Ganita Manjari Class 9 Chapter 5 Solutions I’m Up and Down and Round and Round

By using Ganita Manjari Class 9 Solutions and Part 1 Class 9 Maths Chapter 5 I’m Up and Down and Round and Round NCERT Solutions, students can improve their problem-solving skills.

I’m Up and Down and Round and Round Class 9 Solutions

Class 9 Ganita Manjari Chapter 5 Solutions

Class 9 Maths Ganita Manjari Chapter 5 Solutions I’m Up and Down and Round and Round

Think and Reflect (NCERT Textbook Page No. 93)

Question 1.
Jamuna has a circular piece of paper. She is trying to locate its centre. Amina gives her a suggestion. She follows the instructions and is thrilled to find that it works. Can you guess what Amina told her?
Solution:
Amina likely suggested a very simple but clever geometric trick:
She told Jamuna to fold the circular paper in half.
The crease formed will always pass through the centre of the circle.
If Jamuna folds it again in another direction, the intersection of the two creases marks the exact centre of the circle.

Think and Reflect (NCERT Textbook Page No. 94)

Question 1.
What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?
Solution:
Square
Rotational Symmetry: A square can be rotated by 90°, 180°, 270°, and 360° and still look the same.
So it has 4 rotational symmetries (including the full 360° rotation).

Reflection Symmetry: A square has 4 lines of symmetry:
Two through opposite sides (vertical and horizontal).
Two through opposite vertices (diagonals).
So it has 4 reflection symmetries.

Regular Pentagon
Rotational symmetries: 5 (72°, 144°, 216°, 288°, 360°)
Reflection symmetries: 5 (each through a vertex and the midpoint of the opposite side).

Regular Hexagon
Rotational symmetries: 6 (60°, 120°, 180°, 240°, 300°, 360°)
Reflection symmetries: 6 (three through opposite vertices, three through midpoints of opposite sides).

Question 2.
What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?
Solution:
The longest chord in a circle is its diameter.
For radius r = 5, diameter = 2r = 10 units.
There is no smallest chord in a circle.

Question 3.
The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?
(Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)
Solution:
The locus of points equidistant from two points A and B is the perpendicular bisector of segment AB.

• Any point on the perpendicular bisector has equal distance to A and B (by congruent right triangles).
• Conversely, if a point is equidistant from A and B, it must lie on the perpendicular bisector.
• So yes, the perpendicular bisector is exactly the locus of points equidistant from two given points.

Think and Reflect (NCERT Textbook Page No. 95)

Question 1.
How many circles pass through two points on a plane?
Solution:
Infinitely many circles can pass through two given points A and B. Because the line segment AB can serve as a chord of infinitely many circles, each with a different centre lying on the perpendicular bisector of AB.

Question 2.
Are there circles of all possible radii passing through A and B? What is the radius of the smallest circle passing through A and B? What is the radius of the largest circle passing through A and B?
Solution:
No, there is a minimum required radius, i.e., a circle must be large enough to reach from point A to B. The circle will not touch both points if the radius is too small.
The smallest radius occurs when A and B are endpoints of a diameter.
In that case, radius = half the distance AB.
The largest radius has no bound — as the centre moves farther away along the perpendicular bisector, the radius increases indefinitely.

Question 3.
As you move away from segment AB along its perpendicular bisector, do the radii of the circles containing A and B increase or decrease?
Solution:
As we move away from AB along its perpendicular bisector, the radius of the circle containing A and B increases.
Near AB, the circle is tighter; farther away, it becomes larger.

Question 4.
As you go along the perpendicular bisector, will the circle drawn from that point through A and B appear more curved or less curved?
Solution:
A circle with a smaller radius appears more curved (sharper arc).
As the radius increases, the circle appears less curved, approaching a straight line in the limit.

Question 5.
You are given two points A and B on a plane. How many squares can you draw on the same plane with A and B on the boundary? How many squares can you draw on the plane with A and B as the corners of the square?
Solution:
A and B on the boundary (but not necessarily corners): Infinitely many squares can be drawn, because we can place A and B on different sides or edges in countless orientations.
A and B as corners of the square: Exactly two possible squares.

