NCERT Solutions Class 9 Mathematics Ganita Manjari Chapter 05 I'm Up and Down, and Round and Round

Question 1. Draw ΔABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Answer: Follow these construction steps: Draw line segment AB = 5 cm. At point A, construct ∠CAB = 70°. At point B, construct ∠CBA = 60°. Allow the two rays to meet at C to form ΔABC. Draw the perpendicular bisector of AB. Draw the perpendicular bisectors of BC and AC. Let all three perpendicular bisectors meet at point O - this is the circumcentre. With centre O and radius OA, draw a circle passing through A, B, and C.

Now calculate: ∠C = 180° - (70° + 60°) = 50°. Since all angles of the triangle are less than 90°, the triangle is acute-angled. In an acute-angled triangle, the circumcentre lies inside the triangle. Therefore, the centre is inside the triangle.
In simple words: When all three angles of a triangle are less than 90 degrees, the centre of the circle around it stays inside the triangle.

Exam Tip: Always calculate the third angle and check if it is less than, equal to, or greater than 90° to determine whether the circumcentre lies inside, on, or outside the triangle.

Question 2. Draw ΔABC with AB = 5 cm, ∠A = 100°, AC = 4 cm. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Answer: Follow these construction steps: Draw line segment AB = 5 cm. At point A, construct ∠BAC = 100°. On the ray from A, mark point C so that AC = 4 cm. Join C to B to form ΔABC. Draw the perpendicular bisector of AB. Draw the perpendicular bisector of AC and BC. Let these bisectors meet at O - this is the circumcentre. With centre O and radius OA, draw the circle through A, B, and C.

Since ∠A = 100°, the triangle is obtuse-angled. In an obtuse-angled triangle, the circumcentre lies outside the triangle. Therefore, the centre is outside the triangle.
In simple words: When one angle of a triangle is bigger than 90 degrees, the centre of the circle around it sits outside the triangle.

Exam Tip: A single angle greater than 90° makes the triangle obtuse, and this determines that the circumcentre must be exterior to the triangle.

Question 3. Draw ΔABC, with AB = 6 cm, BC = 7 cm and CA = 7 cm. Draw the circumcircle of ΔABC. Let the circumcentre be O. Measure OA, OB, OC.
Answer: Follow these construction steps: Draw AB = 6 cm. With centre A and radius 7 cm, draw an arc. With centre B and radius 7 cm, draw another arc that cuts the first arc at C. Join AC and BC to form ΔABC. Draw the perpendicular bisector of AB. Draw the perpendicular bisectors of AC and BC. Let all these bisectors meet at O - this is the circumcentre. With centre O and radius OA, draw the circumcircle.

Observation: Since O is the circumcentre, it is equidistant from all three vertices. Therefore, OA = OB = OC. For the triangle with sides 7 cm, 7 cm, 6 cm, when you measure OA, OB, and OC, you will find that all three distances are approximately equal to 4 cm.
In simple words: The centre of a circle around a triangle is always the same distance from each corner of the triangle.

Exam Tip: Remember that the circumcentre is equidistant from all three vertices - this property must hold true regardless of the triangle's shape or size.

Question 4. What is the least possible radius of a circle through two points A and B?
Answer: For all circles passing through two fixed points A and B, the centre must lie on the perpendicular bisector of AB. The smallest circle occurs when the centre is positioned at the midpoint of AB. In this case, AB becomes the diameter of the circle. Therefore, the least possible radius equals half of AB - that is, Radius = AB/2. Hence, the smallest possible radius of a circle through points A and B is half the distance between them.
In simple words: The smallest circle you can draw through two points has those two points as the ends of its diameter.

Exam Tip: The key insight is that the smallest circle has the shortest possible distance from its centre to either point, which occurs when that centre is at the midpoint of the chord AB.

Question 1. Show that the triangle formed by a chord and the centre of the circle is isosceles.
Answer: Let AB be a chord of a circle with centre O. Join OA and OB. In triangle OAB, OA and OB are both radii of the same circle, so OA = OB. Since two sides of the triangle are equal, triangle OAB is isosceles. Hence proved.
In simple words: Any triangle made by joining the two ends of a chord to the centre of the circle must have two equal sides - the two radii.

Exam Tip: This is a fundamental property - always recognise that radii to any two points on a circle are equal, which immediately makes any such triangle isosceles.

Question 2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Answer: Let AB and PQ be two equal chords of the same circle with centre O. The triangles formed are OAB and OPQ. In triangles OAB and OPQ: OA and OP are both radii, so OA = OP. OB and OQ are both radii, so OB = OQ. The chords are given as equal, so AB = PQ. By the SSS (Side-Side-Side) congruence criterion, triangle OAB is congruent to triangle OPQ. Therefore, if two such isosceles triangles have equal base length, they are congruent to each other. Hence proved.
In simple words: If two chords of the same circle have equal length, then the triangles they form with the centre are exactly the same shape and size.

Exam Tip: Use the SSS criterion by identifying all three pairs of equal sides - the two radii in each triangle (which are always equal to each other) plus the equal chords themselves.

Question 1. Can you explain why the converse to Theorem 4 is true, i.e., why does the perpendicular from the centre of a circle to a chord of the circle bisect the chord?
Answer: Converse of Theorem 4: The perpendicular from the centre of a circle to a chord of the circle bisects the chord. Let AB be a chord of a circle with centre O, and let OM be perpendicular to AB. We need to prove: AM = BM.

In triangles OMA and OMB: OA and OB are both radii of the same circle, so OA = OB. OM is common to both triangles. ∠OMA and ∠OMB are both 90°, since OM is perpendicular to AB. By the RHS (Right angle-Hypotenuse-Side) congruence criterion, triangle OMA is congruent to triangle OMB. Therefore, AM = BM by Corresponding Parts of Congruent Triangles (CPCT). This means M is the midpoint of AB. Therefore, the perpendicular from the centre of a circle to a chord bisects the chord. Hence proved.
In simple words: A line drawn from the centre of a circle at a right angle to a chord cuts that chord into two equal halves.

Exam Tip: Recognise the RHS congruence setup - two right-angled triangles that share a common side (the perpendicular) and have equal hypotenuses (the radii) must be congruent.

