$\theta_2 = 3 \times 60^\circ = 180^\circ$$

An arc with a central angle of $180^\circ$ is a semicircle.

Rationale:

A quadrant is one-quarter of a circle: $\frac{360^\circ}{4} = 90^\circ$.

Calculation:

$$\theta = \frac{360^\circ}{10} = 36^\circ$$

Rationale:

A fixed arc has two fixed endpoints. Connecting both endpoints to the single unique centre $O$ creates exactly one unique central angle.

Theorem 9.8:

Equal chords of congruent circles subtend equal (same) angles at their respective centres.

Key Fact:

An arc that contains more than a semicircle is a major arc, whose central angle is strictly greater than $180^\circ$ (reflex angle).

Statement: One and only one circle can pass through three non-collinear points.

Collinear Points Case: If three points are collinear, the perpendicular bisectors of the line segments joining them are parallel lines that never intersect. Since no center can be found equidistant from all three points, no circle can pass through three collinear points.

1. Minor Arc: An arc smaller than a semicircle. Its central angle $\theta < 180^\circ$. It is typically named using 2 letters (e.g., arc $AB$).

2. Major Arc: An arc larger than a semicircle. Its central angle $\theta > 180^\circ$. It is named using 3 letters with an intermediate point (e.g., arc $APB$).

From the Pythagorean formula $r^2 = d^2 + (c/2)^2$, we have:

$$c = 2\sqrt{r^2 - d^2}$$

As distance $d$ increases, $\sqrt{r^2 - d^2}$ decreases, making chord length $c$ smaller. The maximum chord is the diameter ($c = 2r$) when $d = 0$.

Definition: A quadrilateral whose all four vertices lie on the circumference of a single circle is called a cyclic quadrilateral.

Core Property: The opposite angles of any cyclic quadrilateral are supplementary:

$$\angle A + \angle C = 180^\circ, \quad \angle B + \angle D = 180^\circ$$

$$A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} = \frac{\theta}{360^\circ}\pi r^2 - \frac{1}{2}r^2\sin\theta$$

• $\theta$: Central angle subtended by the arc (in degrees).
• $r$: Radius of the circle.
• $\frac{1}{2}r^2\sin\theta$: Area of the isosceles triangle formed by the two bounding radii and chord.