• One square with AB as one side.
• Another square with AB as the diagonal.

Ex 5.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 5.1 Solutions

Exercise 5.1 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 5.1 Solutions

Question 1.
Draw ΔABC with AB = 5 cm, ∠A = 76° and ∠B = 60°. Draw the circumcircle of ΔABC. Is the centre inside or outside the triangle?
Solution:
Construction Steps:
1. Draw ΔABC with AB = 5 cm. At A, make ∠A = 70° and at B, make ∠B = 60° and ∠C = 180° – 70° – 60° = 50°
2. Draw the perpendicular bisector of AB.
3. Draw the perpendicular bisectors of AC and BC. Let the point of intersection of the perpendicular bisectors be O.
4. With O as center and OA as radius, draw a circle. Then, this circle is the required circumcircle.
All angles of ΔABC are acute. The center is inside 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?
Solution:
Construction Steps:
1. Draw ΔABC with AB = 5 cm. At A, make ∠A = 100° and AC = 4 cm.
2. Draw the perpendicular bisector of BC.
3. Draw the perpendicular bisectors of CA and AB. Let the point of intersection of the perpendicular bisectors be O.
4. With O as center and OA as radius, draw a circle. Then, this circle is the required circumcircle.
The center is outside 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.
Solution:
Construction Steps:
1. Draw ΔABC with AB = 6 cm, BC = 7 cm, and AC = 7 cm.
2. Draw the perpendicular bisector of AB.
3. Draw the perpendicular bisectors of AC and BC.
Let the point of intersection of the perpendicular bisectors be O.
4. With O as center and OA as radius, draw a circle.
Then, this circle is the required circumcircle.
OA = OB = OC = 3.8 cm.

Question 4.
What is the least possible radius of a circle through two points A and B?
Solution:
Let M be the midpoint of AB.
AM = \(\frac {1}{2}\)AB
AM is the least possible radius.

Ex 5.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 5.2 Solutions

Exercise 5.2 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 5.2 Solutions

Question 1.
Show that the triangle formed by a chord and the centre of the circle is isosceles.
Solution:
Given: A circle with centre O and chord AB.
Join OA and OB.
To prove: ΔOAB is isosceles.
Consider ΔAOB.
OA = OB (radii of the circle)
Since two sides OA and OB of ΔOAB are equal, the triangle is isosceles.
Then ΔAOB is an isosceles triangle.

Question 2.
Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Solution:
Given: A circle with centre O and equal chords AB and CD.
To prove: ΔAOB ≅ ΔCDE
Join OA, OB, OC, and OD.

Then, OA = OC (radii of the circle)
OB = OD (radii of the circle)
AB = CD (given)
ΔAOB ≅ ΔCOD (By SSS)

Ex 5.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 5.3 Solutions

Exercise 5.3 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 5.3 Solutions

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?
(Hint: Use Theorem 4. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM.)
Solution:
Converse statement: The perpendicular to a chord from the centre of the circle bisects the chord.
Given: A circle with centre O and chord AB.
OP is a perpendicular to the chord from O.
To prove: AP = PB (P is the midpoint of AB).

OA = OB (each = r)
OP = OP (Common)
∠OPA = ∠OPB (each = 90°)
ΔOPA ≅ ΔOPB (By RHS)
AP = PB (By CPCT)
Hence, OP bisects AB.

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.
Solution:
Given: A circle with centre O.
ΔABC, with AB = AC.
To show: Altitude from A to BC passes through O.
Construction: Join AO and extend it to meet BC at D.
Draw OL ⊥ AB and OM ⊥ AC.

Proof: OL ⊥ AB (by construction)
⇒ AL = \(\frac {1}{2}\)AB
Similarly, AM = \(\frac {1}{2}\)AC
AB = AC (Given)
⇒ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)AC
⇒ AL = AM
∠ALO = ∠AMO (each = 90°)
AO = AO (Common)
Then, ΔALO ≅ ΔAMO
⇒ ∠1 = ∠2 (CPCTC)
In ΔABD and ΔACD
AB = AC (Given)
∠1 = ∠2 (Shown)
AD = AD (Common)
⇒ ΔABD ≅ ΔACD (SAS)
⇒ ∠ADB = ∠ADC (CPCTC)
∠ADB = ∠ADC = \(\frac {1}{2}\) × 180° = 90°
⇒ AD ⊥ BC
⇒ AD is an altitude of ΔABC, from A on BC.
O lies on AD.
Hence, O lies on altitude AD.

