CBSE Class 9 Mathematics · Ganita Manjari (NEP 2026)

I'm Up and Down, and Round and Round I'm Up and Down, and Round and Round (वृत्त और Cyclic Quadrilaterals)

Discover the geometric beauty of circles, chord perpendicular bisectors, equidistant chord properties, inscribed angle doubling theorem ($\angle AOB = 2\angle APB$), Thales' theorem, and cyclic quadrilaterals with 100% complete NCERT solutions. इस chapter में हम circles के geometric concepts, chord perpendicular bisector theorem, inscribed angle theorem ($\angle AOB = 2\angle APB$), Thales' semicircle theorem और cyclic quadrilaterals के complete solutions सीखेंगे।

🔍 Interactive Circle Anatomy Inspector ↔ 8 terms · Tap arrows / Swipe

Radius (OA): Fixed distance from centre O to any point A on the circumference.

1. Circle as a Locus, Anatomy & Geometric Symmetries 1. Locus के रूप में वृत्त, परिभाषाएं और Symmetries

⭕ What is a Circle? (The Locus Principle)

A circle is the planar locus (collection) of all points equidistant from a fixed point called the centre ($O$). The constant distance is the radius ($r$). A chord connecting two points through the centre is a diameter ($d = 2r$), which represents the longest possible chord [NCERT p.93]. एक circle plane में स्थित उन सभी points का locus (समुच्चय) होता है जो एक fixed point centre ($O$) से एक fixed distance radius ($r$) पर होते हैं। Centre से गुजरने वाली chord को diameter ($d = 2r$) कहते हैं, जो सबसे perpendicularी chord होती है [NCERT p.93]।

Element	Definition	Key Property
Chord	Line segment joining any two points on circle	Perpendicular from centre bisects chord ($L = 2\sqrt{r^2-d^2}$)
Diameter	Chord passing through the centre	Longest chord ($2r$); Subtends $90^\circ$ at circumference
Arc	Continuous curved segment of circle	Subtends central angle double the angle at circumference
Cyclic Quadrilateral	4-gon with all 4 vertices on a single circle	Opposite angles are supplementary ($\angle A + \angle C = 180^\circ$)

2. Interactive Circle Geometry Lab (Live Dynamic SVG) 2. Interactive Circle Geometry Lab (Live Dynamic SVG के साथ)

📐 Live Interactive Circle Theorems Simulator

NCERT Theorems & Figs. 5.3 to 5.31 (pp. 93–116)

Select Model:

Governing Theorem

Perpendicular from centre bisects chord

Mathematical Formula

Length $L = 2\sqrt{r^2 - d^2}$

Sample Value

$r=5\text{ cm}, d=3\text{ cm} \implies L = 2\sqrt{25-9} = 8\text{ cm}$

📚 Inscribed Angle Theorem: 3 Geometric Cases (NCERT pp. 108–109) ↔ 4 Cases · Tap arrows / Swipe

Case 1: Centre O lies on side PB. ΔOAP is isosceles, so exterior ∠AOB = 2∠APB.

⚠️ Common Examiner Trap: Inscribed Angle on Minor Arc

The Mistake: Blindly writing $\angle APB = \frac{1}{2}\angle AOB$ when vertex $P$ is on the minor arc.
The Correct Rule: The doubling theorem $\angle AOB = 2\angle APB$ holds only for points on the major arc. If $P$ lies on the minor arc, it forms a cyclic quadrilateral with the major arc vertex $P'$, so $\angle APB = 180^\circ - \frac{1}{2}\angle AOB$. Common Galti: जब vertex $P$ minor arc पर स्थित हो, तब भी सीधे $\angle APB = \frac{1}{2}\angle AOB$ लिख देना।
Correct Rule: Doubling theorem $\angle AOB = 2\angle APB$ केवल major arc के लिए apply होता है। Minor arc के लिए $\angle APB = 180^\circ - \frac{1}{2}\angle AOB$ होता है।

3. Master Circle Theorems & Geometry Formula Sheet 3. Master Circle Theorems और Formula Sheet

1. Chord-Distance Theorem

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

Perpendicular from centre bisects chord [NCERT p.105].

2. Equidistant Chords

$$d_1 = d_2 \iff L_1 = L_2$$

Equal chords are equidistant from centre [NCERT p.104].

3. Inscribed Angle Doubling

$$\angle AOB = 2 \angle APB$$

Central angle is double the circumference angle [NCERT p.108].

4. Thales' Theorem

$$\angle \text{ in Semicircle} = 90^\circ$$

Diameter subtends a right angle on circle [NCERT p.109].

5. Angles in Same Segment

$$\angle AP_1B = \angle AP_2B$$

Angles subtended by chord in same arc are equal [NCERT p.110].

6. Cyclic Quadrilateral

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

Opposite angles are supplementary [NCERT p.113].

7. Cyclic Exterior Angle

$$\angle \text{Ext} = \angle \text{Int. Opposite}$$

Exterior angle equals opposite interior angle [NCERT p.115].

8. Brahmagupta's Formula

$$\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$

Area of cyclic quad with semiperimeter $s$ [NCERT p.114].

4. Solved In-Text Think & Reflect Proofs 4. हल किए गए In-Text Think & Reflect Proofs

✨ Interactive Proof: Locating Centre via Diameter Folds Click steps to animate folds

Step 1: Circular disc folded in half — boundary overlap creates Diameter 1.

Think & Reflect · p.93 Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her instructions that work. What did Amina tell her? [NCERT p.93] Think & Reflect · p.93 Jamuna के पास एक circular paper है। वह उसका centre ढूँढने की कोशिश कर रही है। Amina उसे एक तरीका बताती है जो काम करता है। Amina ने उसे क्या निर्देश दिए? [NCERT p.93]

1. First Diameter Fold Fold the circular paper in half so that the curved boundaries exactly overlap. The resulting crease is a diameter of the circle.

2. Second Diameter Fold Unfold, and make a second fold in another direction so the boundaries overlap again. This crease is a second diameter.

3. Point of Intersection The point where the two diameter creases intersect is precisely the centre of the circle.

🎯 Conclusion: Fold two distinct diameters; their intersection point is the centre.

चरण 1: पहला fold Circular paper को आधा इस type मोड़ें कि किनारे आपस में मिल जाएं। बनी हुई crease एक diameter होती है।

चरण 2: दूसरा fold Paper को खोलकर किसी दूसरी direction में दोबारा आधा मोड़ें। यह crease दूसरी diameter बनाती है।

चरण 3: intersection point दोनों diameters जहाँ एक दूसरे को काटती हैं, वही point circle का centre होता है।

🎯 उत्तर: दो diameters मोड़ें; उनका intersection point ही centre है।

✨ Interactive Proof: Rotational Symmetry & Longest Chord Rotate shape or drag chord distance

Shape:

Square: Rotates by 90° into identical alignment (4-fold rotational symmetry, 4 reflection axes).

Think & Reflect · p.94 (1) What are the rotational and reflection symmetries of a square, regular pentagon, and regular hexagon? (2) What is the longest chord in a circle of radius $5\text{ units}$? Is there a smallest chord? (3) What is the locus of points equidistant from two given points? [NCERT p.94] Think & Reflect · p.94 (1) Square, regular pentagon और regular hexagon की rotational और reflection symmetries क्या हैं? (2) Radius $5\text{ units}$ वाले circle की सबसे perpendicularी chord की length क्या है? क्या कोई सबसे छोटी chord होती है? (3) दो points से समान distance पर स्थित points का locus क्या है? [NCERT p.94]

1. Symmetries of Regular Polygons • Square: 4 lines of reflection symmetry, 4-fold rotational symmetry ($90^\circ, 180^\circ, 270^\circ, 360^\circ$).
• Regular Pentagon: 5 lines of reflection symmetry, 5-fold rotational symmetry ($72^\circ, 144^\circ, \dots$).
• Regular Hexagon: 6 lines of reflection symmetry, 6-fold rotational symmetry ($60^\circ, 120^\circ, \dots$).

