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NCERT Class 9 Maths Chapter 5 I am Up and Down and Round and Round
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I am Up and Down and Round and Round
Chapter – 5
| Exercise Set 5.1 |
1. Draw ∆ABC with AB = 5 cm, ∠A = 70° and ∠B = 60°. Draw the circumcircle of ∆ABC. Is the centre inside or outside the triangle?
Ans: In ∆ABC, AB = 5 cm, ∠A = 70°, ∠B = 60°
Steps of Construction:
1. Draw a line segment AB with a length of 5 cm.
2. Using a protractor, draw an angle of 70° at point A.
3. Using a protractor, draw an angle of 60° at point B.
4. The intersection of the two arms of these angles will be point C.
5. Draw perpendicular bisectors of AB, BC, and AC.
6. The point where all three perpendicular bisectors meet is the circumcentre O.
7. Taking O as centre and OA as radius, draw the circumcircle of ∆ABC.
The circumcentre (the centre of the circumcircle) is inside the triangle.
2. Draw ∆ABC with AB = 5 cm, ∠A = 100°, and AC = 4 cm. Draw the circumcircle of ∆ABC. Is the centre inside or outside the triangle?
Ans: In triangle ABC where ∠A=100∘, the circumcentre O is located outside the triangle because it is obtuse-angled.
3. Draw ∆ABC, with AB = 6 cm, BC = 7 cm, and CA = 7 cm. Draw the circumcircle of ∆ABC. Let the circumcentre be O. Measure OA, OB, and OC.
Ans: In ∆ABC, AB = 6 cm, BC = 7 cm, CA = 7 cm
Steps of Construction:
1. Draw a line segment AB with a length of 6 cm.
2. Taking centre as point A, draw an arc of radius 7 cm, and taking centre as point B, draw an arc of radius 7 cm. Mark the point of intersection of these as point C.
3. Join the line segments AC and BC.
4. Draw perpendicular bisectors of AB, BC, and AC.
5. The point where all three perpendicular bisectors meet is the circumcentre O.
6. Using a compass, take the circumcentre as the centre O and draw a circle of radius OA that passes through points A, B, and C. This is the circumcircle of ∆ABC.
Since the circumcentre is equidistant from all three vertices, OA, OB, and OC will all have the same length. This distance is the radius of the circumcircle.
OA = OB = OC = 3.85 cm approx.
4. What is the least possible radius of a circle through two points A and B?
Ans: The smallest circle passing through two points A and B uses the segment AB as the diameter. The radius is exactly half the distance between the two points (r = d/2).
| Exercise Set 5.2 |
1. Show that the triangle formed by a chord and the centre of the circle is isosceles.
Ans: Given: A circle with centre O and chord AB of the circle. Points A and B are on the circumference of the circle.
To prove: ∆OAB is isosceles, i.e., OA = OB.
Proof: In a circle, the distance from the centre O to any point on the circle is always the radius of the circle.
Therefore, OA and OB are both radii of the circle.
Since OA and OB are both radii of the same circle, we can say that OA = OB
This means the two sides of ∆OAB are equal in length.
Since OA = OB, ∆OAB has two equal sides, making it an isosceles triangle.
Therefore, the triangle formed by a chord AB and the centre O of the circle is isosceles because the two sides OA and OB are equal in length.
2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.
Ans: Given: Two isosceles triangles, OAB and OCD, are formed by the chords AB and CD of the same circle, with O as the centre.
The two triangles have equal base lengths, i.e., AB = CD.
To prove: ∆OAB ≅ ∆OCD
Proof: In ∆OAB and ∆OCD,
OA = OD (Radii of the circles are equal)
AB = CD (Equal base length)
OB = OC (Radii of the circles are equal)
∆OAB ≅ ∆OCD (By SSS congruence criterion)
Therefore, if two isosceles triangles formed by the centre of the circle and the chords AB and CD have equal base lengths, then they are congruent to each other.
| Exercise Set 5.3 |
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 Fig. 5.12. You are told that ∠CMA = ∠CMB = 90°. You need to show that AM = BM)
Ans: Show that if CM⊥AB, then AM = BM.
Proof:
Consider right-angled triangles ∆CMA and ∆CMB from
Hypotenuse: CA = CB (Both are radii of the circle).
Side: CM = CM (Common side to both triangles).
Right Angle: ∠CMA = ∠CMB = 90∘ (Given).
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.
Ans: Given: ABC is the isosceles triangle inscribed in the circle with centre O.
AM is the altitude from A to BC.
