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?
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?
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.
4. What is the least possible radius of a circle through two points A and B?

Accepted Answer

1. To draw ΔABC with AB = 5 cm, ∠A = 70°, and ∠B = 60°:
- Draw a line segment AB = 5 cm.
- At point A, construct an angle of 70°.
- At point B, construct an angle of 60°.
- The intersection of these two rays is point C.
- Draw the circumcircle by finding the perpendicular bisectors of at least two sides and locating their intersection (circumcentre).
- Since both angles are acute and sum less than 180°, the circumcentre lies inside the triangle.

2. To draw ΔABC with AB = 5 cm, ∠A = 100°, and AC = 4 cm:
- Draw AB = 5 cm.
- At A, construct an angle of 100°.
- From A along this ray, mark point C such that AC = 4 cm.
- Join BC.
- Draw the circumcircle by constructing perpendicular bisectors of two sides.
- Since ∠A is obtuse, the circumcentre lies outside the triangle.

3. To draw ΔABC with AB = 6 cm, BC = 7 cm, and CA = 7 cm:
- Draw BC = 7 cm.
- With B as centre and radius 6 cm, draw an arc.
- With C as centre and radius 7 cm, draw another arc intersecting the previous arc at A.
- Join AB and AC.
- Construct perpendicular bisectors of two sides to find circumcentre O.
- Measure OA, OB, and OC; all should be equal (radius of circumcircle).

4. The least possible radius of a circle passing through two points A and B is half the distance AB, i.e., the radius of the circle with AB as diameter. — Step-by-step construction and reasoning:
1. Use ruler and protractor to draw sides and angles.
2. Use compass to draw arcs for locating points.
3. Construct perpendicular bisectors to find circumcentre.
4. Measure distances from circumcentre to vertices to confirm radius.
5. For two points, the smallest circle passing through both has AB as diameter, so radius = AB/2.

Question 2

Think, Draw and Infer

1. A, B and C are three collinear points. Can you find a point P such that PA = PB = PC? What can you say about the perpendicular bisectors of AB and BC? Draw and check. Can you show that for three collinear points A, B and C, the perpendicular bisector of AB and BC are parallel? Is it possible for a circle to pass through collinear points? Can you draw a line that cuts a given circle in three distinct points?

2. The circumcircle of a given ΔABC is drawn. Can there be other triangles congruent to ΔABC that share the same circumcircle?

Accepted Answer

1. For three collinear points A, B, and C:
- It is not possible to find a point P such that PA = PB = PC unless all three points coincide.
- The perpendicular bisectors of AB and BC are lines perpendicular to AB and BC at their midpoints.
- Since A, B, and C are collinear, these perpendicular bisectors are parallel lines.
- Because the perpendicular bisectors do not intersect, there is no unique circumcentre.
- Hence, no circle can pass through three collinear points.
- A line cannot cut a circle in three distinct points; maximum intersection points are two.

2. Yes, other triangles congruent to ΔABC can share the same circumcircle if they are rotations or reflections of ΔABC about the circumcentre, as the circumcircle depends only on the vertices lying on it, not on the triangle's orientation. — 1. By definition, the circumcentre is the intersection of perpendicular bisectors.
- For collinear points, perpendicular bisectors are parallel, so no intersection.
- No circle can pass through three collinear points.
- A line can intersect a circle at most at two points.
2. Congruent triangles can be positioned differently on the same circumcircle, sharing the same circumcentre and radius.

Question 3

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 OA and OB, making it an isosceles triangle. — 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). Hence, the triangle is isosceles.

Question 4

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 the two isosceles triangles be OAB and OCD, where O is the centre of the circle, and AB and CD are chords with equal length (AB = CD). Since OA = OB and OC = OD (radii of the circle), and AB = CD (given), by the SSS congruence criterion, triangle OAB is congruent to triangle OCD. — Both triangles have two equal sides (radii) and equal base (chord length). By SSS congruence, the triangles are congruent.

Question 5

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 circle with centre C and a chord AB. Let M be the foot of the perpendicular from C to AB.
We are given that ∠CMA = ∠CMB = 90°, meaning CM is perpendicular to AB.
To prove: AM = BM.