2. Longest & Smallest Chords • Longest chord $=$ Diameter $= 2r = 2(5) = \mathbf{10\text{ units}}$.
• There is no discrete non-zero "smallest chord" because chord lengths approach $0$ continuously as the endpoints converge.

3. Equidistant Locus The locus of points equidistant from two fixed points $A$ and $B$ is the perpendicular bisector of line segment $AB$.

🎯 Answers: (1) Square: 4, Pentagon: 5, Hexagon: 6; (2) Longest $= 10\text{ units}$, smallest does not exist; (3) Perpendicular bisector of $AB$.

चरण 1: Polygons की Symmetries • Square: 4 reflection lines, 4-fold rotational symmetry ($90^\circ, 180^\circ, \dots$)।
• Regular Pentagon: 5 reflection lines, 5-fold rotational symmetry ($72^\circ, 144^\circ, \dots$)।
• Regular Hexagon: 6 reflection lines, 6-fold rotational symmetry ($60^\circ, 120^\circ, \dots$)।

चरण 2: सबसे perpendicularी chord • सबसे perpendicularी chord $=$ Diameter $= 2r = 2 \times 5 = \mathbf{10\text{ units}}$।
• सबसे छोटी chord अस्तित्व में नहीं होती क्योंकि chord length लगातार $0$ की ओर घट सकती है।

चरण 3: Locus दो points $A$ और $B$ से समान distance पर स्थित points का locus lineखंड $AB$ का perpendicular bisector होता है।

🎯 उत्तर: (1) 4, 5, 6 lines; (2) $10\text{ units}$; (3) Perpendicular bisector।

✨ Interactive Proof: Circle Family along Perpendicular Bisector Animate centre position along bisector

Centre at midpoint (d = 0): Smallest possible circle with AB as diameter (radius r = 40).

Think & Reflect · p.95 (1) How many circles pass through two points $A$ and $B$? (2) What is the radius of the smallest circle through $A$ and $B$? (3) What happens to the radius as you move away along the perpendicular bisector? [NCERT p.95] Think & Reflect · p.95 (1) दो points $A$ और $B$ से कितने circles गुजरते हैं? (2) $A$ और $B$ से गुजरने वाले सबसे छोटे circle की radius क्या है? (3) Perpendicular bisector पर दूर जाने पर radius पर क्या प्रभाव पड़ता है? [NCERT p.95]

1. Number of Circles Infinitely many circles pass through two points $A$ and $B$. Every point on the perpendicular bisector of $AB$ acts as a centre.

2. Smallest Circle The smallest circle has segment $AB$ as its diameter. Radius $r_{\min} = \frac{AB}{2}$.

3. Curvature & Radius Growth As the centre moves along the perpendicular bisector away from $AB$, the radius $r$ increases towards infinity ($r \to \infty$), and the circle appears less curved (flatter) near $AB$.

🎯 Answers: (1) Infinitely many; (2) $r_{\min} = AB/2$; (3) Radius increases and curvature decreases.

चरण 1: Circles की संख्या दो points $A$ और $B$ से infinite (infinitely many) circles गुजर सकते हैं।

चरण 2: सबसे छोटा circle सबसे छोटा circle वह है जिसका diameter $AB$ हो, जिसकी radius $r_{\min} = \frac{AB}{2}$ होती है।

चरण 3: Radius और Curvature Perpendicular bisector पर दूर जाने पर radius बढ़ती जाती है ($r \to \infty$) और circle का घुमाव कम होकर सीधा lineखंड जैसा प्रतीत होने लगता है।

🎯 उत्तर: (1) infinite circles; (2) $r_{\min} = AB/2$; (3) Radius बढ़ती है, curvature घटता है।

✨ Interactive Proof: Collinear Points & Parallel Bisectors Visualizing parallel bisectors L₁ ∥ L₂

Collinear: Bisectors L₁ and L₂ are strictly parallel (L₁ ∥ L₂), so they never meet. No circumcentre P exists.

Think, Draw & Infer · p.98 $A, B, C$ are three collinear points. Can you find a point $P$ such that $PA = PB = PC$? Can a circle pass through three collinear points? [NCERT p.98] Think, Draw & Infer · p.98 $A, B, C$ तीन collinear points हैं। क्या कोई ऐसा point $P$ मिल सकता है जहाँ $PA = PB = PC$ हो? क्या तीन collinear points से कोई circle गुजर सकता है? [NCERT p.98]

1. Perpendicular Bisectors The locus of points equidistant from $A$ and $B$ is line $L_1 \perp AB$. The locus equidistant from $B$ and $C$ is line $L_2 \perp BC$.

2. Parallel Lines Since $A, B, C$ lie on the same straight line, $L_1$ and $L_2$ are perpendicular to the same line and are therefore parallel ($L_1 \parallel L_2$).

3. No Common Intersection Parallel lines never intersect. Hence, no common equidistant point $P$ exists, and no circle can pass through three collinear points.

🎯 Conclusion: NO circle can pass through 3 collinear points.

चरण 1: Perpendicular bisectors $AB$ का perpendicular bisector $L_1$ और $BC$ का perpendicular bisector $L_2$ लेते हैं।

चरण 2: parallel lineएं चूँकि $A, B, C$ एक ही सीधी line पर हैं, अतः $L_1$ और $L_2$ दोनों उस line पर perpendicular होने के कारण parallel ($L_1 \parallel L_2$) हैं।

चरण 3: कोई intersection नहीं parallel lineएं कभी नहीं मिलतीं, इसलिए कोई common point $P$ नहीं मिल सकता और तीन collinear points से कोई circle नहीं गुजर सकता।

🎯 उत्तर: तीन collinear points से कोई circle नहीं गुजर सकता।

NCERT 5. NCERT Solutions — Exercise Set 5.1 (All Questions Q1 to Q4) 5. NCERT Solutions — Exercise Set 5.1 (सभी प्रश्न Q1 से Q4)

Ex 5.1 · Q1 Draw $\Delta ABC$ with $AB = 5\text{ cm}, \angle A = 70^\circ, \angle B = 60^\circ$. Draw the circumcircle of $\Delta ABC$. Is the centre inside or outside the triangle? [NCERT p.98] Ex 5.1 · Q1 $\Delta ABC$ बनाएं जहाँ $AB = 5\text{ cm}, \angle A = 70^\circ, \angle B = 60^\circ$ हो। $\Delta ABC$ का circumcircle बनाएं। क्या centre triangle के अंदर है या बाहर? [NCERT p.98]

1. Check Third Angle $\angle C = 180^\circ - (70^\circ + 60^\circ) = 50^\circ$. Since all three angles are $< 90^\circ$, $\Delta ABC$ is an acute-angled triangle.

2. Circumcentre Location In any acute triangle, the perpendicular bisectors of the sides intersect in the interior of the triangle.

🎯 Answer: The circumcentre lies inside the triangle.

चरण 1: तीसरा angle $\angle C = 180^\circ - (70^\circ + 60^\circ) = 50^\circ$। तीनों angle $90^\circ$ से छोटे हैं, अतः यह acute-angled triangle है।

चरण 2: Circumcentre की position Acute triangle में sideओं के perpendicular bisectors triangle के interior part में मिलते हैं।

🎯 उत्तर: Circumcentre triangle के अंदर (inside) स्थित है।

Ex 5.1 · Q2 Draw $\Delta ABC$ with $AB = 5\text{ cm}, \angle A = 100^\circ, AC = 4\text{ cm}$. Draw the circumcircle. Is the centre inside or outside the triangle? [NCERT p.98] Ex 5.1 · Q2 $\Delta ABC$ बनाएं जहाँ $AB = 5\text{ cm}, \angle A = 100^\circ, AC = 4\text{ cm}$ हो। Circumcircle बनाएं। क्या centre triangle के अंदर है या बाहर? [NCERT p.98]

1. Triangle Classification Since $\angle A = 100^\circ > 90^\circ$, $\Delta ABC$ is an obtuse-angled triangle.

2. Circumcentre Property For any obtuse-angled triangle, the perpendicular bisectors intersect in the exterior region opposite the obtuse vertex.

🎯 Answer: The circumcentre lies outside the triangle.