To prove: Altitude AM passes through the centre, i.e., AOM is the straight line
Proof: In ∆AMB and ∆AMC,
AM = AM (Common)
AB = AC (∆ABC is the isosceles triangle)
∠AMB = ∠AMC = 90° (AM is the altitude to BC)
∆AMB ≅ ∆AMC (By RHS congruence criterion)
BM = CM (CPCT)
By theorem, the line joining the centre of a circle (O) and the midpoint (M) of a chord of the circle is perpendicular to the chord (BC).
Therefore, the line OM is perpendicular to BC.
Since OM is perpendicular to BC and AM is perpendicular to BC, the two lines AM and OM must coincide because both must pass through the midpoint M and be perpendicular to BC.
Thus, AM and OM are the same line; AOM is a straight line.
Therefore, the altitude AM passes through the centre O of the circle.
Hence, the altitude from A to BC passes through the centre of the circle.
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.
Ans: Given: Radius of the circle = 5 cm, AB = 6 cm, CD = 8 cm
To find: Length of MN
Let N be the midpoint of chord AB.
Since the perpendicular bisects the chord.
AN = 6/2 = 3 cm
Using a theorem, the line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.
In a right-angled ∆ANO,
OA2 = ON2 + AN2 (by Baudhayana-Pythagoras theorem)
⇒ 52 = ON2 + 32
⇒ 25 = ON2 + 9
⇒ ON2 = 16
⇒ ON = 4 cm
Let M be the midpoint of chord CD.
Since the perpendicular bisects the chord.
CM = 8/2 = 4 cm
Using a theorem, the line joining the centre of a circle and the midpoint of a chord of the circle is perpendicular to the chord.
In a right-angled ∆CMO,
OC2 = OM2 + CM2 (by Baudhayana-Pythagoras theorem)
⇒ 52 = OM2 + 42
⇒ 25 = OM2 + 16
⇒ OM2 = 9
⇒ OM = 3 cm
Length of MN = OM + ON
= 3 cm + 4 cm
= 7 cm
Therefore, the distance between the midpoints of the chord is 7 cm.
| Exercise Set 5.4 |
1. Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.
Ans: Let a circle have centre C and radius r. Consider two chords AB and DE of equal length (AB=DE=L). Let CM and CN be perpendiculars from the centre to these chords.
1. Bisect the Chords: By Theorem 5, the perpendicular from the centre bisects the chord. So, the half-lengths are:
AM = L/2 and DN = L/2
2. Apply the Theorem: In right-angled ∆CMA and ∆CND:
For ∆CMA: r2 = CM2 + AM2
⟹ CM2 = r2 – (L/2)2
For ∆CND: r2 = CN2 + DN2
⟹ CN2 = r2 – (L/2)2
3. Compare Results: Since both CM2 and CN2 equal r2 – (L/2)2, it follows that: CM2 = CN2 ⟹ CM = CN.
2. Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.
Ans: Given: Circle with centre C, CE ⊥ AB, CH ⊥ GF, and CE = CH.
To prove: AB = GF
Proof: Using the theorem, equal chords of a circle subtend equal angles at the centre of the circle.
∠BCA = ∠GCF
In ∆BCA and ∆GCF
GC = CF, CF = CA (radius of circle)
∠BCA = ∠GCF
∴ ∆BCA = ∆GCF by SAS congruence criterion
Thus, GF = AB (by CPCT) (Proved)
3. Solve the previous question using the Baudhayana-Pythagoras theorem.
Ans: Given: Circle with centre C, CE ⊥ AB, CH ⊥ GF, and CE = CH.
To prove: AB = GF
Proof: Baudhayana Pythagoras theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
Let Radii CA = CG = r, and lengths CE = CH = d
In the right-angled ∆CEA,
CA2 = CE2 + AE2 (by Baudhayana-Pythagoras theorem)
EA2 = CA2 – CE2
EA2 = r2 – d2 ……(i)
In the right-angled ∆CHG,
CG2 = CH2 + GH2 (by Baudhayana-Pythagoras theorem)
GH2 = CG2 – CH2
GH2 = r2 – d2 …..(ii)
From (i) and (ii)
EA2 = GH2
⇒ EA = GH
A perpendicular from the centre of a circle to a chord bisects the chord.
∴ AB = 2 × EA
GF = 2 × HG
Since EA = GH, we have AB = GF (proved).
| Exercise Set 5.5 |
1. Find the length of the chord of a circle where the radius is 7 cm, and perpendicular distance is 6 cm.
Ans: Let r = 7 cm be the radius and d = 6 cm be the perpendicular distance from the centre to the chord.