Question 2. An isosceles triangle ABC is inscribed in a circle, with AB = AC. Show that the altitude from A to BC passes through the centre of the circle.
Answer: Given: Triangle ABC is inscribed in a circle and AB = AC. Since AB = AC, triangle ABC is isosceles with base BC. Let AD be the altitude from A to BC (perpendicular to BC).

In triangles ABD and ACD: AB and AC are equal (given). AD is common to both triangles. ∠ADB and ∠ADC are both 90°, since AD is perpendicular to BC. By the RHS congruence criterion, triangle ABD is congruent to triangle ACD. Therefore, BD = DC by CPCT. This means AD bisects BC. Since AD is both perpendicular to BC and also bisects BC, AD is the perpendicular bisector of chord BC. The centre of the circle must lie on the perpendicular bisector of every chord. Therefore, AD passes through the centre of the circle. Hence proved.
In simple words: In an isosceles triangle inscribed in a circle, the height from the unequal angle to the base always passes through the circle's centre.

Exam Tip: The key is recognising that the altitude in an isosceles triangle from the vertex angle is also the perpendicular bisector of the base, which must pass through the circumcentre.

Question 3. Two parallel chords of lengths 6 cm and 8 cm are on opposite sides of the centre of a circle. If the radius of the circle is 5 cm, find the distance between the midpoints of the chords.
Answer: Given: Radius of the circle is 5 cm. The two chords have lengths 6 cm and 8 cm and lie on opposite sides of the centre.

The perpendicular from the centre O to a chord bisects the chord. For the 6 cm chord AB, the perpendicular meets it at M, so AM = 3 cm. For the 8 cm chord CD, the perpendicular meets it at N, so CN = 4 cm. Let d₁ be the distance from O to M, and d₂ be the distance from O to N.

In right triangle OAM, using the Pythagorean theorem: OM² + AM² = OA² ⇒ d₁² + 3² = 5² ⇒ d₁² + 9 = 25 ⇒ d₁² = 16 ⇒ d₁ = 4 cm.

In right triangle OCN, using the Pythagorean theorem: ON² + CN² = OC² ⇒ d₂² + 4² = 5² ⇒ d₂² + 16 = 25 ⇒ d₂² = 9 ⇒ d₂ = 3 cm.

Since the chords are on opposite sides of the centre and their midpoints lie on the same line through the centre, the distance between the midpoints = d₁ + d₂ = 4 + 3 = 7 cm.
In simple words: Find how far each chord's midpoint is from the centre, then add those distances because the chords sit on opposite sides.

Exam Tip: Always use the Pythagorean theorem with the right triangle formed by the radius, the perpendicular distance, and half the chord length.

Question 1. Use the Baudhāyana–Pythagoras theorem to show why Theorem 6 must be true.
Answer: Theorem 6 states: Chords of a circle having the same length are at the same distance from the centre. Let AB and CD be two equal chords of a circle with centre O. Let OM be perpendicular to AB and ON be perpendicular to CD. The perpendicular from the centre to a chord bisects the chord, so AM = MB = AB/2 and CN = ND = CD/2. Since AB = CD, we have AM = CN. Also, OA = OC = radius of the circle.

Now apply the Baudhāyana–Pythagoras theorem in right triangles OMA and ONC: In triangle OMA: OA² = OM² + AM². In triangle ONC: OC² = ON² + CN². Since OA = OC and AM = CN, we have OM² + AM² = ON² + CN². Therefore, OM² = ON², which means OM = ON. Hence, the equal chords AB and CD are equidistant from the centre. Thus, Theorem 6 is proved.
In simple words: Any two chords of a circle that are the same length must sit the same distance away from the circle's centre.

Exam Tip: The Pythagorean relationship between the radius, the perpendicular distance, and half the chord length is the foundation - equal chords lead to equal perpendicular distances through this relationship.

Question 2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GF, and CE = CH, show that AB = GF.
Answer: Given: CE ⊥ AB, CH ⊥ GF, and CE = CH. We need to prove: AB = GF.

Since CE is perpendicular to chord AB, the perpendicular from the centre to a chord bisects the chord. Therefore, AE = EB. Similarly, since CH is perpendicular to chord GF, we have GH = HF.

In triangles CEA and CHG: CA and CG are both radii of the same circle, so CA = CG. CE and CH are given as equal. ∠CEA and ∠CHG are both 90°. By the RHS congruence criterion, triangle CEA is congruent to triangle CHG. Therefore, AE = GH by CPCT. Since AE = EB and GH = HF, we have 2AE = 2GH, which means AB = GF (because AB = 2AE and GF = 2GH). Hence proved.
In simple words: If two perpendiculars from the circle's centre to two chords are equal in length, then the chords themselves must be equal.

Exam Tip: Use RHS congruence by pairing the radii as hypotenuses, the perpendiculars as equal sides, and the right angles to establish that half-lengths of the chords must be equal.

Question 3. Solve the previous question using the Baudhāyana–Pythagoras theorem.
Answer: Given: CE ⊥ AB, CH ⊥ GF, and CE = CH. We need to prove: AB = GF.

Since CE is perpendicular to chord AB, it bisects AB. Therefore, E is the midpoint of AB, so AE = AB/2. Similarly, since CH is perpendicular to chord GF, H is the midpoint of GF, so GH = GF/2.

Now apply the Baudhāyana–Pythagoras theorem in right triangles CEA and CHG. In triangle CEA: CA² = CE² + AE². In triangle CHG: CG² = CH² + GH². Since CA = CG (both radii of the same circle) and CE = CH (given), we have CE² + AE² = CH² + GH². Simplifying: AE² = GH² (since CE = CH). Therefore, AE = GH. Since AB/2 = GF/2, we conclude that AB = GF. Hence proved.
In simple words: Using the Pythagorean theorem, equal perpendicular distances from the centre, combined with equal radii, force the half-lengths of the chords to be equal, making the full chords equal.

Exam Tip: This approach emphasises the Pythagorean relationship: when the radius is the hypotenuse, the perpendicular and half-chord are the two perpendicular sides. Equal perpendiculars and equal hypotenuses force equal half-chords.