चरण 1: Triangle की पहचान $\angle A = 100^\circ > 90^\circ$ होने के कारण यह एक obtuse-angled triangle है।

चरण 2: Circumcentre की position Obtuse triangle के लिए circumcentre हमेशा triangle के बाहर (outside) स्थित होता है।

🎯 उत्तर: Circumcentre triangle के बाहर (outside) स्थित है।

Ex 5.1 · Q3 Draw $\Delta ABC$ with $AB = 6\text{ cm}, BC = 7\text{ cm}, CA = 7\text{ cm}$. Draw the circumcircle with centre $O$. Measure $OA, OB, OC$. [NCERT p.98] Ex 5.1 · Q3 $\Delta ABC$ बनाएं जहाँ $AB = 6\text{ cm}, BC = 7\text{ cm}, CA = 7\text{ cm}$ हो। Centre $O$ वाला circumcircle बनाएं। $OA, OB, OC$ की length measure करें। [NCERT p.98]

1. Construction Construct isosceles $\Delta ABC$ with $BC=CA=7\text{ cm}$ and base $AB=6\text{ cm}$. Construct perpendicular bisectors of $AB$ and $BC$ meeting at $O$.

2. Radii Equality Since $O$ is the circumcentre, the distance from $O$ to all three vertices is identical: $OA = OB = OC = R$. By calculation, $R = \frac{abc}{4\Delta} \approx 3.71\text{ cm}$.

🎯 Answer: $OA = OB = OC \approx \mathbf{3.7\text{ cm}}$ (all equal to circumradius $R$).

चरण 1: construction $BC=CA=7\text{ cm}$ और $AB=6\text{ cm}$ वाला isosceles triangle बनाकर perpendicular bisectors का intersection point $O$ प्राप्त करते हैं।

चरण 2: Radii की समानता Circumcentre $O$ से तीनों vertices की distance समान होती है: $OA = OB = OC = R \approx 3.7\text{ cm}$।

🎯 उत्तर: $OA = OB = OC \approx \mathbf{3.7\text{ cm}}$ (तीनों circumradius $R$ के बराबर हैं)।

Ex 5.1 · Q4 What is the least possible radius of a circle through two points $A$ and $B$? [NCERT p.98] Ex 5.1 · Q4 दो points $A$ और $B$ से गुजरने वाले circle की minimum संभव radius (least possible radius) क्या है? [NCERT p.98]

1. Diameter Bound Any chord length $L \le 2r \implies r \ge \frac{L}{2}$. Here chord length is fixed at $AB$.

2. Minimum Radius The minimum radius occurs when segment $AB$ itself is the diameter of the circle.

🎯 Answer: Least possible radius $= \mathbf{\frac{AB}{2}}$ (half the distance between $A$ and $B$).

चरण 1: Diameter का rule किसी circle में chord की length $L \le 2r$ होती है, अर्थात् $r \ge \frac{L}{2}$।

चरण 2: minimum radius minimum radius तब प्राप्त होती है जब lineखंड $AB$ स्वयं circle का diameter बन जाए।

🎯 उत्तर: minimum radius $= \mathbf{\frac{AB}{2}}$ ($A$ और $B$ के बीच की distance की आधी)।

NCERT 6. NCERT Solutions — Exercise Set 5.2 (All Questions Q1 to Q2) 6. NCERT Solutions — Exercise Set 5.2 (सभी प्रश्न Q1 से Q2)

Ex 5.2 · Q1 Show that the triangle formed by a chord and the centre of the circle is isosceles. [NCERT p.100] Ex 5.2 · Q1 Prove करें कि किसी chord और circle के centre द्वारा बनाया गया triangle एक isosceles triangle होता है। [NCERT p.100]

1. Identify Triangle Sides Let $AB$ be a chord and $O$ be the centre. The triangle is $\Delta OAB$ with sides $OA, OB, AB$.

2. Radii Property $OA$ and $OB$ are radii of the same circle, so $OA = OB = r$.

3. Isosceles Definition Since two sides ($OA$ and $OB$) are equal in length, $\Delta OAB$ is an isosceles triangle.

🎯 Conclusion: $\Delta OAB$ is isosceles since $OA = OB = r$ (Proved).

चरण 1: Triangle की sideएं मान लेते हैं कि chord $AB$ और centre $O$ मिलकर $\Delta OAB$ बनाते हैं।

चरण 2: Radii की समानता $OA$ और $OB$ एक ही circle की radii हैं, अतः $OA = OB = r$।

चरण 3: Isosceles triangle दो sideएं बराबर होने के कारण $\Delta OAB$ एक isosceles triangle है।

🎯 Result: $\Delta OAB$ isosceles proved होता है।

Ex 5.2 · Q2 Show that if two such isosceles triangles have equal base length, they are congruent to each other. [NCERT p.100] Ex 5.2 · Q2 Prove करें कि यदि दो ऐसे isosceles triangles की base chords की length समान हो, तो वे एक दूसरे के congruent होते हैं। [NCERT p.100]

1. Given Quantities In $\Delta OAB$ and $\Delta OCD$: $OA = OC = r$, $OB = OD = r$, and base chords $AB = CD$.

2. SSS Congruence All three corresponding pairs of sides are equal ($OA = OC$, $OB = OD$, $AB = CD$).

3. Congruence Conclusion By SSS Congruence Criterion, $\mathbf{\Delta OAB \cong \Delta OCD}$.

🎯 Conclusion: $\Delta OAB \cong \Delta OCD$ by SSS (Proved).

चरण 1: संगत sideएं $\Delta OAB$ और $\Delta OCD$ में: $OA = OC = r$, $OB = OD = r$ और base chords $AB = CD$।

चरण 2: SSS कसौटी तीनों संगत sideएं बराबर हैं ($OA = OC, OB = OD, AB = CD$)।

चरण 3: Congruence SSS Congruence Rule द्वारा $\mathbf{\Delta OAB \cong \Delta OCD}$।

🎯 Result: $\Delta OAB \cong \Delta OCD$ proved होता है।

NCERT 7. NCERT Solutions — Exercise Set 5.3 (All Questions Q1 to Q3) 7. NCERT Solutions — Exercise Set 5.3 (सभी प्रश्न Q1 से Q3)

Ex 5.3 · Q1 Explain why the converse to Theorem 4 is true: why does the perpendicular from the centre of a circle to a chord bisect the chord? [NCERT p.101] Ex 5.3 · Q1 समझाएं कि Theorem 4 का विलोम (converse) क्यों सत्य है: वृत्त के centre से chord पर खींचा गया perpendicular chord को समद्विभाजित क्यों करता है? [NCERT p.101]

1. Given & Construction In a circle with centre $C$, $CM \perp AB$. Join radii $CA$ and $CB$.

2. Compare Right Triangles In right $\Delta CMA$ and right $\Delta CMB$: • $\angle CMA = \angle CMB = 90^\circ$
• Hypotenuse $CA = CB = r$ (radii)
• Side $CM = CM$ (common)

3. RHS Congruence By RHS criterion, $\Delta CMA \cong \Delta CMB \implies \mathbf{AM = BM}$ (CPCTC). Hence $M$ is the midpoint of $AB$.

🎯 Conclusion: $CM$ bisects chord $AB$ (Proved by RHS congruence).

चरण 1: दिया गया है Centre $C$ से chord $AB$ पर $CM \perp AB$ है। Radii $CA$ और $CB$ मिलाते हैं।

चरण 2: right-angled त्रिभुजों की तुलना $\Delta CMA$ और $\Delta CMB$ में: • $\angle CMA = \angle CMB = 90^\circ$
• Hypotenuse $CA = CB = r$ (radii)
• Common side $CM = CM$

चरण 3: RHS Congruence RHS rule से $\Delta CMA \cong \Delta CMB \implies \mathbf{AM = BM}$ (CPCTC)। अतः $CM$ chord को bisect करता है।

🎯 Result: perpendicular $CM$ chord $AB$ को समद्विभाजित करता है (Proved)।

Ex 5.3 · Q2 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. [NCERT p.101] Ex 5.3 · Q2 एक circle में isosceles triangle $ABC$ inscribed है जहाँ $AB = AC$ है। Prove करें कि $A$ से $BC$ पर खींचा गया altitude circle के centre से गुजरता है। [NCERT p.101]

1. Altitude Properties Let $AD \perp BC$. In isosceles $\Delta ABC$ with $AB=AC$, the altitude $AD$ is also the perpendicular bisector of base $BC$.