Using the Baudhāyana-Pythagoras theorem, we first find half the length of the chord (a):
a2 = r2 – d2
a2 = 72 – 62 = 49 – 36 = 13
a = √13 cm
Since the perpendicular from the centre bisects the chord, the total length is 2a:
Length of Chord = 2√13 cm (approx. 7.21 cm).
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
Ans: Given: d be the perpendicular distance from the centre of the circle to the chord, and r be the radius of the circle.
To prove: Length of the chord =
Proof: OA = r, OM = d
In the right-angled ∆OMA,
OA2 = OM2 + AM2 (by Baudhayana-Pythagoras theorem)
⇒ r2 = d2 + AM2
⇒ AM2 = r2 – d2
⇒ AM = √r2 – d2
Since the perpendicular distance bisects the chord,
AB = 2 × AM = 2√r2 – d2
Therefore, the length of the chord is
3. In a circle, if the distance of chord AB from the centre is twice the distance of another chord CD from the centre, can we conclude that CD = 2AB? Give reasons for your answer.
Ans: No, we cannot conclude that CD = 2AB. While a chord closer to the centre is indeed longer, the relationship between its length and distance is not linear.
Reasons:
The ratio of these lengths depends on the radius r. For example, if r = 5 and x = 1:
Here, CD is clearly not twice the length of AB. Doubling the distance does not result in halving the chord length.
| Exercise Set 5.6 |
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?
Ans: In a circle with centre O and radius OA = OB = 12 cm, if OA and OB are both equal then triangle OAB is an isosceles triangle.
The sum of angles in any triangle is 180°.
In an isosceles triangle, the angles opposite equal sides are also equal (∠OAB = ∠OBA).
Let ∠OAB = ∠OBA = x.
So, x + x + 60°= 180°
⇒ 2x = 120°
⇒ x = 60°
Since all three interior angles are 60°, triangle AOB is an equilateral triangle.
In an equilateral triangle, all three sides are equal in length:
OA = OB = AB = 12 cm
Therefore, the length of chord AB is 12 cm.
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?
Ans: (i) No, because all angles subtended by an arc on the same side of the circle are equal.
So, if X and Y are on the same side of chord AB, then ∠AXB = ∠AYB.
Therefore, they can’t be different.
(ii) No, it is not necessarily true. X and Y could lie on opposite sides of the circle if AB is the diameter.
In that case, both angles will be 90°, regardless of which side X and Y are on.
(iii) Yes, the circle through A, B, and X also passes through Y, provided X and Y are on the same side of AB.
3. Find x in Fig. 5.26.
Ans: The quadrilateral ABCD in your image is cyclic because all its vertices lie on the circle. In a cyclic quadrilateral, opposite angles add up to 180∘.
Since ∠D = 100° and ∠B = x are opposite angles, x + 100°= 180°.
Therefore, the value of x is 80°.
| END-OF-CHAPTER EXERCISES |
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?
Ans: Given: Radius = OA = 13 cm,
Distance of chord from centre = OP = 5 cm
To find: Length of chord, i.e., AB
In the right-angled ∆OPA,
OA2 = OP2 + AP2 (by Baudhayana-Pythagoras theorem)
⇒ 132 = 52 + AP2
⇒ 169 = 25 + AP2
⇒ AP2 = 169 – 25 = 144
⇒ AP = √144 = 12 cm
Since the perpendicular distance bisects the chord,
AB = 2 × AP = 2 × 12 = 24 cm
Therefore, the length of the chord is 24 cm.
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?
Ans: In a circle, the angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circle.
So,
Angle at centre = 2 × Angle at circumference
Angle at centre = 2 × Angle at circumference
Given:
Angle at centre = 70°
Angle at circumference = 70°/2 = 35°
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.
Ans: Given: Diameter = AC = 26 cm, Chord AB = 24 cm
To find: Distance between the length of the chord and the centre i.e., OP
Radius (OA) = 26/2 = 13 cm
A perpendicular line drawn from the centre of a circle to a chord bisects the chord.
AP = 24/2 = 12 cm
In the right-angled ∆OPA,
⇒ OA2 = OP2 + AP2 (by Baudhayana-Pythagoras theorem)
⇒ 132 = OP2 + 122
⇒ 169 = OP2 + 144
⇒ OP2 = 169 – 144 = 25
⇒ OP = 5 cm
Therefore, the distance from the centre of the circle to the chord is 5 cm.