Question 1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.
Answer: Let the chord be AB and O be the centre of the circle. Let OM be the perpendicular from O to chord AB. Given: Radius, OA = OB = 7 cm Perpendicular distance, OM = 6 cm We know that the perpendicular from the centre to a chord bisects the chord. So, AM = MB

In right triangle OMA, OA² = OM² + AM²

\( \implies 7² = 6² + AM² \)

\( \implies 49 = 36 + AM² \)

\( \implies AM² = 13 \)

\( \implies AM = \sqrt{13} \)

Therefore, chord AB = 2 × AM = 2√13 cm. Hence, the length of the chord is 2√13 cm.
In simple words: A perpendicular line from the circle's centre to a chord cuts the chord into two equal parts. Using the Pythagorean theorem with the radius and perpendicular distance, we can find half the chord length, then double it to get the full chord length.

Exam Tip: Always use the property that a perpendicular from the centre bisects the chord, then apply the Pythagorean theorem on the right triangle formed.

Question 2. Explain why the following statement is true: If the perpendicular distance of a chord from the centre is d and the radius is r, then the chord length is 2√(r² - d²).
Answer: Let AB be a chord of a circle with centre O. Let OM be the perpendicular from O to AB. Given: Radius = r Perpendicular distance from centre to chord = d Since the perpendicular from the centre to a chord bisects the chord, M is the midpoint of AB. So, AM = MB = half of the chord length.

In right triangle OMA, By the Baudhāyana-Pythagoras theorem: OA² = OM² + AM²

\( \implies r² = d² + AM² \)

\( \implies AM² = r² - d² \)

\( \implies AM = \sqrt{r² - d²} \)

But chord length AB = 2 × AM. So, AB = 2√(r² - d²). Hence, the chord length is 2√(r² - d²).
In simple words: When you draw a line from the circle's centre straight down to any chord, it always splits that chord into two matching pieces. You can work out the chord's full length by using the right angle and the Pythagorean rule on the triangle it makes.

Exam Tip: This is a key formula derivation - examiners expect you to show the Pythagorean theorem step by step and clearly state the bisection property.

Question 3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, then can we conclude that CD = 2AB? Give reasons for your answer.
Answer: No, we cannot conclude that CD = 2AB. Reason: The length of a chord depends on the formula: Chord length = 2√(r² - d²). This relation is not directly proportional to the distance from the centre. Let the distance of chord CD from the centre be d. Then the distance of chord AB from the centre is 2d. So, AB = 2√(r² - (2d)²) = 2√(r² - 4d²) and CD = 2√(r² - d²). Clearly, CD is not equal to 2AB in general.

For example: Let radius r = 5 cm and d = 2 cm. Then distance of AB from centre = 4 cm.

Now, CD = 2√(5² - 2²) = 2√(25 - 4) = 2√21

AB = 2√(5² - 4²) = 2√(25 - 16) = 2√9 = 6

Now, 2AB = 12 but CD = 2√21 ≈ 9.17, which is not equal to 12. Therefore, the statement is false. Hence, we cannot conclude that CD = 2AB.
In simple words: Even though one chord is farther from the centre than another, the closer chord is not always twice as long. The distance and chord length do not have a simple doubling relationship.

Exam Tip: Always verify claims with concrete numerical examples before agreeing with a statement. This shows critical thinking and prevents wrong assumptions about proportional relationships.

Exercise Set 5.6

Question 1. In a circle with centre O, the central angle AOB is 60°. If the radius of the circle is 12 cm, what is the length of the chord AB?
Answer: Given: Radius OA = OB = 12 cm ∠AOB = 60°. In triangle AOB: OA = OB. So, ∠OAB = ∠OBA [Angles opposite to equal sides of triangle]. Let ∠OAB = ∠OBA = x.

In triangle OAB, ∠AOB + ∠OAB + ∠OBA = 180° [Angles sum property of the triangle]

\( \implies 60° + x + x = 180° \)

\( \implies 2x = 180° - 60° = 120° \)

\( \implies x = 120°/2 = 60° \)

\( \implies \angle OAB = \angle OBA = 60° \)

Thus, triangle AOB is equilateral. Therefore, AB = OA = 12 cm. Hence, the length of chord AB is 12 cm.
In simple words: When the angle at the centre is 60 degrees and the two radii are equal, the triangle turns out to be equilateral - all three sides and angles are the same.

Exam Tip: Recognize that when the central angle is 60° and the triangle is isosceles (two radii equal), you can use the angle sum property to prove all angles are 60°, making it equilateral.

Question 2. Let A and B be two points on a circle with centre O.
(i) Are there points X, Y on the circle, on the same side of AB, such that ∠AXB is different from ∠AYB?
Answer: No. Reason: If X and Y lie on the same side of chord AB, then they lie on the same arc AB. Angles subtended by the same chord in the same segment of a circle are equal. Therefore, ∠AXB = ∠AYB. So, there are no such points X and Y on the same side of AB for which the angles are different.
In simple words: All points on the same arc of a circle subtend the same angle at the chord. You cannot find two different angles on the same side.

Exam Tip: Remember that angles in the same segment are always equal - this is a fundamental circle theorem that rules out different angles on the same side.

Question 2(ii). Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
Answer: No, this is not always true. Reason: Equal angles can also be subtended by the same chord AB at points on opposite arcs. So, even if ∠AXB = ∠AYB, X and Y need not lie on the same side of AB. Therefore, the statement is false.
In simple words: Two points can subtend equal angles to a chord even when they are on opposite sides of the circle - one on each arc.

Exam Tip: Angles in the same segment are equal, but equal angles do not guarantee that points lie in the same segment. Opposite arcs can subtend equal angles too.

Question 2(iii). If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
Answer: Yes. Reason: If ∠AXB = ∠AYB, then the line segment AB subtends equal angles at X and Y. By the converse theorem of concyclicity: If a line segment subtends equal angles at two points on the same side, then the four points are concyclic. Hence, A, B, X and Y lie on the same circle. Therefore, the circle through A, B and X also passes through Y.
In simple words: If two points make the same angle with a line segment, then all four points (the two endpoints of the segment and the two points) must lie on a single circle.

Exam Tip: The converse of the circle theorem states that equal angles subtended at two points on the same side imply concyclicity - this is crucial for proving four points lie on a circle.