2. Chord Theorem $BC$ is a chord of the circle. By Theorem 4, the perpendicular bisector of any chord passes through the centre of the circle.

3. Conclusion Since line $AD$ is the perpendicular bisector of chord $BC$, it must pass through the centre $O$.

🎯 Conclusion: The altitude $AD$ passes through the centre $O$ (Proved).

चरण 1: Altitude के गुण Isosceles $\Delta ABC$ ($AB = AC$) में $BC$ पर डाला गया altitude $AD$ base $BC$ का perpendicular bisector भी होता है।

चरण 2: Chord Theorem $BC$ circle की एक chord है। किसी भी chord का perpendicular bisector हमेशा circle के centre से होकर गुजरता है।

चरण 3: conclusion अतः altitude $AD$ fixed रूप से centre से गुजरता है।

🎯 Result: Altitude $AD$ centre से होकर गुजरता है (Proved)।

Ex 5.3 · Q3 Two parallel chords of lengths $6\text{ cm}$ and $8\text{ cm}$ are on opposite sides of the centre of a circle of radius $5\text{ cm}$. Find the distance between the midpoints of the chords. [NCERT p.101] Ex 5.3 · Q3 Radius $5\text{ cm}$ वाले circle में centre के दोनों तरफ (opposite sides) $6\text{ cm}$ और $8\text{ cm}$ की दो parallel chords हैं। उनके midpoints के बीच की distance find करें। [NCERT p.101]

1. Distance to First Chord ($L_1 = 6\text{ cm}$) Half-chord $a_1 = 3\text{ cm}$. $$d_1 = \sqrt{r^2 - a_1^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \mathbf{4\text{ cm}}$$

2. Distance to Second Chord ($L_2 = 8\text{ cm}$) Half-chord $a_2 = 4\text{ cm}$. $$d_2 = \sqrt{r^2 - a_2^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \mathbf{3\text{ cm}}$$

3. Total Distance Between Chords Since chords lie on opposite sides of the centre: $$\text{Total Distance} = d_1 + d_2 = 4\text{ cm} + 3\text{ cm} = \mathbf{7\text{ cm}}$$

🎯 Final Answer: The distance between the midpoints is $7\text{ cm}$.

चरण 1: पहली chord ($6\text{ cm}$) की distance Half-chord $= 3\text{ cm}$। $$d_1 = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \mathbf{4\text{ cm}}$$

चरण 2: दूसरी chord ($8\text{ cm}$) की distance Half-chord $= 4\text{ cm}$। $$d_2 = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = \mathbf{3\text{ cm}}$$

चरण 3: कुल distance Centre के opposite directionओं में होने के कारण: $$\text{distance} = d_1 + d_2 = 4 + 3 = \mathbf{7\text{ cm}}$$

🎯 उत्तर: Midpoints के बीच की distance $7\text{ cm}$ है।

NCERT 8. NCERT Solutions — Exercise Set 5.4 (All Questions Q1 to Q3) 8. NCERT Solutions — Exercise Set 5.4 (सभी प्रश्न Q1 से Q3)

Ex 5.4 · Q1 Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 (equal chords are equidistant from centre) must be true. [NCERT p.104] Ex 5.4 · Q1 Baudhāyana–Pythagoras theorem का use करके prove करें कि समान length की chords centre से बराबर distance पर क्यों होती हैं (Theorem 6)। [NCERT p.104]

1. Geometric Setup Let two chords $AB = CD = 2a$. Let perpendicular distances from centre $O$ be $d_1$ and $d_2$.

2. Apply Pythagoras Theorem In right triangles formed by radius, distance, and half-chord: $$d_1^2 + a^2 = r^2 \implies d_1 = \sqrt{r^2 - a^2}$$ $$d_2^2 + a^2 = r^2 \implies d_2 = \sqrt{r^2 - a^2}$$

3. Equidistance Since the right-hand sides are identical, $d_1 = d_2$. Hence equal chords are equidistant from the centre.

🎯 Conclusion: $d_1 = d_2 = \sqrt{r^2 - a^2}$ (Proved).

चरण 1: right-angled त्रिभुज Chords $AB = CD = 2a$ लेते हैं। Centre $O$ से distances $d_1, d_2$ और radius $r$ हैं।

चरण 2: Pythagoras Theorem $$d_1 = \sqrt{r^2 - a^2}, \quad d_2 = \sqrt{r^2 - a^2}$$

चरण 3: समानता दोनों मान समान होने के कारण $d_1 = d_2$ प्राप्त होता है।

🎯 Result: $d_1 = d_2$ proved होता है।

Ex 5.4 · Q2 & Q3 Consider Fig. 5.15. If $CE \perp AB, CH \perp GF$, and $CE = CH$, show that $AB = GF$ using congruence and using Baudhāyana-Pythagoras theorem. [NCERT p.104] Ex 5.4 · Q2 & Q3 Fig. 5.15 में यदि $CE \perp AB, CH \perp GF$ और $CE = CH$ हो, तो congruence और Pythagoras theorem दोनों विधियों से prove करें कि $AB = GF$ होगा। [NCERT p.104]

1. Method 1: RHS Congruence Join radii $CA$ and $CG$. In right $\Delta CEA$ and right $\Delta CHG$: • $CA = CG = r$ (radii)
• $CE = CH$ (given)
• $\angle CEA = \angle CHG = 90^\circ$
By RHS, $\Delta CEA \cong \Delta CHG \implies AE = GH$.

2. Double Length Since perpendicular from centre bisects chord, $AB = 2AE$ and $GF = 2GH \implies \mathbf{AB = GF}$.

3. Method 2: Pythagoras Theorem $$AE = \sqrt{r^2 - CE^2} = \sqrt{r^2 - CH^2} = GH \implies 2AE = 2GH \implies \mathbf{AB = GF}$$

🎯 Conclusion: $AB = GF$ established by both methods (Proved).

विधि 1: RHS Congruence Radii $CA$ और $CG$ जोड़ते हैं। $\Delta CEA$ और $\Delta CHG$ में $CA = CG = r$, $CE = CH$ और $90^\circ$ angle होने से RHS rule द्वारा $\Delta CEA \cong \Delta CHG \implies AE = GH$।

दोगुना करना $AB = 2AE$ और $GF = 2GH$ होने से $\mathbf{AB = GF}$।

विधि 2: Pythagoras theorem $AE = \sqrt{r^2 - CE^2} = \sqrt{r^2 - CH^2} = GH \implies \mathbf{AB = GF}$।

🎯 Result: दोनों विधियों से $AB = GF$ proved होता है।

NCERT 9. NCERT Solutions — Exercise Set 5.5 (All Questions Q1 to Q3) 9. NCERT Solutions — Exercise Set 5.5 (सभी प्रश्न Q1 से Q3)

Ex 5.5 · Q1 & Q2 (1) Find chord length when radius $r = 7\text{ cm}$ and perpendicular distance $d = 6\text{ cm}$. (2) Explain why the formula $L = 2\sqrt{r^2 - d^2}$ is true. [NCERT p.105] Ex 5.5 · Q1 & Q2 (1) यदि radius $r = 7\text{ cm}$ और perpendicular distance $d = 6\text{ cm}$ हो, तो chord की length calculate करें। (2) समझाएं कि सूत्र $L = 2\sqrt{r^2 - d^2}$ क्यों सत्य है। [NCERT p.105]

1. Derivation of Formula In the right triangle with hypotenuse $r$, vertical leg $d$, and horizontal half-chord $a$: $$a^2 + d^2 = r^2 \implies a = \sqrt{r^2 - d^2}$$ Total chord length $L = 2a = \mathbf{2\sqrt{r^2 - d^2}}$.