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.
Solution:
Given: Circle with centre 0 and radius 5 cm.
AB = 6 cm, CD = 8 cm, AB || CD.
AB and CD lie on either side of O.
M is the midpoint of AB.
N is the midpoint of CD.
MN is the required distance between the midpoints of the chords.

In ΔOMB, M is the midpoint of AB.
Then OM ⊥ AB.
In ΔOMB,
OM² + 3² = 5² (By Baudhayana Theorem)
⇒ OM² = 25 – 9 = 16 = 4²
⇒ OM = 4 cm
In ΔOND, N is the midpoint of CD.
Then ON ⊥ CD.
In ΔOND,
ON² + 4² = 5²
⇒ ON² + 16 = 25
⇒ ON² = 25 – 16 = 9 = 3²
⇒ ON = 3 cm
Hence, MN = OM + ON
= 4 cm + 3 cm
= 7 cm

Ex 5.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 5.4 Solutions

Exercise 5.4 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 5.4 Solutions

Question 1.
Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.
Solution:
Given: Circle with centre O. Equal chords AB and CD.
To Show: AB and CD are equidistant from the centre of the circle.
Construction: Draw OL ⊥ AB and OM ⊥ CD.
Join OB and OC.

Proof: AB = CD (Given)
⇒ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)CD
Let \(\frac {1}{2}\)AB = \(\frac {1}{2}\)CD = x
Also, OB = OC = r
In ΔOLB, OL² + x² = r² (By Baudhayana Theorem)
⇒ OL² = r² – x²
In ΔOMC, OM² + x² = r² (by Baudhayana Theorem)
⇒ OM² = r² – x²
Now OL² = OM² (each = r² – x²)
⇒ OL = OM
⇒ AB and CD are equidistant from the centre of the circle.
or Equal chords are equidistant from the centre of the circle.

Question 2.
Consider Fig. (below). If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.

Solution:
Given: Circle with centre C having chords AB and GF.
CE ⊥ AB; CH ⊥ GF and CE = CH
To show: AB = GF

Proof: CE = CH (Given)
∠1 = ∠2 (each = 90°)
CB = CG (each = r)
⇒ ΔCEB ≅ ΔCHG (RHS)
⇒ BE = GH (CPCTC)
But BE = \(\frac {1}{2}\)AB and GH = \(\frac {1}{2}\)GF
⇒ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)GF
⇒ AB = GF

Question 3.
Solve the previous question using the Baudhayana-Pythagoras theorem.

Solution:
Given: Circle with centre C.
Chords AB and GF.

CE ⊥ AB; CH ⊥ GF, CE = CH.
To Show: AB = GF
Proof: CE = CH = x (Suppose)
CB = CG = r (radii)
In ΔCEB,
x² + BE² = r² (Baudhayana theorem)
⇒ BE² = r² – x²
In ΔCHG,
x² + GH² = r² (Baudhayana theorem)
⇒ GH² = r² – x²
Now, BE² = GH² (each = r² – x²)
⇒ BE = GH
but BE = \(\frac {1}{2}\)AB and GH = \(\frac {1}{2}\)GF
⇒ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)GF
⇒ AB = GF

Ex 5.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 5.5 Solutions

Exercise 5.5 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 5.5 Solutions

Question 1.
Find the length of the chord of a circle where the radius is 7 cm and the perpendicular distance is 6 cm.
Solution:
Given: Circle with centre O.
Chord AB, r = 7 cm, OM ⊥ AB, OM = 6 cm

Required: l(AB).
In ΔOMB, ∠OMB = 90°
Then, OM² + MB² = OB² (by Baudhayana theorem)
⇒ 6² + MB² = 7²
⇒ MB² = 49 – 36 = 13
⇒ MB = √13 cm
∴ AB = 2 × MB = 2√13 cm

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 \(2 \sqrt{\left(r^2-d^2\right)}\).
Solution:
Given: Circle with centre O and radius r.
Chord AB, OM ⊥ AB; OM = d.