Question 3. Find x in Fig. 5.26.
Answer: In the figure, A, D, C and B are points on the same circle. So, quadrilateral ADCB is a cyclic quadrilateral. Given: ∠ADC = 100°. In a cyclic quadrilateral, opposite angles are supplementary. Therefore, ∠ABC + ∠ADC = 180°

\( \implies x + 100° = 180° \)

\( \implies x = 80° \)

Hence, x = 80°.
In simple words: In any quadrilateral drawn inside a circle, the two angles that sit across from each other always add up to 180 degrees.

Exam Tip: The supplementary property of opposite angles in a cyclic quadrilateral is the quickest way to find unknown angles - always add them to 180°.

End-of-Chapter Exercises

Question 1. In a circle, a chord is 5 cm away from the centre. If the radius of the circle is 13 cm, what is the length of the chord?
Answer: Let AB be the chord and O be the centre. Let OM be the perpendicular from O to AB. Given: OM = 5 cm, OA = 13 cm. Since the perpendicular from the centre to a chord bisects the chord, AM = MB.

In right triangle OMA, OA² = OM² + AM²

\( \implies 13² = 5² + AM² \)

\( \implies 169 = 25 + AM² \)

\( \implies AM² = 144 \)

\( \implies AM = 12 \) cm

Therefore, AB = 2 × AM = 2 × 12 = 24 cm. Hence, the length of the chord is 24 cm.
In simple words: Use the Pythagorean theorem on the right triangle formed by the radius, perpendicular distance, and half-chord. Then double the half-chord to get the full length.

Exam Tip: Set up the Pythagorean equation carefully: radius² = perpendicular distance² + (half chord)². Solve for the half-chord, then multiply by 2.

Question 2. An arc of a circle subtends an angle of 70° at the centre. What is the measure of the angle subtended by the arc at a point on the circle?
Answer: We know: Angle subtended by an arc at the centre = 2 × angle subtended by the same arc at a point on the circle. Given: Angle at the centre = 70°. So, angle at the circle = 70°/2 = 35°. Hence, the required angle is 35°.
In simple words: The angle at the centre is always twice the angle at any point on the circle, so divide the centre angle by 2 to get the angle at the circle.

Exam Tip: This is the angle-at-centre theorem - always remember that the central angle is exactly double the inscribed angle subtending the same arc.

Question 3. The diameter of a circle is 26 cm. A chord of length 24 cm is drawn in the circle. Find the distance from the centre of the circle to the chord.
Answer: Diameter = 26 cm. So, radius = 13 cm. Chord AB length = 24 cm. Half of chord AM = 12 cm. Let OM be the perpendicular distance from the centre O to the chord AB. Then M is the midpoint of the chord.

In right triangle OMA, OA² = OM² + AM²

\( \implies 13² = OM² + 12² \)

\( \implies 169 = OM² + 144 \)

\( \implies OM² = 25 \)

\( \implies OM = 5 \) cm

Hence, the distance from the centre to the chord is 5 cm.
In simple words: Rearrange the Pythagorean theorem to solve for the perpendicular distance instead of the half-chord - subtract the half-chord squared from the radius squared, then take the square root.

Exam Tip: Identify which measurement you need to find (distance, chord, or radius), then rearrange the Pythagorean relation accordingly before solving.

Question 4. A circle has a radius of 15 cm. A chord is drawn. The distance from the centre of the circle to the chord is 9 cm. What is the length of the chord?
Answer: Let AB be the chord and O be the centre. Let OM be the perpendicular from O to AB. Given: OA = 15 cm, OM = 9 cm. Since the perpendicular from the centre to a chord bisects the chord, AM = MB.

In right triangle OMA, OA² = OM² + AM²

\( \implies 15² = 9² + AM² \)

\( \implies 225 = 81 + AM² \)

\( \implies AM² = 144 \)

\( \implies AM = 12 \) cm

Therefore, AB = 2 × AM = 24 cm. Hence, the length of the chord is 24 cm.
In simple words: Once you know the radius and the perpendicular distance, use the Pythagorean theorem to find half the chord. Then multiply by 2 to get the full chord length.

Exam Tip: This is a direct application of the Pythagorean theorem - always solve for the missing side by rearranging the formula before substituting numbers.

Question 5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Answer: Let AB be a chord of a circle with centre O. Let M be the midpoint of AB. We have to show that OM is perpendicular to AB, which means the perpendicular bisector of AB passes through O.

So, in triangles OMA and OMB:
OA = OB [Radii of the same circle]
AM = MB [M is the midpoint of AB]
OM = OM [Common]

Therefore, by SSS congruence △OMA ≅ △OMB. Hence, ∠OMA = ∠OMB [CPCT]. But these two angles form a linear pair, so ∠OMA + ∠OMB = 180°. Since they are equal, ∠OMA = ∠OMB = 90°. Therefore, OM ⊥ AB. So, the line through O and M is the perpendicular bisector of chord AB. Hence proved.
In simple words: The two triangles formed by the centre, the chord, and the perpendicular are identical (congruent), so the perpendicular angle must be 90 degrees.

Exam Tip: Use SSS congruence to prove the triangles are identical, then CPCT to show the angles are equal, finally use the linear pair property to conclude the angle is 90°.

Question 6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of ∠ACB? Explain your reasoning.
Answer: AB is a diameter. The angle subtended by a diameter at any point on the circle is 90°. Therefore, ∠ACB = 90°. Hence, the measure of ∠ACB is 90°.
In simple words: When a chord passes through the centre of a circle (that is, it is a diameter), any point on the circle forms a right angle with the two ends of that diameter.

Exam Tip: This is Thales' theorem - a diameter always subtends a right angle at any point on the circle. It is one of the most important and frequently tested circle theorems.

Question 7. ABCD is a cyclic quadrilateral inscribed in a circle. If ∠A measures 75°, what is the measure of ∠C? If ∠B measures 110°, what is the measure of ∠D?
Answer: In a cyclic quadrilateral, opposite angles are supplementary. So, ∠A + ∠C = 180°

\( \implies 75° + \angle C = 180° \)

\( \implies \angle C = 105° \)

Also, ∠B + ∠D = 180°

\( \implies 110° + \angle D = 180° \)

\( \implies \angle D = 70° \)

Hence, ∠C = 105° and ∠D = 70°.
In simple words: Opposite angles in a cyclic quadrilateral always add up to 180 degrees, so subtract the given angle from 180 to find its opposite.