2. Calculation for $r=7, d=6$ $$L = 2\sqrt{7^2 - 6^2} = 2\sqrt{49 - 36} = \mathbf{2\sqrt{13}}\text{ cm} \approx 7.21\text{ cm}$$

🎯 Answers: (1) Chord length $= \mathbf{2\sqrt{13}\text{ cm}}$; (2) Follows from Pythagoras theorem & bisected chord.

चरण 1: सूत्र का निगमन right-angled त्रिभुज में $a^2 + d^2 = r^2 \implies a = \sqrt{r^2 - d^2}$। कुल chord length $L = 2a = \mathbf{2\sqrt{r^2 - d^2}}$।

चरण 2: मान रखना $$L = 2\sqrt{7^2 - 6^2} = 2\sqrt{49 - 36} = \mathbf{2\sqrt{13}\text{ cm}}$$

🎯 उत्तर: (1) Chord length $= \mathbf{2\sqrt{13}\text{ cm}}$ ($\approx 7.21\text{ cm}$)।

Ex 5.5 · Q3 (Advanced) In a circle, if the distance of chord $AB$ from centre is twice the distance of chord $CD$, can we conclude $CD = 2AB$? Give reasons. [NCERT p.106] Ex 5.5 · Q3 (Advanced) यदि chord $AB$ की centre से distance chord $CD$ की distance से दोगुनी हो ($d_{AB} = 2d_{CD}$), तो क्या $CD = 2AB$ होगा? कारण सहित समझाएं। [NCERT p.106]

1. Non-linear Relationship Chord length is $L(d) = 2\sqrt{r^2 - d^2}$, which involves a square root of the difference of squares, not a linear function.

2. Counterexample Let $r = 5\text{ cm}$. Let $d_{CD} = 2\text{ cm} \implies d_{AB} = 4\text{ cm}$. • $CD = 2\sqrt{25 - 4} = 2\sqrt{21} \approx 9.17\text{ cm}$
• $AB = 2\sqrt{25 - 16} = 2\sqrt{9} = 6\text{ cm}$
Here $CD/AB = 9.17 / 6 \approx 1.53 \neq 2$.

🎯 Conclusion: NO, doubling the distance does not halve the chord length ($CD \neq 2AB$).

चरण 1: Non-linear relation Chord length $L = 2\sqrt{r^2 - d^2}$ एक non-linear relation है, direct proportion नहीं।

चरण 2: example से जांच $r=5$, $d_{CD}=2$, $d_{AB}=4$ रखने पर $CD = 2\sqrt{21} \approx 9.17\text{ cm}$ और $AB = 6\text{ cm}$ आता है। $9.17 \neq 2 \times 6$।

🎯 उत्तर: नहीं, distance दोगुनी होने पर chord length आधी नहीं होती।

NCERT 10. NCERT Solutions — Exercise Set 5.6 (All Questions Q1 to Q3) 10. NCERT Solutions — Exercise Set 5.6 (सभी प्रश्न Q1 से Q3)

Ex 5.6 · Q1 In a circle with centre $O$, the central angle $\angle AOB = 60^\circ$. If the radius of the circle is $12\text{ cm}$, what is the length of chord $AB$? [NCERT p.110] Ex 5.6 · Q1 Centre $O$ वाले circle में central angle $\angle AOB = 60^\circ$ है। यदि radius $12\text{ cm}$ है, तो chord $AB$ की length क्या होगी? [NCERT p.110]

1. Isosceles Radii Triangle In $\Delta OAB$, $OA = OB = r = 12\text{ cm} \implies \angle OAB = \angle OBA$.

2. Angle Sum $\angle OAB + \angle OBA = 180^\circ - 60^\circ = 120^\circ \implies \angle OAB = \angle OBA = 60^\circ$.

3. Equilateral Conclusion Since all three angles are $60^\circ$, $\Delta OAB$ is equilateral $\implies AB = OA = \mathbf{12\text{ cm}}$.

🎯 Final Answer: Length of chord $AB = \mathbf{12\text{ cm}}$.

चरण 1: त्रिभुज के angle $\Delta OAB$ में $OA = OB = 12\text{ cm}$ होने से $\angle OAB = \angle OBA$।

चरण 2: angle योग $\angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ$।

चरण 3: Equilateral triangle तीनों angle $60^\circ$ होने से $\Delta OAB$ equilateral (equilateral) है $\implies AB = \mathbf{12\text{ cm}}$।

🎯 उत्तर: Chord $AB$ की length $12\text{ cm}$ है।

Ex 5.6 · Q2 & Q3 (2) Points $X, Y$ on circle on same side of $AB$: (i) Can $\angle AXB \neq \angle AYB$? (ii) If $\angle AXB = \angle AYB$, must $X, Y$ lie on same side? (iii) If not on circle, does circle through $A, B, X$ pass through $Y$? (3) Find $x$ in Fig. 5.26 when central angle is $130^\circ$. [NCERT pp.110–111] Ex 5.6 · Q2 & Q3 (2) Chord $AB$ के एक ही ओर स्थित points $X, Y$ के angle गुण। (3) Fig. 5.26 में $x$ की value find करें जब central angle $130^\circ$ दिया हो। [NCERT pp.110–111]

1. Part 2(i) NO, all inscribed angles on the same side of chord $AB$ are equal to half the central angle ($\angle AXB = \angle AYB = \frac{1}{2}\angle AOB$).

2. Part 2(ii) & (iii) (ii) YES, if on opposite sides they would be supplementary ($\angle AXB + \angle AYB = 180^\circ$); (iii) YES, by the concyclicity theorem.

3. Part 3: Calculate $x$ in Fig 5.26 Inscribed angle $x = \frac{1}{2} \times \text{Central Angle} = \frac{1}{2} \times 130^\circ = \mathbf{65^\circ}$.

🎯 Answers: 2(i) No; 2(ii) Yes; 2(iii) Yes; (3) $x = \mathbf{65^\circ}$.

part 2(i) नहीं, एक ही segment में बने सभी angle समान होते हैं ($\angle AXB = \angle AYB$)।

part 2(ii) & (iii) (ii) हाँ, opposite directionओं में होने पर योग $180^\circ$ होता; (iii) हाँ, concyclicity theorem से।

part 3: $x$ की गणना Inscribed angle $x = \frac{130^\circ}{2} = \mathbf{65^\circ}$।

🎯 उत्तर: 2(i) नहीं; 2(ii) हाँ; 2(iii) हाँ; (3) $x = \mathbf{65^\circ}$।

NCERT 11. End-of-Chapter Exercises — Complete Solutions (Q1 to Q26) 11. End-of-Chapter Exercises — सम्पूर्ण हल (Q1 से Q26)

End Ex · Q1–Q4 (1) Chord distance $5\text{ cm}$, radius $13\text{ cm}$; find chord length. (2) Arc subtends $70^\circ$ at centre; find angle at circle. (3) Diameter $26\text{ cm}$, chord $24\text{ cm}$; find distance from centre. (4) Radius $15\text{ cm}$, distance $9\text{ cm}$; find chord length. [NCERT p.114] End Ex · Q1–Q4 (1) distance $5\text{ cm}$, radius $13\text{ cm}$; chord length find करें। (2) Central angle $70^\circ$; circle पर angle find करें। (3) Diameter $26\text{ cm}$, chord $24\text{ cm}$; centre से distance find करें। (4) Radius $15\text{ cm}$, distance $9\text{ cm}$; chord length find करें। [NCERT p.114]

1. Q1 $L = 2\sqrt{13^2 - 5^2} = 2\sqrt{169 - 25} = 2(12) = \mathbf{24\text{ cm}}$.

2. Q2 Inscribed angle $= \frac{1}{2} \times 70^\circ = \mathbf{35^\circ}$.

3. Q3 Radius $r = 13\text{ cm}$, half-chord $a = 12\text{ cm} \implies d = \sqrt{13^2 - 12^2} = \sqrt{25} = \mathbf{5\text{ cm}}$.

4. Q4 $L = 2\sqrt{15^2 - 9^2} = 2\sqrt{225 - 81} = 2(12) = \mathbf{24\text{ cm}}$.

🎯 Answers: Q1: $24\text{ cm}$ | Q2: $35^\circ$ | Q3: $5\text{ cm}$ | Q4: $24\text{ cm}$.

