Solutions - Chapter 5: I'm Up and Down, and Round and Round

Chapter 5: Solutions & Answers

Exercise Set 5.1

1. Draw \(\Delta ABC\) with \(AB = 5\text{ cm}\), \(\angle A = 70^\circ\), and \(\angle B = 60^\circ\). Draw the circumcircle of \(\Delta ABC\). Is the centre inside or outside the triangle?

Answer: Inside the triangle.

Reasoning: The third angle \(\angle C = 180^\circ - (70^\circ + 60^\circ) = 50^\circ\). Since all three angles are acute (\(< 90^\circ\)), \(\Delta ABC\) is an acute-angled triangle. As established in the chapter, the circumcentre of an acute-angled triangle always lies inside the triangle.

2. Draw \(\Delta ABC\) with \(AB = 5\text{ cm}\), \(\angle A = 100^\circ\), \(AC = 4\text{ cm}\). Draw the circumcircle of \(\Delta ABC\). Is the centre inside or outside the triangle?

Answer: Outside the triangle.

Reasoning: Since \(\angle A = 100^\circ > 90^\circ\), \(\Delta ABC\) is an obtuse-angled triangle. The circumcentre of an obtuse-angled triangle always lies outside the triangle.

3. Draw \(\Delta ABC\), with \(AB = 6\text{ cm}\), \(BC = 7\text{ cm}\), and \(CA = 7\text{ cm}\). Draw the circumcircle of \(\Delta ABC\). Let the circumcentre be \(O\). Measure \(OA, OB, OC\).

Answer: \(OA = OB = OC \approx 3.7\text{ cm}\) (or measured as per construction scale).

Reasoning: Since \(O\) is the circumcentre, it is equidistant from all three vertices \(A, B,\) and \(C\). Thus, \(OA = OB = OC = \text{circumradius}\).

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

Answer: \(\frac{1}{2} AB\)

Reasoning: For a circle passing through \(A\) and \(B\), the segment \(AB\) is a chord. The smallest circle through \(A\) and \(B\) is the one where \(AB\) forms the diameter of the circle. Therefore, its radius is half the length of \(AB\).

Exercise Set 5.2

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

Proof: Let \(AB\) be a chord of a circle with centre \(O\). Join \(OA\) and \(OB\). In \(\Delta OAB\), \(OA\) and \(OB\) are radii of the same circle. Therefore, \(OA = OB\). Since two sides of \(\Delta OAB\) are equal, \(\Delta OAB\) is an isosceles triangle.

2. Show that if two such isosceles triangles have equal base length, they are congruent to each other.

Proof: Let \(\Delta OAB\) and \(\Delta CDE\) be formed by chords \(AB\) and \(DE\) in circles with radius \(r\). We are given that \(AB = DE\). In \(\Delta OAB\) and \(\Delta CDE\):

\(OA = CD = r\)
\(OB = CE = r\)
\(AB = DE\) (Given)

By SSS Congruence Criterion, \(\Delta OAB \cong \Delta CDE\).

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?

Proof: Let \(C\) be the centre, \(AB\) be the chord, and \(CM \perp AB\). Join \(CA\) and \(CB\). In right-angled triangles \(\Delta CMA\) and \(\Delta CMB\):

\(\angle CMA = \angle CMB = 90^\circ\) (Given)
\(CA = CB = r\) (Hypotenuse, Radii)
\(CM = CM\) (Common side)

By RHS Congruence Criterion, \(\Delta CMA \cong \Delta CMB\). Hence, \(AM = BM\) (by CPCT). Therefore, the perpendicular bisects the chord.

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.

Proof: In an isosceles triangle \(ABC\) where \(AB = AC\), the altitude from \(A\) to \(BC\) is also the perpendicular bisector of the base \(BC\). Since the perpendicular bisector of any chord of a circle passes through its centre, the perpendicular bisector of chord \(BC\) (which is the altitude from \(A\)) must pass 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.