Proof:
Triangles CMA and CMB are right triangles with right angles at M.
CM is common to both triangles.
CA = CB since both are radii of the circle.
Therefore, by RHS (Right angle-Hypotenuse-Side) congruence criterion, ΔCMA ≅ ΔCMB.
Hence, corresponding parts are equal, so AM = BM.

Thus, the perpendicular from the centre of a circle to a chord bisects the chord. — The key idea is to use congruent triangles formed by the radius and the perpendicular from the centre to the chord. Since the two triangles share the radius and the perpendicular, and have right angles, they are congruent by RHS criterion, which implies the chord is bisected.

Question 6

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 inscribed in a circle with AB = AC (isosceles triangle).
To prove: The altitude from A to BC passes through the centre of the circle.

Proof:
Since AB = AC, triangle ABC is isosceles with vertex A.
The altitude from A to BC is also the median and angle bisector of angle A.
Let O be the centre of the circle.
Since the triangle is inscribed, O lies on the perpendicular bisector of BC.
The altitude from A to BC is perpendicular to BC.
Because AB = AC, the perpendicular bisector of BC passes through A.
Therefore, the altitude from A to BC passes through O, the centre of the circle.

Hence, the altitude from A to BC passes through the centre of the circle. — In an isosceles triangle inscribed in a circle, the altitude from the vertex angle is also the median and angle bisector, which aligns with the perpendicular bisector of the base. Since the centre of the circle lies on the perpendicular bisector of any chord, the altitude from A to BC passes through the centre.

Question 7

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
Chords AB and CD are parallel and on opposite sides of the centre O.

Let the distances of chords AB and CD from the centre O be d1 and d2 respectively.

Using the perpendicular from the centre to the chord bisects the chord (Theorem 5), let E and F be midpoints of AB and CD respectively.

In right triangle OEA:
OE = d1 (distance from centre to chord AB)
EA = half of AB = 3 cm
Using Pythagoras theorem:
OE^2 + EA^2 = OA^2
=> d1^2 + 3^2 = 5^2
=> d1^2 + 9 = 25
=> d1^2 = 16
=> d1 = 4 cm

Similarly, in right triangle OFC:
OF = d2 (distance from centre to chord CD)
FC = half of CD = 4 cm
Using Pythagoras theorem:
OF^2 + FC^2 = OC^2
=> d2^2 + 4^2 = 5^2
=> d2^2 + 16 = 25
=> d2^2 = 9
=> d2 = 3 cm

Since the chords are on opposite sides of the centre, the distance between midpoints E and F is d1 + d2 = 4 + 3 = 7 cm.

Answer: The distance between the midpoints of the chords is 7 cm. — The perpendicular from the centre to a chord bisects the chord. Using the right triangles formed by the radius, half the chord, and the distance from the centre to the chord, we find the distances of each chord from the centre. Since the chords are on opposite sides, the distance between their midpoints is the sum of these distances.

Question 8

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 the perpendicular from O meet AB at point M. Then, triangles OMA and OMB are right triangles with OM as one leg and AM and BM as the other legs respectively. Since OA = OB (radii of the circle) and OM is common, by the Pythagoras theorem, AM = BM. Hence, the perpendicular from the centre bisects the chord, proving Theorem 6. — Draw the circle with centre O and chord AB. Drop perpendicular OM from O to AB. Triangles OMA and OMB are right angled at M. OA = OB (radii). OM common. By Pythagoras theorem, OA^2 = OM^2 + AM^2 and OB^2 = OM^2 + BM^2. Since OA = OB, AM^2 = BM^2, so AM = BM. Thus, the perpendicular bisects the chord.

Question 9

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 CE ⟂ AB and CH ⟂ GH with CE = CH, we need to prove AB = GF. Since CE and CH are perpendicular distances from the centre to chords AB and GH respectively and are equal, by Theorem 7 (Chords equidistant from the centre are equal), AB = GF. — In Fig. 5.15, CE and CH are perpendiculars from the centre to chords AB and GH respectively. Given CE = CH, the chords AB and GH are equidistant from the centre. By Theorem 7, chords equidistant from the centre are equal in length. Hence, AB = GF.