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?
Ans: Given: Radius = OA = 15 cm,
Distance of the chord from centre = OP = 9 cm
To find: Length of chord, i.e., AB
In the right-angled ∆OPA,
OA2 = OP2 + AP2 (by Baudhayana-Pythagoras theorem)
⇒ 152 = 92 + AP2
⇒ 225 = 81 + AP2
⇒ AP2 = 225 – 81 = 144
⇒ AP = √144 = 12 cm
Since the perpendicular distance bisects the chord,
AB = 2 × AP = 2 × 12 = 24 cm
Therefore, the length of the chord is 24 cm.
5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.
Ans: Given: A circle with centre O and a chord AB. Let M be the midpoint of AB, and OM ⊥ AB.
To Prove: The perpendicular bisector of AB passes through O.
Construction: Join OA and OB.
Congruence: In triangles ∆OMA and ∆OMB:
- OA = OB (Radii of the same circle)
- AM=BM (M is the midpoint)
- OM=OM (Common side)
Conclusion: By SSS Congruence, ∆OMA ≅ ∆OMB. Thus, ∠OMA = ∠OMB. Since they form a linear pair and are equal, both are 90°.
This confirms that the line from the centre to the midpoint is perpendicular, so the perpendicular bisector must pass through the centre.
6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of the ∠ACB? Explain your reasoning.
Ans: Given: AB is the diameter of the circle.
C is a point on the circumference of the circle.
∠ACB is the angle subtended by the diameter AB at the point C on the circumference.
To find: The measure of ∠ACB.
Since AB is the diameter of the circle and C is a point on the circumference, the angle ∠ACB is subtended by the diameter AB.
By the corollary of Theorem 9, the angle subtended by the diameter at any point on the circumference of the circle is always a right angle.
Therefore, the measure of ∠ACB is 90°.
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?
Ans: To find ∠C:
Since ∠A and ∠C are opposite angles:
∠A + ∠C = 180°
75° + ∠C = 180°
∠C = 180° – 75° = 105°
To find ∠D:
Since ∠B and ∠D are opposite angles:
∠B + ∠D = 180°
110° + ∠D = 180°
∠D = 180° – 110° = 70°
The measure of ∠C is 105° and the measure of ∠D is 70°.
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.
Ans: Quadrilateral PQRS is inscribed in a circle.
∠P = (2x + 10)° and ∠R = (3x – 20)°
In a cyclic quadrilateral, opposite angles add up to 180°.
∠P + ∠R = 180°
⇒ (2x + 10) + (3x – 20) = 180
⇒ 2x + 10 + 3x – 20 = 180
⇒ 5x – 10 = 180
⇒ 5x = 190
⇒ x = 38
∠P = 2x + 10 = 2(38) + 10 = 76 + 10 = 86°
∠R = 3x – 20 = 3(38) – 20 = 114 – 20 = 94°
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.
Ans: Given: The length of the chord AB = 16 cm, the perpendicular distance from the centre of the circle O to the chord AB = 6 cm.
To find: Radius of the circle (r).
P is the midpoint of chord AB.
Since OP is the perpendicular distance from the centre to the chord, P divides the chord AB into two equal parts.
Therefore, the length of AP = 16/2 = 8 cm.
In a right-angled ∆OPA,
OA2 = OP2 + AP2 (by Baudhayana-Pythagoras theorem)
⇒ r2 = 62 + 82
⇒ r2 = 36 + 64 = 100
⇒ r = 10 cm
Therefore, the radius of the circle is 10 cm.
10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.
Ans: Let ABCD be the cyclic quadrilateral with AB = BC = 5, CD = DA = 12
Since there are two pairs of adjacent equal sides, ABCD is a kite. Join BD.
In a kite, the triangles formed by joining the vertices between unequal sides are congruent:
∠ABD ≅ ∠CBD
So, ∠BAD = ∠BCD
But in a cyclic quadrilateral, opposite angles are supplementary:
∠BAD + ∠BCD = 180°
Since they are equal,
∠BAD = ∠BCD = 90°
Thus, ∆ABD is a right triangle with legs AB = 5, AD = 12
Hence, Area of ∆ABD = 12 × 5 × 12 = 30
Since ∠ABD ≅ ∠CBD,
Area of ∆CBD = 30 square units
Therefore, the area of quadrilateral ABCD = 30 + 30 = 60 square units.
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?