Solution: Let \(O\) be the centre and \(r = 5\text{ cm}\). For chord \(AB = 6\text{ cm}\), its midpoint \(M\) gives \(AM = 3\text{ cm}\). Distance from centre \(OM = \sqrt{r^2 - AM^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = 4\text{ cm}\).
For chord \(CD = 8\text{ cm}\), its midpoint \(N\) gives \(CN = 4\text{ cm}\). Distance from centre \(ON = \sqrt{r^2 - CN^2} = \sqrt{5^2 - 4^2} = \sqrt{25 - 16} = 3\text{ cm}\).
Since the chords lie on opposite sides of the centre, distance \(MN = OM + ON = 4\text{ cm} + 3\text{ cm} = 7\text{ cm}\).

Exercise Set 5.4

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

Proof: Let two chords be \(AB = 2a\) and \(FG = 2a\) in a circle of radius \(r\) with centre \(C\). Perpendicular distances from \(C\) to chords \(AB\) and \(FG\) are \(CE\) and \(CH\) respectively. Since perpendicular bisects the chord, half lengths are \(AE = a\) and \(FH = a\). Applying Pythagoras Theorem: \(CE^2 = CA^2 - AE^2 = r^2 - a^2\) \(CH^2 = CF^2 - FH^2 = r^2 - a^2\) Since \(CE^2 = CH^2\), we get \(CE = CH\). Thus, equal chords are equidistant from the centre.

2. If \(CE \perp AB\), \(CH \perp GH\), and \(CE = CH\), show that \(AB = GF\).

Proof: In right triangles \(\Delta CEA\) and \(\Delta CHF\):

\(\angle CEA = \angle CHF = 90^\circ\)
\(CA = CF = r\) (Hypotenuse)
\(CE = CH\) (Given)

By RHS Congruence Criterion, \(\Delta CEA \cong \Delta CHF\). Therefore, \(AE = FH\) (CPCT). Since perpendicular from centre bisects the chord, \(AB = 2AE\) and \(GF = 2FH\). Thus, \(AB = GF\).

3. Solve the previous question using the Baudhāyana-Pythagoras theorem.

Proof: In right triangles \(\Delta CEA\) and \(\Delta CHF\): \(AE^2 = CA^2 - CE^2 = r^2 - d^2\) \(FH^2 = CF^2 - CH^2 = r^2 - d^2\) (since \(CE = CH = d\)) Thus, \(AE = FH \implies 2AE = 2FH \implies AB = GF\).

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.

Solution: Using the formula \(L = 2\sqrt{r^2 - d^2}\): \(L = 2\sqrt{7^2 - 6^2} = 2\sqrt{49 - 36} = 2\sqrt{13}\text{ cm} \approx 7.21\text{ cm}\).

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 \(2\sqrt{r^2-d^2}\).

Explanation: Let \(O\) be the centre, \(AB\) be the chord, and \(OM \perp AB\) of length \(d\). Join \(OA = r\). In right triangle \(\Delta OMA\), by Pythagoras theorem, \(AM^2 = OA^2 - OM^2 = r^2 - d^2 \implies AM = \sqrt{r^2 - d^2}\). Since the perpendicular from centre bisects the chord, length of chord \(AB = 2 \times AM = 2\sqrt{r^2 - d^2}\).

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.

Answer: No, we cannot conclude that \(CD = 2AB\).

Reason: Let distance of \(CD\) be \(d\), so distance of \(AB\) is \(2d\). Length of \(CD = 2\sqrt{r^2 - d^2}\) and Length of \(AB = 2\sqrt{r^2 - 4d^2}\). Ratio \(\frac{CD}{AB} = \frac{\sqrt{r^2 - d^2}}{\sqrt{r^2 - 4d^2}} \neq 2\). The relationship between chord length and distance is non-linear (involves square roots), so doubling the distance does not halve the chord length.

Exercise Set 5.6

1. In a circle with centre \(O\), the central angle \(AOB\) is \(60^\circ\). If the radius of the circle is 12 cm, what is the length of the chord \(AB\)?