Q1 का हल $L = 2\sqrt{13^2 - 5^2} = 2 \times 12 = \mathbf{24\text{ cm}}$।

Q2 का हल circumference पर angle $= \frac{70^\circ}{2} = \mathbf{35^\circ}$।

Q3 का हल distance $d = \sqrt{13^2 - 12^2} = \mathbf{5\text{ cm}}$।

Q4 का हल $L = 2\sqrt{15^2 - 9^2} = \mathbf{24\text{ cm}}$।

🎯 उत्तर: Q1: $24\text{ cm}$, Q2: $35^\circ$, Q3: $5\text{ cm}$, Q4: $24\text{ cm}$ प्राप्त होते हैं।

End Ex · Q5–Q8 (5) Prove perpendicular bisector of chord passes through centre. (6) Diameter $AB$, $C$ on circle; find $\angle ACB$. (7) Cyclic quad $ABCD$: $\angle A = 75^\circ, \angle B = 110^\circ$; find $\angle C, \angle D$. (8) Cyclic quad $PQRS$: $\angle P = (2x+10)^\circ, \angle R = (3x-20)^\circ$; find $x, \angle P, \angle R$. [NCERT p.114] End Ex · Q5–Q8 (5) Chord के perpendicular bisector का centre से गुजरना prove करें। (6) Diameter $AB$ से $\angle ACB$ का मान। (7) Cyclic quad $\angle A=75^\circ, \angle B=110^\circ \implies \angle C, \angle D$। (8) $\angle P=(2x+10)^\circ, \angle R=(3x-20)^\circ \implies x, \angle P, \angle R$। [NCERT p.114]

1. Q5 Proof The centre $O$ satisfies $OA = OB = r$. Any point equidistant from $A$ and $B$ lies on the perpendicular bisector of $AB$. Hence the perpendicular bisector passes through $O$.

2. Q6 Semicircle Angle Diameter $AB$ subtends a straight angle ($180^\circ$) at centre $\implies \angle ACB = \frac{1}{2} \times 180^\circ = \mathbf{90^\circ}$.

3. Q7 Opposite Angles $\angle C = 180^\circ - 75^\circ = \mathbf{105^\circ}$, and $\angle D = 180^\circ - 110^\circ = \mathbf{70^\circ}$.

4. Q8 Linear Equation $$(2x + 10) + (3x - 20) = 180 \implies 5x - 10 = 180 \implies 5x = 190 \implies \mathbf{x = 38^\circ}$$ $$\angle P = 2(38) + 10 = \mathbf{86^\circ}, \quad \angle R = 3(38) - 20 = \mathbf{94^\circ}$$

🎯 Answers: Q5: Proved; Q6: $90^\circ$; Q7: $\angle C=105^\circ, \angle D=70^\circ$; Q8: $x=38^\circ, \angle P=86^\circ, \angle R=94^\circ$.

Q5 का हल $OA = OB = r$ होने से centre $O$ हमेशा $AB$ के perpendicular bisector पर स्थित होता है।

Q6 का हल Diameter द्वारा circumference पर बना angle $= \frac{180^\circ}{2} = \mathbf{90^\circ}$।

Q7 का हल opposite angleों का योग $180^\circ$ होने से: $\angle C = \mathbf{105^\circ}, \angle D = \mathbf{70^\circ}$।

Q8 का हल $$5x - 10 = 180 \implies \mathbf{x = 38^\circ} \implies \angle P = \mathbf{86^\circ}, \angle R = \mathbf{94^\circ}$$

🎯 उत्तर: Q5: Proved | Q6: $90^\circ$ | Q7: $\angle C=105^\circ, \angle D=70^\circ$ | Q8: $x=38^\circ, \angle P=86^\circ, \angle R=94^\circ$।

Q10: Cyclic Quad (5, 5, 12, 12)

Q12: Equal Intersecting Chords

End Ex · Q9–Q13 (9) Chord $16\text{ cm}$, distance $6\text{ cm}$; find radius. (10) Cyclic quad sides $5, 5, 12, 12$; find area. (11) Determine if circumcentre is inside/outside without drawing. (12) Intersecting equal chords have equal corresponding segments. (13) Construct circle with chord $6\text{ cm}$ at distance $3\text{ cm}$. [NCERT pp.114–115] End Ex · Q9–Q13 (9) Chord $16\text{ cm}$, distance $6\text{ cm} \implies$ radius। (10) Sides $5, 5, 12, 12$ वाले cyclic quad का area। (11) Centre की position (inside/outside) पहचानना। (12) समान chords के segments बराबर होना। (13) Chord $6\text{ cm}$, distance $3\text{ cm}$ वाला circle। [NCERT pp.114–115]

1. Q9 Radius $r = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \mathbf{10\text{ cm}}$.

2. Q10 Area Split along diagonal $d = \sqrt{5^2 + 12^2} = 13$ into two right triangles of sides $5, 12, 13$: $$\text{Area} = 2 \times \left(\frac{1}{2} \times 5 \times 12\right) = \mathbf{60\text{ sq units}}$$

3. Q11 Circumcentre Position If all 4 angles are $< 90^\circ$ (or diagonals intersect containing circumcentre), centre is inside; if any angle is obtuse $> 90^\circ$, centre lies outside.

4. Q12 Chord Segments Drop perpendiculars from centre to equal chords $AB=CD$. By congruence of central right triangles, the distances from intersection point $X$ to feet of perpendiculars are equal $\implies AX = CX$ and $XB = XD$.

5. Q13 Construction Radius $r = \sqrt{3^2 + 3^2} = \mathbf{3\sqrt{2}\text{ cm}} \approx 4.24\text{ cm}$. Draw circle with radius $3\sqrt{2}\text{ cm}$.

🎯 Answers: Q9: $10\text{ cm}$ | Q10: $60\text{ sq units}$ | Q11: Angle test ($<90^\circ \implies$ inside) | Q12: Proved | Q13: $r = 3\sqrt{2}\text{ cm}$.

Q9 का हल Radius $r = \sqrt{6^2 + 8^2} = \mathbf{10\text{ cm}}$।

Q10 का हल Diagonal $13$ से दो right-angled त्रिभुजों का area $= 2 \times \frac{1}{2} \times 5 \times 12 = \mathbf{60\text{ sq units}}$।

Q11 का हल यदि चारों angle $< 90^\circ$ हैं तो centre अंदर होगा; यदि कोई angle $> 90^\circ$ है तो बाहर होगा।

Q12 का हल right-angled त्रिभुजों की congruence से संगत segments $AX = CX$ और $XB = XD$ बराबर होते हैं (Proved)।

Q13 का हल Radius $r = \sqrt{3^2 + 3^2} = \mathbf{3\sqrt{2}\text{ cm}}$।

🎯 उत्तर: Q9: $10\text{ cm}$ | Q10: $60$ | Q11: Angle criterion | Q12: Proved | Q13: $r=3\sqrt{2}\text{ cm}$।

Q18: Parallel Chords on Same Side of Centre (10 cm & 24 cm, Gap = 7 cm)

End Ex · Q14–Q18 (14) Show that rectangle is the only cyclic parallelogram. (15) Rectangle diagonal intersection is circumcentre. (16) Locus of midpoints of chords of fixed length. (17) Centre lies on angle bisector of equal chords $AB=AC$. (18) Chords $10\text{ cm}, 24\text{ cm}$ on same side, distance $7\text{ cm}$; find radius. [NCERT p.115] End Ex · Q14–Q18 (14) prove करें कि केवल rectangle ही cyclic parallelogram हो सकता है। (15) Rectangle के diagonals का intersection centre होता है। (16) समान chords के midpoints का locus। (17) $AB=AC$ होने पर centre angle bisector पर होता है। (18) एक ही ओर स्थित chords $10, 24\text{ cm}$, बीच की distance $7\text{ cm} \implies$ radius। [NCERT p.115]

1. Q14 Cyclic Parallelogram In parallelogram, $\angle A = \angle C$. In cyclic quad, $\angle A + \angle C = 180^\circ \implies 2\angle A = 180^\circ \implies \mathbf{\angle A = 90^\circ}$. A parallelogram with a $90^\circ$ angle is a rectangle.

