Q1. 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:

Given: Radius (r) = 13 cm, distance from centre to chord (d) = 5 cm.

Using the formula: length of chord = \( 2\sqrt{r^2 - d^2} \)

= \( 2\sqrt{13^2 - 5^2} = 2\sqrt{169 - 25} = 2\sqrt{144} = 2 \times 12 = 24 \text{ cm} \)

Therefore, the length of the chord is 24 cm.

Q2. An arc of a circle subtends an angle of \( 70^{\circ} \) at the centre. What is the measure of the angle subtended by the arc at a point on the circle?

Answer:

The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.

Given central angle = 70°.

Angle at the circumference = \( \frac{70°}{2} = 35° \)

Therefore, the angle subtended by the arc at a point on the circle is 35°.

Q3. 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:

Given: Diameter = 26 cm, so radius (r) = 13 cm. Chord length = 24 cm.

Let the distance from the centre to the chord be d.

Using the formula: chord length = \( 2\sqrt{r^2 - d^2} \)

24 = \( 2\sqrt{13^2 - d^2} \)

12 = \( \sqrt{169 - d^2} \)

144 = 169 - d²

d² = 25

d = 5 cm

Therefore, the distance from the centre to the chord is 5 cm.

Q4. 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:

Given: Radius (r) = 15 cm, distance from centre to chord (d) = 9 cm.

Using the formula: length of chord = \( 2\sqrt{r^2 - d^2} \)

= \( 2\sqrt{15^2 - 9^2} = 2\sqrt{225 - 81} = 2\sqrt{144} = 2 \times 12 = 24 \text{ cm} \)

Therefore, the length of the chord is 24 cm.

Q5. 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 the perpendicular bisector of AB be line l, intersecting AB at M.

Since M is the midpoint of AB, AM = MB.

Since l is perpendicular to AB, \( \angle OMA = \angle OMB = 90° \).

In triangles OMA and OMB:

• OM = OM (common side)
• AM = MB (M is the midpoint of AB)
• \( \angle OMA = \angle OMB = 90° \)

By the SAS congruence criterion, \( \triangle OMA \cong \triangle OMB \).

Therefore, OA = OB (corresponding parts of congruent triangles).

Since OA and OB are both radii of the circle, O is equidistant from A and B. Therefore, O lies on the perpendicular bisector of AB.

Hence, the perpendicular bisector of a chord passes through the centre of the circle.

Q6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of the \( \angle ACB \)? Explain your reasoning.

Answer:

The angle subtended by a diameter at any point on the circle is 90°.

Therefore, \( \angle ACB = 90° \).

Reason: The arc AB (not containing C) is a semicircle. The angle subtended by this arc at the centre is 180° (a straight angle). The angle subtended by the same arc at any point on the remaining part of the circle is half of 180°, which is 90°.

Q7. ABCD is a cyclic quadrilateral inscribed in a circle. If \( \angle A \) measures \( 75^{\circ} \), what is the measure of \( \angle C \)? If \( \angle B \) measures \( 110^{\circ} \), what is the measure of \( \angle D \)?

Answer:

In a cyclic quadrilateral, opposite angles are supplementary (sum to 180°).

\( \angle A + \angle C = 180° \)

75° + \( \angle C = 180° \)

\( \angle C = 180° - 75° = 105° \)

\( \angle B + \angle D = 180° \)

110° + \( \angle D = 180° \)

\( \angle D = 180° - 110° = 70° \)

Therefore, \( \angle C = 105° \) and \( \angle D = 70° \).

Q8. Quadrilateral PQRS is inscribed in a circle. If \( \angle P = (2x + 10)^{\circ} \) and \( \angle R = (3x - 20)^{\circ} \), find the value of \( x \) and the measures of \( \angle P \) and \( \angle R \).

Answer:

In a cyclic quadrilateral, opposite angles are supplementary.

\( \angle P + \angle R = 180° \)

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

5x - 10 = 180

5x = 190

x = 38

\( \angle P = 2(38) + 10 = 76 + 10 = 86° \)

\( \angle R = 3(38) - 20 = 114 - 20 = 94° \)

Therefore, x = 38, \( \angle P = 86° \), and \( \angle R = 94° \).

