Question 1

Show that the triangle formed by a chord and the centre of the circle is isosceles.

Accepted Answer

Consider a circle with centre O and a chord AB. Join OA and OB. Since OA and OB are radii of the circle, OA = OB. Therefore, triangle OAB has two equal sides and is isosceles. — In a circle, all radii are equal. The triangle formed by joining the centre to the endpoints of a chord has two sides equal (the radii), so it is 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.

Accepted Answer

Let two isosceles triangles OAB and OCD be formed by chords AB and CD respectively, with OA = OB and OC = OD (radii). If AB = CD (equal bases), then by the Side-Side-Side (SSS) congruence criterion, triangles OAB and OCD are congruent because OA = OC (radii), OB = OD (radii), and AB = CD (given). — Both triangles have two equal sides (radii) and equal base length, so by SSS congruence, they are congruent.

Question 3

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.)

Accepted Answer

Solution:

Given: A chord AB of a circle with centre C, and a perpendicular CM from C to AB.

To prove: CM bisects AB, i.e., AM = BM.

Proof:
1. Since CM is perpendicular to AB, ∠CMA = ∠CMB = 90°.
2. In triangles CMA and CMB:
   - CM = CM (common side)
   - CA = CB (radii of the circle)
   - ∠CMA = ∠CMB = 90°
3. By RHS congruence criterion, ΔCMA ≅ ΔCMB.
4. Therefore, corresponding parts are equal, so AM = BM.

Hence, the perpendicular from the centre of a circle to a chord bisects the chord. — The proof uses the RHS congruence criterion for right triangles formed by the perpendicular from the centre to the chord. Since the two triangles share the perpendicular segment and have equal radii, the triangles are congruent, implying the chord is bisected.

Question 4

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.

Accepted Answer

Solution:

Given: Triangle ABC is isosceles with AB = AC, inscribed in a circle with centre O.

To prove: The altitude from A to BC passes through the centre O.

Proof:
1. Since AB = AC, triangle ABC is isosceles with vertex A.
2. The altitude from A to BC is also the median and angle bisector of angle A.
3. Let the altitude from A meet BC at D.
4. Since ABC is inscribed in the circle, points A, B, C lie on the circle.
5. The perpendicular bisector of chord BC passes through the centre O.
6. Since D is the midpoint of BC (because altitude in isosceles triangle), the line AD is the perpendicular bisector of BC.
7. Therefore, the altitude from A passes through the centre O.

Hence proved. — In an isosceles triangle inscribed in a circle, the altitude from the apex is also the perpendicular bisector of the base chord, which passes through the centre of the circle.

Question 5

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.

Accepted Answer

Solution:

Given:
Length of chord 1, AB = 6 cm
Length of chord 2, CD = 8 cm
Radius of circle, r = 5 cm

To find: Distance between midpoints of chords AB and CD.

Let O be the centre of the circle.
Let M and N be the midpoints of chords AB and CD respectively.

Since chords are parallel and on opposite sides of the centre, OM and ON are perpendicular distances from centre to chords.

Using the right triangle formed by radius and half chord:

For chord AB:
AM = AB/2 = 3 cm
OM = √(r² - AM²) = √(5² - 3²) = √(25 - 9) = √16 = 4 cm

For chord CD:
CN = CD/2 = 4 cm
ON = √(r² - CN²) = √(25 - 16) = √9 = 3 cm

Since chords are on opposite sides of centre, distance between midpoints M and N = OM + ON = 4 + 3 = 7 cm

Answer: 7 cm — The distance from the centre to a chord is given by the perpendicular from the centre to the chord. Using Pythagoras theorem in the right triangle formed by radius, half chord, and distance from centre to chord, we find these distances and add them since chords are on opposite sides.

Question 6

Use the Baudhayana-Pythagoras theorem to show why Theorem 6 must be true.

Accepted Answer

Theorem 6 states that the perpendicular drawn from the centre of a circle to a chord bisects the chord. To prove this using the Baudhayana-Pythagoras theorem, consider a circle with centre O and chord AB. Let OC be the perpendicular from O to AB, meeting AB at C. We need to show that AC = CB.

In triangles OAC and OBC:
- OA = OB (radii of the circle)
- OC = OC (common side)
- Angle OCA = Angle OCB = 90° (since OC is perpendicular)

By the Pythagoras theorem:
OA² = OC² + AC²
OB² = OC² + BC²
Since OA = OB, we have AC² = BC², so AC = BC.