Exam Tip: The supplementary property is the foundation for all cyclic quadrilateral problems - memorize it and apply it directly without extra steps.

Question 8. Quadrilateral PQRS is inscribed in a circle. If ∠P = (2x + 10)° and ∠R = (3x - 20)°, find the value of x and the measures of ∠P and ∠R.
Answer: Since PQRS is a cyclic quadrilateral, opposite angles are supplementary. So, ∠P + ∠R = 180°

\( \implies (2x + 10) + (3x - 20) = 180 \)

\( \implies 5x - 10 = 180 \)

\( \implies 5x = 190 \)

\( \implies x = 38 \)

Now, ∠P = 2x + 10 = 2(38) + 10 = 86°

∠R = 3x - 20 = 3(38) - 20 = 94°

Hence, x = 38, ∠P = 86° and ∠R = 94°.
In simple words: Set the sum of opposite angles equal to 180 degrees, solve the equation for x, then substitute back to find each angle.

Exam Tip: Always expand the brackets and combine like terms carefully when working with algebraic angle expressions in cyclic quadrilaterals.

Question 9. The distance of a chord of length 16 cm from the centre of a circle is 6 cm. Find the radius of the circle.
Answer: Chord length AB = 16 cm. Half chord AM = MB = 8 cm. Distance from centre to chord OM = 6 cm. Let r be the radius.

In the right triangle OAM, OA² = AM² + OM²

\( \implies r² = 8² + 6² \)

\( \implies r² = 64 + 36 \)

\( \implies r² = 100 \)

\( \implies r = 10 \) cm

Hence, the radius of the circle is 10 cm.
In simple words: Use the Pythagorean theorem with the perpendicular distance and half the chord as the two shorter sides, then solve for the radius (the hypotenuse).

Exam Tip: In problems where you need to find the radius, the radius becomes the hypotenuse of the right triangle - apply the Pythagorean theorem accordingly.

Question 10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Answer: Since the cyclic quadrilateral has sides 5, 5, 12, 12, its semiperimeter is s = (5 + 5 + 12 + 12)/2 = 34/2 = 17. For a cyclic quadrilateral, area is given by Brahmagupta's formula:

Area = √[(s - a)(s - b)(s - c)(s - d)]

Substituting the values, we have:

Area = √[(17 - 5)(17 - 5)(17 - 12)(17 - 12)]

\( = \sqrt{(12)(12)(5)(5)} \)

\( = \sqrt{3600} \)

\( = 60 \)

Hence, the area of the cyclic quadrilateral is 60 square units.
In simple words: For any quadrilateral drawn inside a circle, multiply together (half the perimeter minus each side), then take the square root - this is Brahmagupta's formula.

Exam Tip: Always calculate the semiperimeter first, then substitute into Brahmagupta's formula - watch for perfect squares under the radical sign, as they simplify nicely.

Question 11. Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside the quadrilateral or outside? What is the best way of finding out?
Answer: In a cyclic quadrilateral, we know that opposite angles are supplementary (their sum is 180°). To find whether the centre of the circumcircle (circumcentre) lies inside or outside the quadrilateral, we can use the angle-at-centre theorem. If all the interior angles of the quadrilateral are less than 90°, then the circumcentre lies inside the quadrilateral. If any one of the interior angles is greater than 90°, then the circumcentre lies outside the quadrilateral. The best way is to examine the interior angles of the quadrilateral - if any angle exceeds 90°, the circumcentre is outside; otherwise, it is inside.
In simple words: Check each angle of the quadrilateral. If all are smaller than 90 degrees, the centre is inside. If any angle is bigger than 90 degrees, the centre is outside.

Exam Tip: Recognize that obtuse angles (greater than 90°) in a cyclic quadrilateral force the circumcentre to lie outside, while acute angles (less than 90°) allow it to lie inside - this is a quick visual test.

Question 12. When two chords intersect, each of them is divided into two line segments. Show that if the intersecting chords are of equal length, then the line segments of one chord are equal to the corresponding line segments of the other chord.
Answer: Let chords AB and CD intersect at point P inside the circle, with O as the centre. From the figure, ON is perpendicular to CD and OM is perpendicular to AB. Since the chords are equal in length, their distances from the centre are also equal. So, OM = ON.
In triangles OPM and OPN, we have: OP = OP (common side), OM = ON (proved above), and ∠OMP = ∠ONP = 90° (perpendiculars). By RHS congruency, triangle OPM is congruent to triangle OPN. Therefore, PM = PN (CPCT).
Since OM is perpendicular to AB, it bisects the chord. So, AM = MB. Similarly, since ON is perpendicular to CD, we have CN = ND. Given that AB = CD, we get BM = DN.
Adding the results PM = PN and BM = DN, we obtain PM + BM = PN + DN, which gives us PB = PD. Subtracting this from AB = CD, we get AP = CP. Hence proved.
In simple words: When two equal chords cross each other inside a circle, the line from the centre to each chord cuts it in half. This means the pieces on one side of the crossing point match the pieces on the other chord.

Exam Tip: Use RHS congruency on the two right triangles formed by the perpendiculars from the centre - this is the key step that equates the segments.

Question 13. Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
Answer: Given: Chord length = 6 cm, Distance from centre to chord = 3 cm.
Steps of Construction:
1. Draw a line segment AB = 6 cm.
2. Find the midpoint M of AB.
3. Draw a perpendicular line to AB at M.
4. On this perpendicular, mark a point O such that OM = 3 cm.
5. Join OA or OB.
6. With centre O and radius OA, draw a circle.
Since OM is perpendicular to AB and passes through its midpoint, AB becomes a chord of the circle. We have AM = 3 cm and OM = 3 cm. Using the Pythagorean theorem: OA² = OM² + AM² = 3² + 3² = 18. Therefore, OA = 3√2 cm. The required circle has centre O and radius 3√2 cm.
In simple words: Draw the chord first, then find its middle point. From that middle point, go straight out for 3 cm to locate the centre. The distance from the centre to the end of the chord is the radius.

Exam Tip: Always mark the distance OM = 3 cm accurately on the perpendicular bisector, and verify your radius using the Pythagorean theorem before drawing.