Q9. 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:

Given: Chord length = 16 cm, distance from centre to chord (d) = 6 cm.

Using the formula: chord length = \( 2\sqrt{r^2 - d^2} \)

16 = \( 2\sqrt{r^2 - 6^2} \)

8 = \( \sqrt{r^2 - 36} \)

64 = r² - 36

r² = 100

r = 10 cm

Therefore, the radius of the circle is 10 cm.

Q10. A cyclic quadrilateral has sides 5, 5, 12, 12 units. Find its area.

Answer:

For a cyclic quadrilateral, the area can be found using Brahmagupta's formula:

Area = \( \sqrt{(s-a)(s-b)(s-c)(s-d)} \), where s is the semi-perimeter.

Here, a = 5, b = 5, c = 12, d = 12.

s = \( \frac{5 + 5 + 12 + 12}{2} = \frac{34}{2} = 17 \)

Area = \( \sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5} = \sqrt{3600} = 60 \) square units.

Therefore, the area of the cyclic quadrilateral is 60 square units.

Q11. Consider a cyclic quadrilateral. Without drawing its circumcircle, how can we find out whether the centre of the circumcircle lies inside or outside the quadrilateral?

Answer:

For a cyclic quadrilateral, the centre of the circumcircle lies:

• Inside the quadrilateral if the quadrilateral is convex and all its interior angles are less than 180° (which is always true for a convex quadrilateral). However, specifically, the circumcentre lies inside if the quadrilateral is formed by four points on the circle that are such that the centre is within the polygon formed by them.
• Outside the quadrilateral if the quadrilateral is concave (one interior angle is greater than 180°). In a concave cyclic quadrilateral, the centre of the circle lies outside the quadrilateral.

Another way: For a cyclic quadrilateral, if one angle is greater than 180° (i.e., the quadrilateral is concave), the centre is outside. If all angles are less than 180°, the centre is inside.

Alternatively, the centre is inside the quadrilateral if the sum of any pair of opposite angles is 180° and the quadrilateral is convex.

Q12. 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 AB and CD be two equal chords of a circle intersecting at P.

We know that AB = CD (given).

Draw perpendiculars from the centre O to the chords AB and CD. Let them meet AB at M and CD at N. Then M and N are the midpoints of AB and CD respectively (Theorem 5).

Since AB = CD, AM = MB = CN = ND.

Now, triangles OMP and ONP are right-angled at M and N.

OM = ON (equal chords are equidistant from the centre, Theorem 6).

OP = OP (common side).

By the RHS congruence criterion, \( \triangle OMP \cong \triangle ONP \).

Therefore, MP = NP.

Now, AP = AM + MP and CP = CN + NP.

Since AM = CN and MP = NP, AP = CP.

Similarly, BP = DP.

Thus, the line segments of one chord are equal to the corresponding line segments of the other chord.

Q13. 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?

Answer:

Steps:

1. Draw a line segment AB of length 6 cm.
2. Draw the perpendicular bisector of AB.
3. Mark a point O on the perpendicular bisector such that the distance from O to AB is 3 cm. (This is the centre of the circle.)
4. With O as centre, draw a circle passing through A and B.

This gives the required circle.

The hint suggests that this is the circumcircle of an isosceles triangle OAB, where OA = OB = radius, and the height from O to AB is 3 cm. The length of AB is 6 cm, so AM = MB = 3 cm.

Using the Baudhayana-Pythagoras theorem, radius = \( \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2} \) cm.

Q14. Show that rectangle is the only parallelogram that can be inscribed in a circle.

Answer:

Let ABCD be a parallelogram inscribed in a circle.

In a parallelogram, opposite angles are equal.

Since ABCD is cyclic, opposite angles are supplementary.

Let \( \angle A = \angle C \) and \( \angle A + \angle C = 180° \).

Therefore, \( 2\angle A = 180° \), so \( \angle A = 90° \).

Similarly, \( \angle B = \angle D = 90° \).

Thus, all angles of the parallelogram are right angles.

A parallelogram with all right angles is a rectangle.

Therefore, the only parallelogram that can be inscribed in a circle is a rectangle.

Q15. 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.

In a rectangle, the diagonals are equal and bisect each other.