To prove: AB = \(2 \sqrt{\left(r^2-d^2\right)}\)
In ΔOMB, ∠OMB = 90°
Then, d² + MB² = r² (by Baudhayana theorem)
⇒ MB² = r² – d²
⇒ MB = \(\sqrt{r^2-d^2}\)
∴ AB = 2MB = \(2 \sqrt{\left(r^2-d^2\right)}\)
Hence proved.

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.
Solution:
Given: Circle with centre O.
Chords AB and CD, OL ⊥ AB, OM ⊥ CD.
OL = 2OM

Required: To check if CD = 2AB
Let OM = x. Then OL = 2x
In ΔOLB, ∠OLB = 90°
⇒ (2x)² + LB² = r²
⇒ LB² = r² – 4x²
⇒ LB = \(\sqrt{r^2-4 x^2}\)
∴ AB = 2LB = \(2 \sqrt{r^2-4 x^2}\)
and 2AB = \(4 \sqrt{r^2-4 x^2}\) ……(1)
In ΔOMD, ∠OMD = 90°
⇒ x² + MD² = r² (by Baudhayana theorem)
⇒ MD² = r² – x²
⇒ MD = \(\sqrt{r^2-x^2}\)
∴ CD = \(2 \sqrt{r^2-x^2}\) ……..(2)
From (1) and (2), CD ≠ 2AB.

Ex 5.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Exercise 5.6 Solutions

Exercise 5.6 Class 9 Ganita Manjari Solutions – Ganita Manjari Class 9 Ex 5.6 Solutions

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?
Solution:
Given: Circle with centre O, radius (r) = 12 cm and chord AB.

Required: l(AB)
In ΔAOB,
OA = OB (each = r)
⇒ ∠OAB = ∠OBA
⇒ ∠OAB = ∠OBA = x° (Suppose)
In ΔOAB,
x° + x° + 60° = 180° (Angle sum property of a Δ)
⇒ 2x° = 120°
⇒ x° = 60°
∴ ∠OAB = ∠OBA = 60°
Also ∠AOB = 60°
Then ΔOAB is an equilateral triangle.
⇒ OA = OB = AB
⇒ AB = 12 cm

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?
(ii) Is it true that if ∠AXB = ∠AYB, then X and Y lie on the same side of the circle?
(iii) If ∠AXB = ∠AYB, and X and Y do not lie on the circle, does the circle through A, B and X also pass through Y?
Solution:

(i) No.
∠AXB = ∠AYB
X and Y are on the same segment.
Angles on the same segment are equal.
(ii) No.
If AB is diameter, then x and y could lie on opposite sides of the circle.
(iii) Yes.
The four points will be concyclic.

Question 3.
Find x in Figure below.

Solution:
ABCD is a cyclic quadrilateral.
∴ ∠B + ∠D = 180°
⇒ x + 100° = 180°
⇒ x = 180° – 100°
⇒ x = 80°

Ganita Manjari Class 9 Maths Chapter 5 End of Chapter Exercise Solutions

I’m Up and Down and Round and Round End of Chapter Exercise Solutions

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?
Solution:
Here, OC = 5 cm, r = 13 cm, ∠OCB = 90°

Then OC² + BC² = OB² (by Baudhayana theorem)
⇒ 5² + BC² = 13²
⇒ BC² = 169 – 25 = 144 = 12²
⇒ BC = 12 cm
∴ AB = 2BC
= 2 × 12 cm
= 24 cm
Hence, length of the chord is 24 cm.

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?
Solution:
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc at any point on the circle outside the arc.

∴ ∠AOB = 2∠ACB
⇒ ∠ACB = \(\frac {1}{2}\)∠AOB
= \(\frac {1}{2}\) × 70°
= 35°
Hence, \(\overparen{A B}\) subtends an angle of 35° at C.