Ans: In a cyclic quadrilateral, the position of the circumcentre depends on its angles. If all interior angles are acute (less than 90°), the circumcentre lies inside the quadrilateral. If any angle is obtuse (greater than 90°), the circumcentre lies outside it. If one angle is exactly 90°, the circumcentre lies on the quadrilateral (on its side). Thus, the easiest way to determine the position is by simply examining the measures of the angles.
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.
Ans: Given: Let chords AB and CD intersect at point E inside a circle with centre O.
Draw perpendiculars OM and ON from O to chords AB and CD, respectively. Join OE.
To prove: AE = CE, BE = DE
Proof: Since equal chords are equidistant from the centre.
OM = ON
Also, perpendiculars from the centre bisect the chords.
AM = MB, CN = ND
In a right-angled ∆OME and ∆ONE,
OM = ON (Equal chords are equidistant)
OE = OE (Common)
∠OME = ∠ONE = 90°
∆OME ≅ ∆ONE (RHS congruence rule)
ME = NE (CPCT)
AE = AM + ME
CE = CN + NE
Since AM = CN and ME = NE (from congruence)
⇒ AE = CE
BE = MB – ME
DE = ND – NE
Since MB = ND and ME = NE
⇒ BE = DE
The line segments of chord AB (AE, BE) are equal to the corresponding segments of chord CD (CE, DE).
Therefore, if two equal chords intersect inside a circle, they are divided into equal corresponding segments.
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?)
Ans: A chord of length 6 cm that is 3 cm from the centre of a circle belongs to a specific circle with a unique radius.
14. Show that rectangle is the only parallelogram that can be inscribed in a circle.
Ans: Given: A parallelogram ABCD is inscribed in a circle.
To prove: The parallelogram ABCD is a rectangle.
Proof: In a circle, the opposite angles of a cyclic quadrilateral are supplementary.
∠A + ∠C = 180° and ∠B + ∠D = 180°
In a parallelogram, the opposite angles are equal.
So, ∠A = ∠C
and ∠B = ∠D
Since ∠A = ∠C
and ∠A + ∠C = 180°
⇒ 2∠A = 180°
⇒ ∠A = 90°
All angles of the parallelogram are right angles.
Therefore, a rectangle is the only parallelogram that can be inscribed in a circle.
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.
Ans: Property of Rectangles: Each interior angle of a rectangle is 90∘. For rectangle ABCD, ∠ABC = 90∘.
Property of Circles: An angle of 90∘ on the circumference is subtended by the diameter. Therefore, the diagonal AC must be a diameter of the circle.
Similarly: Since ∠BCD = 90∘, the diagonal BD must also be a diameter of the circle.
Centre Definition: By definition, all diameters of a circle intersect at exactly one point: the centre (O).
Conclusion:
Since both diagonals AC and BD are diameters, their point of intersection is necessarily the centre of the circle.
16. Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?
Ans: If you draw many chords of the same length inside a circle, and mark the middle point of each chord, all those midpoints will form a smaller circle inside the big one. This smaller circle has the same centre as the big circle, so it is called a smaller concentric circle.
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”.
Ans: Given: A circle with centre O,
Chords AB = AC (congruent)
To prove: The centre O lies on the angle bisector of ∠BAC, i.e., ∠BAO = ∠OAC
Proof: In a circle, all radii are equal.
So, OA = OB = OC
In triangles ∆AOB and ∆AOC,
OA = OA (Common side)
OB = OC (Radii of the circle)
AB = AC (Given congruent chords)
∴ ∆AOB ≅ ∆AOC (By SSS congruence rule)
⇒ ∠BAO = ∠CAO (CPCT)
This means line AO bisects ∠BAC.
Therefore, the centre of the circle lies on the angle bisector of ∠BAC.
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.
Ans: To find the radius r, we use the perpendicular distance from the centre to both chords and the Baudhāyana-Pythagoras theorem.
Step-by-Step Calculation:
Identify Half-Chords: Let the chords be AB = 24 cm and CD = 10 cm. Their half lengths are 12 cm and 5 cm.
Set up Variables: Let x be the distance from the centre to the 24 cm chord. The distance to the 10 cm chord is x + 7.
Form Equations: Using r2 = a2 + d2:
1. r2 = 122 + x2 = 144 + x2
2. r2 = 52 + (x + 7)2 = 25 + x2 + 14x + 49 = x2 + 14x + 74
Solve for x: 144 + x2 = x2 + 14x + 74
144 = 14x + 74 ⇒ 14x = 70 ⇒ x = 5 cm
Calculate r:
r2 = 144 + 52 = 144 + 25 = 169 ⇒ r = 13 cm
The radius of the circle is 13 cm.