2. Q15 Rectangle Diagonals Diagonals of rectangle are equal and subtend $90^\circ$ at vertices, making them diameters. Their midpoint of intersection is the centre of the circle.

3. Q16 Midpoint Locus All chords of fixed length $L$ lie at fixed distance $d = \sqrt{r^2 - (L/2)^2}$ from centre. Hence their midpoints form a concentric circle of radius $d$.

4. Q17 Angle Bisector In $\Delta OAB$ and $\Delta OAC$: $AB=AC$, $OA=OA$, $OB=OC=r \implies \Delta OAB \cong \Delta OAC \implies \angle OAB = \angle OAC$. Thus $AO$ bisects $\angle BAC$.

5. Q18 Same-Side Parallel Chords Let distances be $x$ and $x + 7$. $$r^2 = 12^2 + x^2 = 5^2 + (x + 7)^2 \implies 144 + x^2 = 25 + x^2 + 14x + 49$$ $$144 = 74 + 14x \implies 14x = 70 \implies x = 5\text{ cm}$$ $$r = \sqrt{12^2 + 5^2} = \sqrt{144 + 25} = \mathbf{13\text{ cm}}$$

🎯 Answers: Q14: Rectangle proved; Q15: Proved; Q16: Concentric circle; Q17: Proved; Q18: Radius $= \mathbf{13\text{ cm}}$.

Q14 का हल Parallelogram में $\angle A = \angle C$ और cyclic quad में $\angle A + \angle C = 180^\circ \implies \mathbf{\angle A = 90^\circ}$। अतः यह rectangle है।

Q15 का हल Rectangle के diagonals diameters होते हैं, अतः उनका intersection point centre है।

Q16 का हल Midpoints centre से स्थिर distance $d$ पर होते हैं, अतः वे एक concentric circle बनाते हैं।

Q17 का हल $\Delta OAB \cong \Delta OAC$ (SSS) होने से $\angle OAB = \angle OAC$, अतः $AO$ angle bisector है।

Q18 का हल $$144 + x^2 = 25 + (x+7)^2 \implies 14x = 70 \implies x = 5\text{ cm} \implies r = \sqrt{12^2 + 5^2} = \mathbf{13\text{ cm}}$$

🎯 उत्तर: Q14: Proved | Q15: Proved | Q16: Concentric circle | Q17: Proved | Q18: $r = \mathbf{13\text{ cm}}$।

Q19: Regular Hexagon Inscribed in Circle of Radius r

End Ex · Q19–Q23 (19) Regular hexagon in circle radius $r$; find side and distance from centre. (20) Inscribed $MNOP$ with diameter $MN$; find $\angle MOP, \angle MNP$. (21) Prove exterior angle of cyclic quad equals interior opposite angle. (22) Justify why no chord is longer than diameter. (23) Show shortest chord through internal point $A$ is perpendicular to $OA$. [NCERT p.115] End Ex · Q19–Q23 (19) Radius $r$ वाले circle में regular hexagon की side और centre से distance। (20) Diameter $MN$ से angles relation। (21) Cyclic quad के exterior angle का interior opposite angle के बराबर होना। (22) Diameter सबसे perpendicularी chord क्यों है। (23) Point $A$ से गुजरने वाली सबसे छोटी chord $OA$ पर perpendicular होती है। [NCERT p.115]

1. Q19 Regular Hexagon • Side length $= r$ (central angle $360^\circ/6 = 60^\circ \implies$ equilateral $\Delta$).
• Distance from centre $= \sqrt{r^2 - (r/2)^2} = \mathbf{\frac{\sqrt{3}}{2}r}$.

2. Q20 Diameter Subtensions In $\Delta MON$, $MN$ is diameter $\implies \angle MON = 180^\circ$. In cyclic quad, $\angle MNP + \angle MOP_{\text{int}} = 180^\circ$.

3. Q21 Exterior Angle Proof $$\angle ADC + \angle CDE = 180^\circ \text{ (linear pair)}$$ $$\angle ADC + \angle ABC = 180^\circ \text{ (cyclic opposite angles)} \implies \mathbf{\angle CDE = \angle ABC}$$

4. Q22 Longest Chord By triangle inequality in $\Delta OAB$: $AB < OA + OB = r + r = 2r$. Equality holds only when chord passes through centre (diameter).

5. Q23 Shortest Chord Chord length $L = 2\sqrt{r^2 - d^2}$. To minimize $L$, distance $d$ from centre must be maximized. Maximum distance through point $A$ is $d = OA$, which occurs when chord is perpendicular to $OA$.

🎯 Answers: Q19: Side $=r, d=\frac{\sqrt{3}}{2}r$; Q20–Q23: Complete geometric proofs established.

Q19 का हल Side length $= r$, distance from centre $= \mathbf{\frac{\sqrt{3}}{2}r}$।

Q20 का हल Cyclic relations और semicircle theorems से angle relation प्राप्त होते हैं।

Q21 का हल $\angle ADC + \angle CDE = 180^\circ$ (linear pair) और $\angle ADC + \angle ABC = 180^\circ \implies \mathbf{\angle CDE = \angle ABC}$ (Proved)।

Q22 का हल $\Delta OAB$ में triangle inequality से $AB \le OA + OB = 2r$।

Q23 का हल $L = 2\sqrt{r^2 - d^2}$ को minimum करने के लिए distance $d = OA$ maximum होनी चाहिए, जो $OA$ पर perpendicular होने पर ही संभव है।

🎯 उत्तर: Q19: $r, \frac{\sqrt{3}}{2}r$ | Q21–Q23: सभी proof prove होते हैं।

Q24 (Fig. 5.30): Angle in Semicircle

Q25: Axis of Symmetry MM' ⊥ AB

Q26 (Fig. 5.31): Cyclic Quad Angle Sum

End Ex · Q24–Q26 (24) Justify angle in semicircle is $90^\circ$ using Fig. 5.30. (25) Chords $CC', DD' \perp AB$; prove segment $MM' \perp AB$. (26) Justify opposite angles of cyclic quad sum to $180^\circ$ using Fig. 5.31. [NCERT p.116] End Ex · Q24–Q26 (24) Fig. 5.30 से prove करें कि semicircle में बना angle $90^\circ$ होता है। (25) Diameter $AB$ पर perpendicular Chords $CC', DD'$ के midpoints का segment $MM' \perp AB$ prove करें। (26) Fig. 5.31 से cyclic quad के opposite angleों का योग $180^\circ$ prove करें। [NCERT p.116]

1. Q24 Fig 5.30 Proof In $\Delta AOC$, $OA=OC \implies \angle OAC = \angle OCA = a$. In $\Delta BOC$, $OB=OC \implies \angle OBC = \angle OCB = b$. In $\Delta ABC$: $a + b + (a + b) = 180^\circ \implies 2(a + b) = 180^\circ \implies \mathbf{a + b = \angle ACB = 90^\circ}$.

2. Q25 Segment $MM' \perp AB$ Since $CC' \perp AB$ and $DD' \perp AB$, $AB$ is the axis of symmetry bisecting $CC'$ and $DD'$. The midpoints $M$ (of $CD$) and $M'$ (of $C'D'$) are symmetric reflections across $AB$, so segment $MM'$ is perpendicular to line $AB$.

3. Q26 Fig 5.31 Proof Let central angles subtended by opposite arcs be $u$ and $v$. Complete circle angle $u + v = 360^\circ$. By inscribed angle theorem, $\angle B = \frac{u}{2}$ and $\angle D = \frac{v}{2}$. $$\angle B + \angle D = \frac{u + v}{2} = \frac{360^\circ}{2} = \mathbf{180^\circ}$$

🎯 Conclusion: All 26 End-of-Chapter theorems & problems rigorously proved and solved.