Question 10

Solve the previous question using the Baudhayana-Pythagoras theorem.

Accepted Answer

To solve using the Baudhayana-Pythagoras theorem, consider triangles formed by the radii and the perpendiculars to the chords. Since CE = CH, and CE ⟂ AB, CH ⟂ GH, the right triangles formed have equal hypotenuses (radii) and one leg equal (CE = CH). By the Pythagoras theorem, the other legs (half the chords) are equal, hence AB = GF. — In triangles CEA and CHG, CE = CH (given), CA = CG (radii), and both have right angles at E and H respectively. By the Pythagoras theorem, AE^2 = CA^2 - CE^2 and GH/2^2 = CG^2 - CH^2. Since CA = CG and CE = CH, AE = GH/2. Thus, AB = 2*AE = 2*GH/2 = GH = GF.

Question 11

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

Accepted Answer

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

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

Calculate:
Length = 2 × sqrt(7² - 6²) = 2 × sqrt(49 - 36) = 2 × sqrt(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, we apply Pythagoras theorem:
Half chord length = sqrt(r² - d²).
Multiply by 2 to get full chord length.

Question 12

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{r^2 - d^2}.

Accepted Answer

Consider a circle with center 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.

OM = d (perpendicular distance from center to chord).

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

Using Pythagoras theorem in triangle OMB:
OB² = OM² + MB²
r² = d² + MB²
=> MB = sqrt(r² - d²)

Since AB = 2 × MB,
Chord length AB = 2 × sqrt(r² - d²).

Hence proved. — The perpendicular from the center to the chord bisects the chord, forming two right triangles.
Applying Pythagoras theorem gives the half chord length as sqrt(r² - d²).
Doubling it gives the full chord length.

Question 13

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 radius of the circle be r.
Let the perpendicular distance of chord AB from the center be d.
Then, the perpendicular distance of chord CD from the center is 2d.

Using the chord length formula:
Length of AB = 2 × sqrt(r² - d²)
Length of CD = 2 × sqrt(r² - (2d)²) = 2 × sqrt(r² - 4d²)

Since sqrt(r² - 4d²) is not equal to 2 × sqrt(r² - d²), we cannot conclude that CD = 2AB.

In fact, as the distance from the center increases, the chord length decreases.
Therefore, the chord closer to the center (AB) is longer than the chord farther (CD).
Hence, CD ≠ 2AB.

So, the statement is false. — The chord length depends on the perpendicular distance from the center.
If one chord is twice as far from the center as another, its length is not twice the other's length.
Because the chord length formula involves the square root of (r² - d²), the relationship is not linear.

Question 14

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. Suppose the angle is θ1. Similarly, measure the angle subtended by arc CLD at centre O, call it θ2. If θ1 < 180°, then arc AKB is a minor arc; if θ1 > 180°, then it is a major arc. Similarly, check θ2 for arc CLD. Based on the measurements, state which arcs are minor and which are major. — The angle subtended by an arc at the centre determines whether the arc is minor or major. Minor arcs subtend angles less than 180°, major arcs subtend angles greater than 180°. By measuring the angles at O for arcs AKB and CLD, we classify them accordingly.

Question 15

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 length of chord AB.

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

Triangle AOB is isosceles with sides OA = OB.

Using the formula for chord length:
Chord length AB = 2 × radius × sin(central angle/2)
= 2 × 12 × sin(60°/2)
= 24 × sin(30°)
= 24 × 1/2
= 12 cm.

Therefore, length of chord AB is 12 cm. — The chord subtends a central angle of 60°. The chord length formula is 2r sin(θ/2). Substituting r=12 cm and θ=60°, we get chord length = 12 cm.

I’m Up and Down, and Round and Round

I’m Up and Down, and Round and Round — Study Notes

I'm Up and Down, and Round and Round

5.1 DEFINITIONS

5.2 SYMMETRIES OF A CIRCLE

Practice Questions — I’m Up and Down, and Round and Round

All 8 Chapters in Mathematics