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.
Solution:
Here, diameter = 26 cm
∴ r = \(\frac {26}{2}\) = 13 cm
and AB = 24 cm (given)
OC ⊥ AB
∴ CB = \(\frac {24}{2}\) = 12 cm
In ∆OCB,
OC² + 12² = 13²
⇒ OC² = 169 – 144 = 25 = 5²
⇒ OC = 5

The chord is at a distance of 5 cm from the centre of the circle.

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?
Solution:
Here, r = 15 cm
AB is a chord; O is the centre of the circle.
OL ⊥ AB, OL = 9 cm (given)
∠OLB = 90°

Now, OL² + LB² = OB²
⇒ 9² + LB² = 15²
⇒ LB² = 225 – 81 = 144 = 12²
⇒ LB = 12
Now AB = 2 × LB
= 2 × 12 cm
= 24 cm
Hence, the length of the chord is 24 cm.

Question 5.
Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Solution:
Given: Circle with centre O, chord AB.
L is the mid-point of AB.
OL is joined.
To Show: The perpendicular bisector of AB passes through O.
Construction: Join OA and OB.

Proof: OA = OB (each = r)
AL = LB (L is the midpoint of AB)
OL = OL (Common)
⇒ ∆OLA ≅ ∆OLB (SSS)
⇒ ∠OLA = ∠OLB (CPCT)
Now, ∠OLA = ∠OLB = x° (Suppose)
Then, x° + x° = 180°
⇒ 2x° = 180°
⇒ x° = 90°
OL ⊥ AB
Also, L is the midpoint of AB (Given)
Hence, OL is ⊥ bisector of AB, or a perpendicular bisector of AB passes through O.

Question 6.
The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
Solution:
AB is a diameter.
⇒ \(\overparen{A C B}\) is a semi-circle.
⇒ ∠ACB = 90° (Angle in a semi-circle is a right angle)

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?
Solution:
ABCD is a cyclic quadrilateral.

∠A = 75°
Now ∠A + ∠C = 180°
⇒ 75° + ∠C = 180°
⇒ ∠C = 180° – 75°
⇒ ∠C = 105°
and ∠B = 110°
∴ ∠B + ∠D = 180°
⇒ 110° + ∠D = 180°
⇒ ∠D = 180° – 110°
⇒ ∠D = 70°

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.
Solution:
PQRS is a cyclic quadrilateral.
Given: ∠P = (2x + 10)° …..(1)
∠R = (3x – 20)° …..(2)

From Theorem 11
⇒ ∠P + ∠R = 180°
⇒ (2x + 10)° + (3x – 20)° = 180°
⇒ 5x – 10° = 180°
⇒ 5x = 190°
⇒ x = 38°
Put the value of x in eqn. (1) and (2) respectively.
∠P = (2 × 38 + 10)° = 86°
∠B = (3 × 38 – 20)° = 94°

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.
Solution:
Given: AB = 16 cm
OC ⊥ AB
OC = 6 cm
Now, CB = \(\frac {1}{2}\) × AB
= \(\frac {1}{2}\) × 16 cm
= 8 cm

In ∆OCB, ∠OCB = 90°
⇒ OC² + CB² = OB²
⇒ 6² + 8² = r²
⇒ 36 + 64 = r2
⇒ r² = 100 = 10²
⇒ r = 10 cm

Question 10.
A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Solution:
Recall the Baudhayana triplet (5, 12, 13)
∠B = 90°, ∠D = 90°
∠B + ∠D = 180°

ABCD is a cyclic quadrilateral.
Ar. ABCD = 2 × Area ABC
= 2 × (\(\frac {1}{2}\) × 5 × 12) sq. units
= 60 sq. units.

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?
Solution:
Let ABCD be a cyclic quadrilateral.
We draw the perpendicular bisectors of any two sides, say AB and BC.
The meeting point of the perpendicular bisectors is the circumcentre.
If the two perpendicular bisectors meet inside the quadrilateral, the circumcentre is an interior point.
If they meet outside the quadrilateral, the circumcentre is an exterior point.

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.
Solution:
Given: Circle with centre O.
AB = CD
AB and CD meet at P.
To Show: AP = CP and PB = PD.