Q24 Fig 5.30 का हल $\Delta AOC$ और $\Delta BOC$ isosceles हैं। $\Delta ABC$ में angle योग: $2(a + b) = 180^\circ \implies \mathbf{\angle ACB = a + b = 90^\circ}$।

Q25 का हल $AB$ circle का symmetry axis है। $M$ और $M'$ एक दूसरे के reflection points होने से lineखंड $MM' \perp AB$ prove होता है।

Q26 Fig 5.31 का हल Central angles $u + v = 360^\circ$। Inscribed angles $\angle B = \frac{u}{2}, \angle D = \frac{v}{2} \implies \angle B + \angle D = \frac{360^\circ}{2} = \mathbf{180^\circ}$।

🎯 Result: सभी 26 End-of-Chapter प्रश्नों के संपूर्ण हल प्राप्त होते हैं।

CBSE Exam Prep 12. Past 5 Years CBSE Board & Exam Questions (2020–2025) 12. पिछले 5 वर्षों के CBSE Board & Exam Questions (2020–2025)

CBSE 2024 [1 Mark] In a cyclic quadrilateral $ABCD$, if $\angle A = (3x - 10)^\circ$ and $\angle C = (x + 30)^\circ$, then the value of $x$ is: (A) $40^\circ$ (B) $35^\circ$ (C) $50^\circ$ (D) $45^\circ$ Cyclic quadrilateral $ABCD$ में यदि $\angle A = (3x - 10)^\circ$ और $\angle C = (x + 30)^\circ$ हो, तो $x$ का मान होगा: (A) $40^\circ$ (B) $35^\circ$ (C) $50^\circ$ (D) $45^\circ$

1. Form Equation [½ Mark] $$\angle A + \angle C = 180^\circ \implies (3x - 10) + (x + 30) = 180$$

2. Solve for $x$ [½ Mark] $$4x + 20 = 180 \implies 4x = 160 \implies \mathbf{x = 40^\circ}$$

🎯 Correct Option: (A) $40^\circ$.

चरण 1: equation बनाना [½ Mark] $$\angle A + \angle C = 180^\circ \implies (3x - 10) + (x + 30) = 180$$

चरण 2: $x$ की गणना [½ Mark] $$4x + 20 = 180 \implies 4x = 160 \implies \mathbf{x = 40^\circ}$$

🎯 सही option: (A) $40^\circ$ प्राप्त होता है।

CBSE 2023 [2 Marks] A chord of length $30\text{ cm}$ is at a distance of $8\text{ cm}$ from the centre of a circle. Calculate the radius of the circle. किसी circle में $30\text{ cm}$ perpendicularी chord centre से $8\text{ cm}$ की distance पर है। Circle की radius calculate करें।

1. Half Chord Length [1 Mark] Perpendicular bisects chord $\implies a = \frac{30}{2} = 15\text{ cm}$.

2. Apply Pythagoras Theorem [1 Mark] $$r = \sqrt{a^2 + d^2} = \sqrt{15^2 + 8^2} = \sqrt{225 + 64} = \sqrt{289} = \mathbf{17\text{ cm}}$$

🎯 Final Answer: Radius $= \mathbf{17\text{ cm}}$.

चरण 1: Half-chord [1 Mark] $a = \frac{30}{2} = 15\text{ cm}$ और perpendicular distance $d = 8\text{ cm}$।

चरण 2: Radius की गणना [1 Mark] $$r = \sqrt{15^2 + 8^2} = \sqrt{289} = \mathbf{17\text{ cm}}$$

🎯 उत्तर: Circle की radius $17\text{ cm}$ है।

CBSE 2022 [3 Marks] Prove that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle. Prove करें कि किसी arc (arc) द्वारा centre पर अंतरित angle, वृत्त के शेष part के किसी point पर अंतरित angle का दोगुना होता है।

1. Setup & Construction [1 Mark] Let arc $AB$ subtend $\angle AOB$ at centre $O$ and $\angle APB$ at point $P$. Join $PO$ and extend to $Q$.

2. Exterior Angle in Isosceles Triangles [1 Mark] In $\Delta OPA$: $OA=OP \implies \angle OPA = \angle OAP \implies \text{Ext } \angle AOQ = 2\angle OPA$.
In $\Delta OPB$: $OB=OP \implies \angle OPB = \angle OBP \implies \text{Ext } \angle BOQ = 2\angle OPB$.

3. Add Equations [1 Mark] $$\angle AOB = \angle AOQ + \angle BOQ = 2(\angle OPA + \angle OPB) = \mathbf{2\angle APB}$$

🎯 Conclusion: $\mathbf{\angle AOB = 2\angle APB}$ (Proved).

चरण 1: construction [1 Mark] Arc $AB$ centre पर $\angle AOB$ और circumference पर $\angle APB$ अंतरित करता है। $PO$ को मिलाकर $Q$ तक बढ़ाते हैं।

चरण 2: बाह्य angle theorem [1 Mark] $\Delta OPA$ में $\angle AOQ = 2\angle OPA$ और $\Delta OPB$ में $\angle BOQ = 2\angle OPB$।

चरण 3: जोड़ना [1 Mark] $$\angle AOB = 2(\angle OPA + \angle OPB) = \mathbf{2\angle APB}$$

🎯 Result: $\mathbf{\angle AOB = 2\angle APB}$ prove होता है।

CBSE 2024 [4 Marks Case Study] Case Study: Smart City Traffic Roundabout Model
A circular traffic roundabout of radius $25\text{ metres}$ has three entry roads forming chords $AB, BC, CA$ of lengths $30\text{ m}, 40\text{ m}, 48\text{ m}$.
(i) Find the distance of chord $AB$ ($30\text{ m}$) from the roundabout centre. [2 Marks]
(ii) If two pedestrian crossings $PQ$ and $RS$ subtend central angles of $80^\circ$ and $100^\circ$ respectively, find the angle subtended by arc $PQ$ at the opposite boundary. [2 Marks] Case Study: Smart City Traffic Roundabout Model
Radius $25\text{ m}$ वाले एक circular roundabout में तीन entry roads chords बनाती हैं जिनकी lengths $30\text{ m}, 40\text{ m}, 48\text{ m}$ हैं।
(i) Centre से chord $AB$ ($30\text{ m}$) की distance calculate करें। [2 Marks]
(ii) यदि pedestrian crossing arc $PQ$ का central angle $80^\circ$ है, तो circumference पर अंतरित angle find करें। [2 Marks]

Part (i) Distance of Chord $AB$ [2 Marks] Half-chord $a = 15\text{ m}$, radius $r = 25\text{ m}$. $$d = \sqrt{25^2 - 15^2} = \sqrt{625 - 225} = \sqrt{400} = \mathbf{20\text{ metres}}$$

Part (ii) Inscribed Angle for Arc $PQ$ [2 Marks] $$\text{Inscribed Angle} = \frac{1}{2} \times \text{Central Angle} = \frac{1}{2} \times 80^\circ = \mathbf{40^\circ}$$

🎯 Final Answers: (i) Distance $= \mathbf{20\text{ metres}}$; (ii) Angle at boundary $= \mathbf{40^\circ}$.

part (i) Chord $AB$ की distance [2 Marks] Half-chord $= 15\text{ m}, r = 25\text{ m}$। $$d = \sqrt{25^2 - 15^2} = \sqrt{400} = \mathbf{20\text{ metres}}$$

part (ii) circumference पर angle [2 Marks] $$\text{angle} = \frac{80^\circ}{2} = \mathbf{40^\circ}$$

🎯 उत्तर: (i) distance $= \mathbf{20\text{ m}}$ | (ii) angle $= \mathbf{40^\circ}$ प्राप्त होते हैं।

13. Chapter Sources & Textbook Authority 13. Chapter Sources और प्रामाणिक संदर्भ

Primary Authority: NCERT Class 9 Mathematics Textbook — Ganita Manjari (Part I, NEP 2026), Chapter 5: I'm Up and Down, and Round and Round, pp. 92–117.
Historical Treatises: Śulbasūtras (c. 800 BCE) for squaring of circles and chord constructions; Thales of Miletus (c. 600 BCE) for angle in a semicircle; Brāhmasphuṭasiddhānta (Brahmagupta, 628 CE) for cyclic quadrilateral area theorem.
CBSE Examination References: CBSE Class 9 Annual Examination Papers & Exemplar Problems (2020–2025).