Question 14. Show that rectangle is the only parallelogram that can be inscribed in a circle.
Answer: Let ABCD be a parallelogram inscribed in a circle. Since it is a cyclic quadrilateral, the sum of opposite angles equals 180 degrees. So, ∠A + ∠C = 180°. However, in a parallelogram, opposite angles are equal, meaning ∠A = ∠C. Substituting this into the cyclic quadrilateral property: ∠A + ∠A = 180°, which gives ∠A = 90°. By the same logic, all angles of the parallelogram must be 90 degrees. A parallelogram with all angles equal to 90 degrees is, by definition, a rectangle. Therefore, a rectangle is the only parallelogram that can be inscribed in a circle. Hence proved.
In simple words: A four-sided shape with opposite sides equal (a parallelogram) can only fit inside a circle if all its corners are right angles, which makes it a rectangle.

Exam Tip: The key insight is combining the cyclic property (opposite angles sum to 180°) with the parallelogram property (opposite angles are equal) - this forces all angles to be 90°.

Question 15. Show that if a rectangle is inscribed in a circle, then the point of intersection of its diagonals must lie at the centre of the circle.
Answer: Let ABCD be a rectangle inscribed in a circle. Let the diagonals AC and BD meet at point O. In a rectangle, the diagonals are equal in length and bisect each other. Therefore, OA = OC and OB = OD. Also, AC = BD. From these two facts, we have OA = OB = OC = OD. This means point O is at the same distance from all four vertices A, B, C, and D. The point inside a circle that is equidistant from all points on the circle is, by definition, the centre of the circle. Therefore, the intersection point of the diagonals of a rectangle is the centre of the circle. Hence proved.
In simple words: In a rectangle, the diagonals cut each other exactly in the middle and are the same length. This middle point ends up being equidistant from all four corners, which is what the centre of a circle must be.

Exam Tip: Show that OA = OB = OC = OD clearly - this single equality is sufficient to prove the point O is the circle's centre.

Question 16. Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Answer: Let the circle have radius r and let each chord have a fixed length x. The perpendicular from the centre to a chord bisects that chord. So for every chord, half the chord length is x/2. Let d represent the distance of the midpoint from the centre. By the Pythagorean theorem: r² = d² + (x/2)². Rearranging, d² = r² - (x/2)². Since both r and x are fixed values, d is also fixed. This means every midpoint sits at the same distance d from the centre. The set of all points at a fixed distance from a given point forms a circle. Therefore, the midpoints of all chords of fixed length form a circle with the same centre as the original circle and with radius equal to d.
In simple words: No matter which way you draw a chord of the same length, its middle point always stays the same distance away from the circle's centre. All these middle points trace out a smaller circle.

Exam Tip: The key is recognizing that d is constant - once you establish this, the result follows directly from the definition of a circle.

Question 17. In a circle with centre O, chords AB and AC are congruent. Explain why this statement is true: "The centre of the circle lies on the angle bisector of ∠BAC".
Answer: Given that AB = AC. Join OA, OB, and OC. In triangles AOB and AOC, we have: OB = OC (both are radii of the same circle), OA = OA (common side), and AB = AC (given). By SSS congruence, triangle AOB is congruent to triangle AOC. Therefore, ∠BAO = ∠OAC. Since these two angles are equal, the line segment AO bisects the angle ∠BAC. Thus, the centre O lies on the angle bisector of ∠BAC. Hence proved.
In simple words: When two chords starting from the same point are equal in length, they make equal angles with the line joining that point to the centre. This line is the angle bisector.

Exam Tip: Use SSS congruence to prove the triangles are equal - this directly gives the angle equality you need.

Question 18. Two parallel chords of lengths 10 cm and 24 cm are on the same side of the centre of a circle. The distance between the chords is 7 cm. Find the radius of the circle.
Answer: Let the radius be r. For chord AB of length 10 cm: half of AB is 5 cm. Let OM = d₁ be its distance from the centre. For chord CD of length 24 cm: half of CD is 12 cm. Let ON = d₂ be its distance from the centre. Since the longer chord is closer to the centre: d₁ - d₂ = 7 ... (1)
Using the Pythagorean theorem in triangle OAM: r² = d₁² + 5² = d₁² + 25
Using the Pythagorean theorem in triangle OCN: r² = d₂² + 12² = d₂² + 144
Equating these: d₁² + 25 = d₂² + 144, which gives d₁² - d₂² = 119. Factoring: (d₁ - d₂)(d₁ + d₂) = 119. Substituting d₁ - d₂ = 7: 7(d₁ + d₂) = 119, so d₁ + d₂ = 17 ... (2)
From equations (1) and (2): 2d₁ = 24, so d₁ = 12 and d₂ = 5.
Therefore, r² = 12² + 5² = 144 + 25 = 169, giving r = 13 cm. The radius of the circle is 13 cm.
In simple words: Use the Pythagorean theorem on both chords to write two equations with the radius. Then solve the system using the fact that the chords are 7 cm apart.

Exam Tip: Set up the two Pythagorean equations carefully, then use the distance condition to create a second equation in terms of d₁ and d₂. Solve the system by elimination.

Question 19. A regular hexagon is inscribed in a circle of radius r. Find the length of the sides of the hexagon and the distance of each side from the centre of the circle.
Answer: A regular hexagon divides the circle into 6 equal central angles. Each central angle is 360°/6 = 60°. Let AB be one side of the hexagon and O be the centre. Then OA = OB = r and ∠AOB = 60°. Since two sides of triangle AOB are equal (OA = OB) and the included angle is 60°, triangle AOB is equilateral. Therefore, AB = r. So each side of the regular hexagon has length r.
To find the distance of each side from the centre, let OM be perpendicular to AB. Since triangle AOB is equilateral, AM = AB/2 = r/2. In right triangle OMA: OA² = OM² + AM² gives r² = OM² + (r/2)². Solving for OM: OM² = r² - r²/4 = 3r²/4, so OM = (√3/2)r.
Therefore, the side length of the hexagon is r and the distance of each side from the centre is (√3/2)r.
In simple words: A regular hexagon inside a circle creates 6 equilateral triangles, each with all sides equal to the radius. The distance from the centre to each side is half the radius times the square root of 3.