Let the diagonals intersect at O. Then OA = OB = OC = OD.

Since ABCD is inscribed in a circle, the circle passes through A, B, C, and D.

Because OA = OB = OC = OD, O is equidistant from all four vertices of the rectangle.

By the definition of a circle, the point equidistant from all points on the circle is the centre.

Therefore, O, the point of intersection of the diagonals, is the centre of the circle.

Q16. Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

Answer:

For a fixed chord length, the distance from the centre to the chord is also fixed (since chord length = \( 2\sqrt{r^2 - d^2} \)).

All chords of a given length are at a fixed distance from the centre.

The midpoints of all such chords will lie at that fixed distance from the centre.

Therefore, the shape formed by the midpoints of all chords of a fixed length is a concentric circle (a circle with the same centre) whose radius is the fixed distance from the centre to the chord.

Q17. 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 \( \angle BAC \)."

Answer:

Given: AB = AC.

Since AB and AC are equal chords, they are equidistant from the centre. Let the perpendiculars from O to AB and AC meet AB and AC at M and N respectively.

Then OM = ON.

In right triangles OAM and OAN:

• OA = OA (common side)
• OM = ON (proved above)

By the RHS congruence criterion, \( \triangle OAM \cong \triangle OAN \).

Therefore, \( \angle OAM = \angle OAN \) (corresponding parts of congruent triangles).

This means that AO bisects \( \angle BAC \).

Hence, the centre O lies on the angle bisector of \( \angle BAC \).

Q18. 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 circle have centre O. Let the chords be AB = 10 cm and CD = 24 cm, on the same side of the centre.

Let the perpendicular distances from O to AB and CD be d₁ and d₂ respectively, with d₁ < d₂ (since AB is shorter, it's farther from the centre).

Using the formula: chord length = \( 2\sqrt{r^2 - d^2} \)

For AB: 10 = \( 2\sqrt{r^2 - d_1^2} \) → 5 = \( \sqrt{r^2 - d_1^2} \) → r² - d₁² = 25 ... (1)

For CD: 24 = \( 2\sqrt{r^2 - d_2^2} \) → 12 = \( \sqrt{r^2 - d_2^2} \) → r² - d₂² = 144 ... (2)

Given that the distance between the chords is 7 cm, d₂ - d₁ = 7 (since they are on the same side).

From (1): d₁² = r² - 25

From (2): d₂² = r² - 144

Subtracting: d₂² - d₁² = (r² - 144) - (r² - 25) = -119

But (d₂ - d₁)(d₂ + d₁) = d₂² - d₁² = -119

Since d₂ - d₁ = 7, we get 7(d₂ + d₁) = -119 → d₂ + d₁ = -17, which is impossible (distances cannot be negative).

Let me reconsider: The lengths are 10 cm and 24 cm, and the distance between the chords is 7 cm. Since they are on the same side, the distance between the chords is d₂ - d₁ = 7.

From the equations:

d₁² = r² - 25

d₂² = r² - 144

Subtracting: d₂² - d₁² = -119

7(d₂ + d₁) = -119 → d₂ + d₁ = -17, which is impossible.

This means that the chord of length 24 cm cannot be on the same side as the chord of length 10 cm with the given distance between them, or my assumption about which chord is closer is wrong.

Since a longer chord is closer to the centre, d₂ < d₁. So d₁ - d₂ = 7.

Then d₁² - d₂² = 119

(d₁ - d₂)(d₁ + d₂) = 119

7(d₁ + d₂) = 119 → d₁ + d₂ = 17

Solving d₁ - d₂ = 7 and d₁ + d₂ = 17, we get d₁ = 12, d₂ = 5.

From r² = d₁² + 25 = 144 + 25 = 169, so r = 13 cm.

Therefore, the radius of the circle is 13 cm.

Q19. 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 inscribed in a circle can be divided into 6 equilateral triangles, each with side length equal to the radius r of the circle.

The side length of the hexagon is equal to the radius of the circle, so side = r.

The distance from the centre to each side is the altitude of one of these equilateral triangles.

In an equilateral triangle of side r, the altitude is \( \frac{\sqrt{3}}{2}r \).

Therefore, the distance of each side from the centre is \( \frac{\sqrt{3}}{2}r \).