Construction: Draw OL ⊥ AB, OM ⊥ CD, and join OP.
Proof: OL = OM (equal chords are equidistant from the centre of the circle)
OP = OP (Common)
∠OLP = ∠OMP (each = 90°)
∆OLP ≅ ∆OMP (RHS)
LP = MP (CPCTC) …..(1)
AB = CD (Given)
⇒ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)CD
⇒ AL = CM ……(2)
Adding (1) and (2)
LP + AL = MP + CM
⇒ AP = CP
Again, AB = CD (Given)
AP = CP (Shown)
Subtracting, AB – AP = CD – CP
⇒ PB = PD

Question 13.
Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.
(Hint: Is it a circumcircle of a suitable triangle?)
Solution:
Construction Steps:
1. Draw AB = 6 cm.
2. Draw the ⊥ bisector of AB, meeting AB at L.
3. With L as centre and radius 3 cm, draw an arc meeting line l at O.
4. With O as centre and OA as radius, draw a circle. This circle is the required circle.

Question 14.
Show that a rectangle is the only parallelogram that can be inscribed in a circle.
Solution:
Let ABCD be a cyclic parallelogram.
Then ∠A = ∠C (Opposite angles of a parallelogram are equal)
Also ∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral are supplementary)
Then, ∠A + ∠A = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°

∴ ABCD is a rectangle.
(If an interior angle of a parallelogram is a right angle, it is a rectangle.)

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.
Solution:
Given: Rectangle ABCD is inscribed in a circle.
AC and BD meet at O.

To Show: O is the centre of the circle.
Proof: ABCD is a rectangle.
Then AC = BD (Diagonals of a rectangle are equal in length)
⇒ \(\frac {1}{2}\)AC = \(\frac {1}{2}\)BD
⇒ AO = OB
Also, AO = OC and BO = OD (Diagonals bisect each other)
⇒ OA = OB = OC = OD
Hence, O is the centre of the circle through A, B, C, and D.

Question 16.
Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Solution:
Chords of equal length are equidistant from the centre of the circle.
Their midpoints are also equidistant from the centre of the circle.
The midpoints taken together are a collection of points equidistant from the centre.
Hence, they form a circle.

Quadrilateral 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”.
Solution:
Given: Circle with centre O.
AB = AC

To Show: AO bisects ∠BAC.
Construction: Draw OL ⊥ AB and OM ⊥ AC
Proof: AB = AC (Given)
⇒ \(\frac {1}{2}\)AB = \(\frac {1}{2}\)AC
⇒ AL = AM
AO = AO (Common)
∠ALO = ∠AMO (each = 90°)
⇒ ∆ALO ≅ ∆AMO (RHS)
⇒ ∠1 = ∠2 (CPCTC)
⇒ AO bisects ∠BAC.

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.
Solution:
Given: Circle with centre O.
AB || CD
AB = 10 cm, CD = 24 cm
OL ⊥ CD, OM ⊥ AB
LM = 7 cm

Required: Radius of circle
MB = \(\frac {1}{2}\) × 10 cm = 5 cm
LD = \(\frac {1}{2}\) × 24 cm = 12 cm
Let OL = x

In ∆OLD, x² + 144 = r² …….(1)
In ∆OMB, (x + 7)² + 25 = r² …….(2)
From (1) and (2),
x² + 144 = (x + 7)² + 25
⇒ x² + 144 = x² + 14x + 49 + 25
⇒ 14x = 70
⇒ x = 5
Then, in ∆OLD,
r² = 5² + 144 = 169 = 13²
∴ r = 13
Hence, the radius of the circle is 13 cm.

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.
Solution:
Consider ∆AOB
∠AOB = \(\frac {1}{6}\) × 360° = 60°
OA = OB (each = r)
Then, ∠OAB = ∠OBA = x° (Suppose)

∴ x + x + 60 = 180
⇒ 2x = 120
⇒ x = 60
∴ ∠OAB = ∠OBA = 60°
Hence, ∆OAB is an equilateral triangle.
Then, AB = OB = OA
∴ AB = r

Now, \(d^2+\left(\frac{r}{2}\right)^2=r^2\)
⇒ \(d^2=r^2-\frac{r^2}{4}=\frac{3}{4} r^2\)
⇒ d = \(\frac{\sqrt{3}}{2} r\)
Hence each is at a distance of \(\frac{\sqrt{3}}{2} r\) units from the centre.