Thus, the perpendicular from the centre bisects the chord, proving Theorem 6. — Using the Pythagoras theorem on the two right triangles formed by the perpendicular from the centre to the chord, and using the equality of radii, we conclude that the two segments of the chord are equal, hence the chord is bisected.

Question 7

Consider Fig. 5.15. If CE is perpendicular to AB, CH is perpendicular to GH, and CE = CH, show that AB = GF.

Accepted Answer

Given that CE ⟂ AB and CH ⟂ GH, and CE = CH.

In triangles CEB and CHG:
- CE = CH (given)
- Angle CEB = Angle CHG = 90° (perpendicular)
- CB = CG (radii of the circle)

By RHS congruence criterion, triangle CEB ≅ triangle CHG.

Therefore, EB = HG.

Since AB = AE + EB and GF = GH + GF (depending on figure), and given the equal segments, it follows that AB = GF. — By showing congruence of the two right triangles formed by the perpendiculars and equal segments, the corresponding chords AB and GF are equal in length.

Question 8

Solve the previous question using the Baudhayana-Pythagoras theorem.

Accepted Answer

Using the Baudhayana-Pythagoras theorem on triangles CEB and CHG:

Since CE = CH and both are perpendiculars,

CE² + EB² = CB²
CH² + HG² = CG²

Given CB = CG (radii), and CE = CH, it follows that EB = HG.

Therefore, AB = AE + EB and GF = GH + FG, and since AE = GF (from figure), AB = GF.

Hence, the chords AB and GF are equal in length. — Applying the Pythagoras theorem to the right triangles formed by the perpendiculars from the centre to the chords, and using equality of radii and perpendiculars, we conclude the chords are equal.

Question 9

Find the length of the chord of a circle where the radius is 7cm and perpendicular distance is 6cm.

Accepted Answer

Given: Radius r = 7 cm, perpendicular distance d = 6 cm from the centre to the chord.

We know the length of the chord = 2 × √(r² - d²).

Calculate:
Length = 2 × √(7² - 6²) = 2 × √(49 - 36) = 2 × √13 = 2 × 3.6055 = 7.211 cm (approx).

Therefore, the length of the chord is approximately 7.21 cm. — Using the right triangle formed by the radius, the perpendicular distance, and half the chord length, apply Pythagoras theorem:
Half chord length = √(r² - d²).
Multiply by 2 to get full chord length.

Question 10

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).

Accepted Answer

Consider a circle with centre O and radius r. Let AB be a chord of the circle, and let the perpendicular from O to AB meet AB at point M.

Since OM is perpendicular to chord AB, triangle OMB is right angled at M.

OM = d (perpendicular distance from centre to chord), OB = r (radius).

By Pythagoras theorem in triangle OMB:
MB = √(OB² - OM²) = √(r² - d²).

Since AB = 2 × MB, chord length = 2√(r² - d²).

Thus, the formula is derived using the right triangle formed by the radius, perpendicular distance, and half the chord. — The perpendicular from the centre to the chord bisects the chord, forming two right triangles. Using Pythagoras theorem in one of these triangles gives half the chord length as √(r² - d²). Doubling it gives the full chord length.

Question 11

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.

Accepted Answer

Let the distance of chord CD from the centre be d.
Then, distance of chord AB from the centre = 2d.

Let the radius of the circle be r.

Length of chord AB = 2√(r² - (2d)²) = 2√(r² - 4d²).
Length of chord CD = 2√(r² - d²).

If CD = 2AB, then:
2√(r² - d²) = 2 × 2√(r² - 4d²) => √(r² - d²) = 2√(r² - 4d²).

Squaring both sides:
(r² - d²) = 4(r² - 4d²) => r² - d² = 4r² - 16d².

Rearranged:
-3r² + 15d² = 0 => 15d² = 3r² => 5d² = r².

This is a specific condition, not generally true.

Therefore, we cannot conclude that CD = 2AB just because the distance of AB from the centre is twice that of CD.

Hence, the statement is false in general. — Using the chord length formula and given distances, we see the relation CD = 2AB leads to a specific condition on r and d, which is not generally true. So the conclusion does not hold in general.

Question 12

A circle with centre O is drawn, and A, B, C, D are points on the circle (see Fig. 5.19). Measure the angles subtended by arc AKB and arc CLD at the centre O. If the angle at the centre is less than 180°, it is a minor arc. If the angle at the centre is greater than 180°, it is a major arc. State whether arcs AKB and CLD are minor arcs or major arcs.