Solution: In \(\Delta OAB\), \(OA = OB = 12\text{ cm}\). So \(\angle OAB = \angle OBA\). Sum of angles in \(\Delta OAB = 180^\circ \implies \angle OAB + \angle OBA = 180^\circ - 60^\circ = 120^\circ\). Since \(\angle OAB = \angle OBA\), each is \(60^\circ\). Thus, \(\Delta OAB\) is an equilateral triangle. Therefore, length of chord \(AB = OA = 12\text{ cm}\).

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

Answers:
(i) No: Angles subtended by the same arc at points on the same side of the circle are equal (\(\angle AXB = \angle AYB\)).
(ii) Yes: Points subtending equal angles from line segment \(AB\) on the same side must lie on the same arc of the circle.
(iii) Yes: By Theorem 10 (Concyclicity), if segment \(AB\) subtends equal angles at \(X\) and \(Y\) on the same side, then \(A, B, X, Y\) lie on the same circle.

3. Find \(x\) in Fig. 5.26 (Central angle \(\angle AOC = 100^\circ\)).

Solution: By Theorem 9, the angle subtended by an arc at the centre is double the angle subtended by it at any point on the circle. \(\angle AOC = 2 \times \angle ABC\) \(100^\circ = 2x \implies x = 50^\circ\).

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?

Solution: \(L = 2\sqrt{r^2 - d^2} = 2\sqrt{13^2 - 5^2} = 2\sqrt{169 - 25} = 2\sqrt{144} = 2 \times 12 = 24\text{ cm}\).

2. 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?

Solution: Angle on circle \(= \frac{1}{2} \times \text{Central Angle} = \frac{70^\circ}{2} = 35^\circ\).

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: Radius \(r = \frac{26}{2} = 13\text{ cm}\). Half chord length \(a = \frac{24}{2} = 12\text{ cm}\). Distance \(d = \sqrt{r^2 - a^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5\text{ cm}\).

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?

Solution: Half chord \(= \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}\). Length of chord \(= 2 \times 12 = 24\text{ cm}\).

5. Prove that the perpendicular bisector of a chord passes through the centre of the circle.

Proof: The locus of all points equidistant from the endpoints of a line segment is its perpendicular bisector. For any chord \(AB\) of a circle, the centre \(O\) is equidistant from \(A\) and \(B\) because \(OA = OB = \text{radius}\). Therefore, the centre \(O\) must lie on the perpendicular bisector of chord \(AB\).

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

Answer: \(\angle ACB = 90^\circ\).

Reasoning: By Corollary to Theorem 9, the angle subtended by a diameter at any point on the circle is a right angle (\(90^\circ\)).

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

Solution: Opposite angles of a cyclic quadrilateral add up to \(180^\circ\).
\(\angle C = 180^\circ - \angle A = 180^\circ - 75^\circ = 105^\circ\).
\(\angle D = 180^\circ - \angle B = 180^\circ - 110^\circ = 70^\circ\).

8. 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 \(\angle R\).

Solution: Since \(PQRS\) is cyclic, \(\angle P + \angle R = 180^\circ\).
\((2x + 10) + (3x - 20) = 180\)
\(5x - 10 = 180 \implies 5x = 190 \implies x = 38\).
\(\angle P = 2(38) + 10 = 76 + 10 = 86^\circ\).
\(\angle R = 3(38) - 20 = 114 - 20 = 94^\circ\).

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: Half chord length \(= 8\text{ cm}\), distance \(d = 6\text{ cm}\). Radius \(r = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10\text{ cm}\).

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

Solution: Using Brahmagupta's Formula for cyclic quadrilateral area \(=\sqrt{(s-a)(s-b)(s-c)(s-d)}\): Semi-perimeter \(s = \frac{5 + 5 + 12 + 12}{2} = 17\). \(\text{Area} = \sqrt{(17-5)(17-5)(17-12)(17-12)} = \sqrt{12 \times 12 \times 5 \times 5} = 12 \times 5 = 60\text{ square units}\).

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?

Answer: Check if any angle of the quadrilateral is obtuse or acute relative to its diagonals, or examine the position of the circumcentre relative to its triangles. Specifically, if all internal angles subtended by diagonals are acute, the centre lies inside. If one of the triangles formed by three vertices is obtuse, its circumcentre lies outside that triangle, which determines its placement relative to the quadrilateral.