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.
Ans: Let ABCDEF be a regular hexagon inscribed in a circle with centre O and radius r.
Since, the hexagon is regular, AB = BC = CD = DE = EF = FA
We know that equal chords of a circle subtend equal angles at the centre.
Therefore, ∠AOB = ∠BOC = ∠COD = ∠DOE = ∠EOF = ∠FOA
These six central angles together make a complete angle around the centre.
Hence, each central angle is 360°/6 = 60°
So, ∠AOB = 60°
Also, OA = OB = r (radii of the same circle)
Thus, in ∠AOB, OA = OB, and ∠AOB = 60°
Therefore, the remaining two angles are equal and together measure = 180° – 60° = 120°
Hence, each remaining angle is 60°.
So, all three angles of ∆AOB are 60°.
Therefore, ∆AOB is equilateral.
Hence, AB = OA = OB = r
Therefore, each side of the regular hexagon is r.
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.
Ans: In a cyclic quadrilateral MNOP where MN is a diameter, ∠MOP and ∠MNP both subtend the same arc MP.
According to the theorem, the angle subtended by an arc at the center is double the angle at the circumference. Therefore, ∠MOP = 2∠MNP. Additionally, since MN is a diameter, ∠MPN must be 90∘, making triangle MPN a right-angled triangle.
The angle at the center ∠MOP is twice the angle ∠MNP because they both subtend the same arc MP.
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).
Ans: Given: In a cyclic quadrilateral ABCD, side CD is produced to a point E.
To prove: ∠ADE = ∠ABC
Proof: ∠CDA + ∠ADE = 180° (Linear Pair)
∠ABC + ∠ADC = 180° (Opposite angles of a cyclic quadrilateral are supplementary.)
Equating both the cases, we get ∠CDA + ∠ADE = ∠ABC + ∠ADC
∠ADE = ∠ABC
22. “There is no chord of a circle that is longer than its diameter.” How do you justify this statement?
Ans: A diameter is the longest chord because it passes through the centre, maximizing the distance between two points on the circle.
Any chord AB and the centre O form ∆OAB. By the triangle inequality, the side AB must be less than OA + OB. Since OA and OB are radii, AB < 2r. Because the diameter equals 2r, no chord can exceed it.
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.
Ans: 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
Contraction: 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
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.
24. How would you use the following figure to justify the statement that the angle in a semicircle is 90°?
Ans: Isosceles Triangles: Segment OA is a radius, just like the segments from O to the base. This creates two isosceles triangles within the larger triangle.
Equal Angles: In the left triangle, the base angles are both a. In the right triangle, the base angles are both b.
Angle Sum Property: For the large triangle, the sum of all interior angles must be 180°.
Calculation:
a + (a + b) + b = 180°
2a + 2b = 180°
2(a + b) = 180°
a + b = 90°
Conclusion:
Since the angle at vertex A is exactly a + b, it must be 90°, proving the theorem for any point on the semicircle.
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.
Ans: Given: A circle with diameter AB.
Two chords CC’ and DD’ are drawn perpendicular to AB.
M and M’ are the midpoints of chords CD and C’D’ respectively.
To Prove: The line segment MM’ joining the midpoints of chords CD and C’D’ is perpendicular to AB.
Proof: Let the diameter AB meet the chords CC’ and DD’ at P and Q, respectively.
Since CC’ ⊥ AB and DD’ ⊥ AB, CC’ || DD’ because lines perpendicular to the same line are parallel.
Now consider quadrilateral CDD’C’.
Its opposite sides CC’ and DD’ are parallel, so CDD’C’ is a trapezium.
Also, M and M’ are given as the midpoints of the non-parallel sides CD and C’D’ respectively.
In a trapezium, the line segment joining the midpoints of the non-parallel sides is parallel to the parallel sides.
Therefore, MM’ || CC’.
But CC’ ⊥ AB.
Hence, any line parallel to CC’ is also perpendicular to AB.
Therefore, MM’ ⊥ AB.
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°?
Ans: The figure uses radii to create four isosceles triangles, proving that opposite angles p + q and u + v sum to 180°.
The radii OA, OB, OC, OD form four isosceles triangles where base angles are equal. The sum of all angles in quadrilateral ABCD is 2p + 2q + 2u + 2v = 360°. Dividing by two shows (p + q) + (u + v) = 180°, proving opposite angles are supplementary.
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