Accepted Answer

To solve this, first measure the angle subtended by arc AKB at the centre O using a protractor or given measurements. If this angle is less than 180°, then arc AKB is a minor arc; if greater than 180°, it is a major arc. Similarly, measure the angle subtended by arc CLD at centre O and classify it as minor or major arc accordingly. The classification depends solely on whether the central angle is less than or greater than 180°. — The definition of minor and major arcs depends on the central angle subtended by the arc. By measuring the central angle for each arc, we determine the type of arc. This exercise helps understand the concept of minor and major arcs in a circle.

Question 13

Activity: Draw a circle and a chord AB. Fix an arc AKB formed by AB and a point K between A, B on the circle. Measure the angle subtended at the centre by arc AKB. Take three points P, Q, R on the circle outside arc AKB. Measure the angles subtended by arc AKB at points P, Q, R. What do you notice? Repeat this activity for a different arc AKB. Based on this activity, we can make a statement about the angles subtended by an arc of a circle at the centre and at a point of the circle outside the arc.

Accepted Answer

Perform the activity by drawing the circle and chord AB, then mark point K between A and B on the circle to form arc AKB. Measure the angle subtended by arc AKB at the centre O. Next, select points P, Q, R on the circle but outside arc AKB and measure the angles subtended by arc AKB at these points. You will notice that all these angles at points P, Q, R are equal. Repeating the activity with a different arc AKB will yield the same observation. This leads to the conclusion that the angle subtended by an arc at the centre is double the angle subtended by the same arc at any point on the circle outside the arc. — This activity demonstrates the key property of angles subtended by arcs: the central angle is twice any angle subtended by the same arc at the circumference outside the arc. This is a fundamental theorem in circle geometry.

Question 14

In a circle with centre O, the central angle AOB is  $60^{\circ}$ . If the radius of the circle is  $12\mathrm{cm}$ , what is the length of the chord AB?

Accepted Answer

Given: Central angle AOB = 60°, radius OA = OB = 12 cm.

We need to find the length of chord AB.

Solution:
In triangle AOB, OA = OB = 12 cm (radii of the circle), and angle AOB = 60°.

Since OA = OB, triangle AOB is isosceles.

Using the formula for chord length:
Chord length AB = 2 × OA × sin(½ × angle AOB)
= 2 × 12 × sin(30°)
= 24 × 0.5 = 12 cm.

Therefore, the length of chord AB is 12 cm. — The chord length in a circle can be found using the formula 2r sin(θ/2), where θ is the central angle subtended by the chord.
Here, θ = 60°, r = 12 cm.
So, chord AB = 2 × 12 × sin(30°) = 24 × 0.5 = 12 cm.

Question 15

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  $\angle A X B$  is different from  $\angle A Y B$ ?
(ii) Is it true that if  $\angle A X B = \angle A Y B$ , then  $X$  and  $Y$  lie on the same side of the circle?
(iii) If  $\angle A X B = \angle A Y B$ , and  $X$  and  $Y$  do not lie on the circle, does the circle through  $A, B$  and  $X$  also pass through  $Y$ ?

Accepted Answer

(i) No. If X and Y are points on the same side of chord AB on the circle, then the angles subtended by chord AB at X and Y are equal. This is a property of angles in the same segment of a circle. Therefore, there cannot be two points X and Y on the same side of AB such that \angle AXB ≠ \angle AYB.

(ii) Yes. If \angle AXB = \angle AYB, then points X and Y lie on the same side of chord AB on the circle. This follows from the property that angles subtended by the same chord in the same segment are equal.

(iii) No. If X and Y do not lie on the circle, then the circle passing through A, B, and X does not necessarily pass through Y. The equality of angles \angle AXB and \angle AYB alone does not guarantee that Y lies on the same circle as A, B, and X. — The angle subtended by a chord at any point on the same segment of the circle is constant. Hence, (i) no two points on the same side of AB can have different angles subtended by AB.

(ii) Equal angles subtended by chord AB imply points lie on the same segment (same side).

(iii) If points are not on the circle, equality of angles does not imply concyclicity.

I’m Up and Down, and

I’m Up and Down, and — Study Notes

Introduction

5.1 DEFINITIONS

5.2 SYMMETRIES OF A CIRCLE

Practice Questions — I’m Up and Down, and

All 8 Chapters in Ganita Manjari (English)