Exam Tip: Recognize that the central angle of 60° combined with equal radii creates an equilateral triangle - this is the fastest way to find the side length.

Question 20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about ∠MOP and ∠MNP? Explain your reasoning.
Answer: Since MNOP is inscribed in a circle, it is a cyclic quadrilateral. In a cyclic quadrilateral, the sum of opposite angles is always 180 degrees. Therefore, ∠MOP + ∠MNP = 180°. This means the two angles are supplementary.
In simple words: When a four-sided shape sits inside a circle, the angles at opposite corners always add up to 180 degrees.

Exam Tip: For cyclic quadrilaterals, always remember that opposite angles are supplementary - this is the defining property.

Question 21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., ∠CDE = ∠ABC, where E is a point on the extension of side CD).
Answer: In cyclic quadrilateral ABCD, the sum of opposite angles is 180 degrees. So ∠ABC + ∠ADC = 180° ... (1). Since E lies on the extension of CD, angles ∠ADC and ∠CDE form a linear pair and are supplementary. So ∠ADC + ∠CDE = 180° ... (2). From equations (1) and (2), we can write: ∠ABC + ∠ADC = ∠ADC + ∠CDE. Subtracting ∠ADC from both sides gives ∠ABC = ∠CDE. Therefore, the exterior angle at any vertex of a cyclic quadrilateral equals the interior angle at the opposite vertex. Hence proved.
In simple words: When you extend a side of a four-sided shape inside a circle, the new angle created outside equals the opposite angle inside the shape.

Exam Tip: Set up two equations using the cyclic quadrilateral property and the linear pair property, then subtract to isolate the angle relationship.

Question 22. There is no chord of a circle that is longer than its diameter. How do you justify this statement?
Answer: A chord is defined as a line segment connecting any two points on a circle. The longest possible chord is the one that passes through the centre of the circle - this is the diameter. For any other chord that does not pass through the centre, the perpendicular distance from the centre to that chord is greater than zero. If the radius is r and the perpendicular distance from the centre to a chord is d, then the chord length is given by: Chord length = 2√(r² - d²). Since d > 0 for any chord not passing through the centre, we have r² - d² < r². Taking square roots: √(r² - d²) < r. Multiplying by 2: 2√(r² - d²) < 2r. But 2r is the diameter. Therefore, every chord other than the diameter is shorter than the diameter. Hence, no chord of a circle is longer than its diameter. Hence proved.
In simple words: The farther a chord is from the centre, the shorter it becomes. The closest a chord can be to the centre is at the centre itself, making it a diameter, which is the longest chord possible.

Exam Tip: Use the formula for chord length in terms of radius and distance from centre - this provides the algebraic proof needed.

Question 23. Let A be any point within a given circle with centre O. Show that the shortest chord of the circle that passes through point A is the one that is perpendicular to OA.
Answer: Let the circle have centre O and let A be a point inside it. Any chord passing through A will have some perpendicular distance from the centre O. From our knowledge of chords, we know that the farther a chord is from the centre, the shorter it becomes. Therefore, among all chords passing through A, the shortest one will be the chord whose distance from O is the largest. Now consider all possible lines passing through point A. The perpendicular distance from O to any line is maximized when that line is perpendicular to OA. This is because the perpendicular from a point to a line gives the shortest distance, and when the line itself is perpendicular to OA, this distance equals the length OA, which is the maximum possible. Therefore, the chord perpendicular to OA has the maximum distance from O among all chords through A. Consequently, this chord is the shortest chord passing through A. Hence proved.
In simple words: To make a chord as short as possible while passing through point A, draw it so it stands at a right angle to the line from A to the centre. This positions the chord as far from the centre as it can be.

Exam Tip: The relationship between chord distance from centre and chord length is crucial - use it to argue why maximum distance gives minimum chord length.

Question 24. How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Answer: Let BC be the diameter of the circle and O be the centre. Point A lies on the semicircle. Join OA. Since O is the centre, OA = OB = OC because all three are radii of the same circle. Triangle AOB is isosceles with OA = OB, so ∠ABO = ∠BAO. Let ∠ABO = a. Then ∠BAO = a as well. Similarly, triangle AOC is isosceles with OA = OC, so ∠ACO = ∠CAO. Let ∠ACO = b. Then ∠CAO = b. Therefore, ∠BAC = ∠BAO + ∠CAO = a + b. Now consider triangle ABC. The sum of its angles is 180 degrees: ∠ABC + ∠BAC + ∠ACB = 180°. Substituting ∠ABC = a, ∠BAC = a + b, and ∠ACB = b: a + (a + b) + b = 180°, which simplifies to 2a + 2b = 180°, or a + b = 90°. Since ∠BAC = a + b, we have ∠BAC = 90°. Therefore, the angle in a semicircle is 90°. Hence proved.
In simple words: When a triangle is drawn inside a circle with one side as the diameter, the angle opposite that diameter is always a right angle.

Exam Tip: Use the isosceles triangle property twice - once for each triangle formed by the two radii - to establish the angle relationships needed.

Question 25. In a circle, two chords CC' and DD' are drawn perpendicular to a diameter AB. Prove that the segment MM' joining the midpoints of the chords CD and C'D' is perpendicular to AB.
Answer: Let AB be the diameter of the circle. Since chords CC' and DD' are perpendicular to AB, the points C and C' are symmetric with respect to the diameter AB. Similarly, D and D' are also symmetric with respect to AB. Let M be the midpoint of chord CD and M' be the midpoint of chord C'D'. Since C corresponds to C' under reflection across AB, and D corresponds to D' under the same reflection, the midpoint of CD must correspond to the midpoint of C'D'. In other words, M and M' are symmetric with respect to the diameter AB. A fundamental property of reflection is that the line segment joining two points that are symmetric about a line is perpendicular to that line. Therefore, MM' is perpendicular to AB. Hence proved.
In simple words: When a diameter acts as a mirror, it reflects the chords symmetrically. The line joining the midpoints of reflected chords must cross the mirror at a right angle.

Exam Tip: Use the symmetry property created by reflection in the diameter - this is the cleanest approach without needing coordinate calculations.