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.
Solution:
Given: MNOP is a cyclic quadrilateral.

MN is a diameter.
∠MOP = ∠MNP (Angles in the same segment are equal.)

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).
Solution:
Given: ABCD is a cyclic quadrilateral, and AD is extended to E.

∠1 + ∠2 = 180° (Opposite angles of a cyclic quadrilateral)
∠2 + ∠3 = 180° (Linear pair)
Then, ∠1 + ∠2 = ∠2 + ∠3
⇒ ∠1 = ∠3
⇒ ∠ABC = ∠CDE

Question 22.
“There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
Solution:
Given: Circle with centre O.
Diameter PQ.
Chords AB and CD.

Starting from AB, as we move towards the centre, the chord becomes longer (CD > AB).
We observe that as the distance from the centre decreases, the length of the chord increases.
When the chord coincides with the diameter, its length becomes maximum.

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.
Solution:
Given: Circle with centre O and point A inside the circle.
Chord PQ passing through A such that OA ⊥ PQ.
Any other chord RS passing through A (not perpendicular to OA).

To prove: PQ < RS
Construction: Draw OM ⊥ RS, where M is a point on RS.
Proof: Since OM ⊥ RS (by construction)
∆OMA is a right-angled triangle with the right angle at M.
In a right-angled triangle, the hypotenuse is the longest side.
In ∆OMA, side OA is the hypotenuse. Therefore, OA > OM.
Of any two chords, the one that is farther from the centre is shorter.
The distance of chord PQ from the centre is OA (since OA ⊥ PQ).
The distance of chord RS from the centre is OM (since OM ⊥ RS).
Since OA > OM, chord PQ is at a greater distance from the centre than chord RS.
PQ is farther from the centre than RS; it must be the shorter chord.
PQ < RS
Therefore, the shortest chord through point A is the chord perpendicular to OA.

Question 24.
How would you use the following figure to justify the statement that the angle in a semicircle is 90°?

Solution:
In ∆AOB, OA = OB (each = r)
Then, ∠1 = ∠2
In the same way, ∠3 = ∠4
Adding ∠1 + ∠3 = ∠2 + ∠4
∠B + ∠C = ∠BAC

In ∆BAC, ∠BAC + (∠B + ∠C) = 180°
⇒ ∠BAC + ∠BAC = 180°
⇒ 2∠BAC = 180°
⇒ ∠BAC = 90°
∴ Angle in a semicircle is a right angle.

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.
Solution:
Given: Circle with O as centre; diameter AB; chords CC’ and DD’ ⊥ AB.
M is the midpoint of CD.
M’ is the midpoint of C’D’.
To Show: MM’ ⊥ AB

Proof: CC’ ⊥ AB
DD’ ⊥ AB (Given)
⇒ CC’ || DD’
⇒ CC’D’D is a trapezium.
M is the midpoint of CD.
M’ is the midpoint of C’D’.
Then MM’ || CC’
⇒ ∠2 = ∠1 (Corresponding angles)
⇒ ∠2 = 90°
⇒ MM’ ⊥ AB.

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°?

Solution:
Given: Circle with centre O.
Cyclic quadrilateral ABCD.

To show: ∠B + ∠D = 180°
∠A + ∠C = 180°
Proof: ∠A + ∠B + ∠C + ∠D = 360°
∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 = 360° ……(1)
In ∆AOD, OA = OD (each = r)
∴ ∠1 = ∠2 = a°
Similarly, ∠3 = ∠4 = b°
∠5 = ∠6 = c°
∠1 = ∠8 = d°
Substituting in (1)
a + a + b + b + c + c + d + d = 360°
⇒ 2(a + b + c + d) = 360°
⇒ (a + b) + (c + d) = 180°
⇒ ∠A + ∠C = 180°
Similarly, ∠B + ∠D = 180°