Q20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about \( \angle MOP \) and \( \angle MNP \)? Explain your reasoning.

Answer:

Since MN is a diameter, \( \angle MON = 180° \) (a straight angle).

\( \angle MOP \) is an angle subtended by arc MP at the centre. Without more information, we cannot determine its exact measure.

However, \( \angle MNP \) is an angle subtended by arc MP at point N on the circle.

By Theorem 9, the angle subtended by an arc at the centre is twice the angle subtended at the circumference.

So, \( \angle MOP = 2 \angle MNP \).

Also, since MN is a diameter, \( \angle MPN = 90° \) (angle in a semicircle).

Q21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., \( \angle CDE = \angle ABC \), where E is a point on the extension of side CD).

Answer:

In a cyclic quadrilateral, the sum of opposite angles is 180°.

\( \angle ABC + \angle ADC = 180° \).

Since E is on the extension of side CD, \( \angle ADC + \angle CDE = 180° \) (linear pair).

From these two equations: \( \angle ABC + \angle ADC = \angle ADC + \angle CDE \).

Therefore, \( \angle ABC = \angle CDE \).

So, the exterior angle at any vertex is equal to the interior opposite angle.

Q22. "There is no chord of a circle that is longer than its diameter." How do you justify this statement?

Answer:

The diameter is the longest chord of a circle.

Let AB be a chord of a circle with centre O. Then OA = OB = r (radius).

In triangle OAB, by the triangle inequality theorem, AB < OA + OB = r + r = 2r.

The diameter is 2r.

Therefore, AB < 2r (if AB is not the diameter).

If AB is a diameter, then AB = 2r.

Thus, no chord can be longer than the diameter.

Q23. 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 radius r. Let A be a point inside the circle. Consider a chord passing through A. Let the perpendicular distance from O to this chord be d.

The length of the chord is \( 2\sqrt{r^2 - d^2} \).

For the chord to be as short as possible, \( \sqrt{r^2 - d^2} \) must be as small as possible, which means d must be as large as possible.

The maximum possible value of d is OA (when the chord is perpendicular to OA).

Therefore, the shortest chord passing through A is the one perpendicular to OA.

Q24. How would you use the following figure to justify the statement that the angle in a semicircle is \( 90^{\circ} \)? [IMAGE NEEDED: Figure 5.30 from the textbook showing a semicircle]

Answer:

In the figure, AB is the diameter. Let C be any point on the circle.

The arc AB (the semicircle) subtends an angle of 180° at the centre O.

By Theorem 9, the angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circle.

Therefore, \( \angle ACB = \frac{1}{2} \times 180° = 90° \).

Thus, the angle in a semicircle is always 90°.

Q25. 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 the circle have centre O and diameter AB. Chords CC' and DD' are perpendicular to AB.

Since CC' is perpendicular to AB, and CD is a chord of the circle, the perpendicular from the centre to a chord bisects the chord.

Let M be the midpoint of CD, and M' be the midpoint of C'D'.

We need to prove that MM' is perpendicular to AB.

Since CC' is perpendicular to AB, C and C' are at equal distances from AB. Similarly, D and D' are at equal distances from AB.

Thus, the midpoints M and M' of CD and C'D' lie on the line parallel to the diameters, and the segment joining them is perpendicular to AB.

This is a known result that follows from the symmetry of the circle and the properties of chords.

Q26. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is \( 180^{\circ} \)? [IMAGE NEEDED: Figure 5.31 from the textbook showing a cyclic quadrilateral]

Answer:

In the figure, ABCD is a cyclic quadrilateral inscribed in a circle.

Consider the angles \( \angle A \) and \( \angle C \).

\( \angle A \) subtends arc BCD, and \( \angle C \) subtends arc BAD.

Together, arcs BCD and BAD make the entire circle.

The angle subtended by the entire circle at the centre is 360°.

By Theorem 9, the angle subtended by an arc at the centre is twice the angle subtended at the circumference.

Therefore, \( \angle A + \angle C = \frac{1}{2} \times 360° = 180° \).

Similarly, \( \angle B + \angle D = 180° \).

Thus, the sum of opposite angles of a cyclic quadrilateral is 180°.