Question 26. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is 180°?
Answer: Let ABCD be a cyclic quadrilateral with centre O. Join OA, OB, OC, and OD. Since all these segments are radii of the same circle, OA = OB = OC = OD. The triangles formed with these radii are therefore isosceles triangles. Let ∠OAB = p and ∠OBA = q. Let ∠ODC = v and ∠OCD = u. Using the angle-sum property in the triangles formed around the centre, the complete angle around O is 360 degrees. A key property relating central angles to inscribed angles is that the angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circumference. Applying this principle to arcs and the angles they subtend: ∠A + ∠C = 180°. Similarly, ∠B + ∠D = 180°. Therefore, the sum of opposite angles of a cyclic quadrilateral is 180 degrees. Hence proved.
In simple words: In a four-sided shape drawn inside a circle, the angles at opposite corners always add to 180 degrees because of the symmetry created by the radii.

Exam Tip: Use the angle at centre - angle at circumference relationship to connect the central angles and angles at vertices of the quadrilateral.

Quick Reference - All 12 Theorems in Class 9 Ganita Manjari Chapter 5

Important Theorems Explained Simply

• Theorem 9 (Central Angle - Double Inscribed Angle): This is the most important result in this chapter. It tells us that if you pick any arc of a circle, the angle it creates at the centre is exactly double the angle it creates at any point on the remaining circle. The proof depends on the exterior angle theorem and the isosceles triangle property (since all radii are equal). The famous corollary - that any angle in a semicircle is 90 degrees - follows directly from this. This is because a diameter subtends a 180 degree angle at the centre, making the inscribed angle exactly 90 degrees.
• Theorem 11 (Cyclic Quadrilateral): This states that opposite angles of any cyclic quadrilateral always add up to 180 degrees. The proof applies Theorem 9 twice - once for each arc - and uses the fact that together both arcs make a complete 360 degree rotation at the centre. Their half-angles therefore sum to 180 degrees.

Theorem 1 (Unique Circumcircle)

Three non-collinear points always determine exactly one circle. The reason is elegant: the centre must be the same distance from all three points, which means it must lie on the perpendicular bisector of each pair. Since two non-parallel lines meet at exactly one point, this gives us a unique centre.

Key Definitions in Ganita Manjari Chapter 5 - Circles

• Circle: All points on a flat surface that are the same distance from a fixed point. That fixed point is called the centre, and the fixed distance is called the radius.
• Locus: The set of all points that fit a certain rule or condition. A circle is an example of a locus - it is made up of all points that are equally far from a fixed centre.
• Chord: A line segment that has both of its endpoints on the circle.
• Diameter: A chord that goes through the centre of the circle. It is the longest chord you can draw and it equals twice the radius.
• Arc: A piece of the circle, starting at one point and ending at another. The bigger piece is called the major arc and the smaller piece is called the minor arc.
• Circumcircle: The single circle that passes through all three corners (vertices) of a triangle. Its centre is called the circumcentre and is found where the perpendicular bisectors of the triangle's three sides cross.
• Concyclic Points: Points that all sit on the same circle.
• Cyclic Quadrilateral: A four-sided shape whose all four corners lie on a single circle. The opposite angles of a cyclic quadrilateral always add up to 180 degrees.

Chapter 5 Ganita Manjari - Exercise-Wise Overview

• Exercise Set 5.1: Drawing circumcircles for triangles with given measurements; checking if the circumcentre falls inside or outside the triangle; finding the smallest possible radius that passes through two given points.
• Exercise Set 5.2: Proving that a triangle made by a chord and the centre has two equal sides (isosceles); proving two such triangles with the same base length are congruent (exactly the same).
• Exercise Set 5.3: Proving the reverse of Theorem 4 (a line from the centre that is perpendicular to a chord cuts the chord in half); altitude of an isosceles triangle inscribed in the circle that passes through the centre; the distance between the middle points of two parallel chords.
• Exercise Set 5.4: Using the Baudhayana-Pythagoras theorem to prove Theorem 6; showing AB equals GF when chords are the same distance from the centre.
• Exercise Set 5.5: Finding how long a chord is when you know the radius and how far it is from the centre; deriving the general formula for chord length: 2√(r² - d²).
• Exercise Set 5.6: Finding chord length using the angle at the centre and the radius; angles in the same part of the circle; finding unknown angles in drawings with circles.

End-of-Chapter Exercises

There are 26 questions that cover all the chapter theorems, angle problems with cyclic quadrilaterals, calculations involving chords and distances, and challenging starred (*) proofs about rectangles drawn inside circles, chords that cross each other, and the shortest chord you can draw through a point inside a circle.

Starred (*) Advanced Questions - For High Achievers

Chapter 5 has several starred questions in the end-of-chapter exercises that go well beyond what is normally expected in exams:

• Q12: When two chords of the same length cross each other, prove that the matching pieces on each side are also equal in length.
• Q14: Prove that a rectangle is the only type of parallelogram that can be drawn inside a circle.
• Q15: Prove that if a rectangle is drawn inside a circle, its two diagonal lines meet at the centre.
• Q16: Find out what shape is formed by connecting the middle points of all chords that have the same fixed length in a given circle.
• Q19: A regular hexagon sits inside a circle that has radius r - find the length of each side and how far each side is from the centre.
• Q23: Show that if you draw a chord through any point inside a circle, the shortest chord is the one that stands at a right angle to the line joining that point to the centre.
• Q25: Two chords are drawn perpendicular to a diameter - prove that the line joining the middle points of two related chords is also perpendicular to that diameter.

These questions are very helpful for preparing for NTSE (National Talent Search Exam), Mathematical Olympiad, and Class 10 advanced level studies.

Frequently Asked Questions (FAQs) - Class 9 Maths Ganita Manjari Chapter 5

Is Class 9 Maths Ganita Manjari Chapter 5 easy or difficult?

The chapter falls in the middle-to-hard range. The beginning sections that cover definitions, symmetry, and basic facts about chords are pretty easy to understand. Things get harder from Section 5.7 onwards, where Theorem 9 about angles in arcs needs you to think through multiple steps carefully. The theorems about cyclic quadrilaterals (four-sided shapes inside circles) in Section 5.8 need strong thinking skills but become less tricky once you fully grasp Theorem 9. If you already understand how to prove triangles are congruent (the same shape and size) from earlier chapters, you will find the way this chapter is set up much simpler to follow.