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.

Proof: Let chords \(AB = CD\) intersect at \(P\). Drop perpendiculars \(OM \perp AB\) and \(ON \perp CD\) from centre \(O\). Since equal chords are equidistant from centre, \(OM = ON\). In right triangles \(\Delta OMP\) and \(\Delta ONP\): \(OP\) is hypotenuse (common), \(OM = ON\). By RHS, \(\Delta OMP \cong \Delta ONP \implies MP = NP\). Since perpendicular bisects chord, \(AM = MB = CN = ND = \frac{1}{2}AB\). Now, \(AP = AM + MP = CN + NP = CP\). And \(PB = AB - AP = CD - CP = PD\). Thus, corresponding segments are equal (\(AP = CP\) and \(PB = PD\)).

13. Draw a circle in which a chord of 6 cm length stands at a distance of 3 cm from the centre.

Construction Step / Calculation: Radius \(r = \sqrt{3^2 + 3^2} = \sqrt{18} = 3\sqrt{2}\text{ cm} \approx 4.24\text{ cm}\). Draw a circle of radius \(3\sqrt{2}\text{ cm}\), draw a perpendicular line segment of 3 cm from centre, and construct a chord of 6 cm perpendicular to it at its end.

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

Proof: Let \(ABCD\) be a parallelogram inscribed in a circle. In a parallelogram, opposite angles are equal: \(\angle A = \angle C\). Since \(ABCD\) is cyclic, opposite angles sum to \(180^\circ\): \(\angle A + \angle C = 180^\circ\). Thus, \(\angle A + \angle A = 180^\circ \implies 2\angle A = 180^\circ \implies \angle A = 90^\circ\). A parallelogram with one angle equal to \(90^\circ\) is a rectangle. Hence, any inscribed parallelogram must be a rectangle.

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.

Proof: Each angle of a rectangle is \(90^\circ\). The diagonal divides the rectangle into right-angled triangles inscribed in the circle. By Corollary to Theorem 9, a chord subtending a \(90^\circ\) angle at the circumference must be a diameter. Thus, both diagonals are diameters of the circle. Since all diameters intersect at the centre, the intersection of the diagonals lies at 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?

Answer: A concentric circle.

Reasoning: By Theorem 6, all chords of equal fixed length are at an equal distance \(d\) from the centre. The locus of midpoints of these chords consists of all points at distance \(d\) from the centre, which forms a smaller circle concentric with the given circle.

17. In a circle with centre \(O\), chords \(AB\) and \(AC\) are congruent. Explain why the centre of the circle lies on the angle bisector of \(\angle BAC\).

Proof: Join \(OB\) and \(OC\). In \(\Delta OAB\) and \(\Delta OAC\):

\(AB = AC\) (Given equal chords)
\(OA = OA\) (Common)
\(OB = OC\) (Radii)

By SSS Congruence, \(\Delta OAB \cong \Delta OAC\). Therefore, \(\angle OAB = \angle OAC\) (CPCT). This shows that \(AO\) bisects \(\angle BAC\), so the centre \(O\) lies on the angle bisector of \(\angle BAC\).

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.

Solution: Let radius be \(r\). Half chord lengths are \(5\text{ cm}\) and \(12\text{ cm}\). Let distance of longer chord (24 cm) from centre be \(x\). Distance of shorter chord (10 cm) from centre is \(x + 7\). Using Pythagoras theorem: \(r^2 = x^2 + 12^2 = x^2 + 144\) \(r^2 = (x + 7)^2 + 5^2 = x^2 + 14x + 49 + 25 = x^2 + 14x + 74\) Equating both: \(x^2 + 144 = x^2 + 14x + 74 \implies 14x = 70 \implies x = 5\text{ cm}\). Radius \(r = \sqrt{5^2 + 144} = \sqrt{25 + 144} = \sqrt{169} = 13\text{ cm}\).

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.

Solution:
1. Side length: A regular hexagon divides the central angle into 6 equal parts of \(\frac{360^\circ}{6} = 60^\circ\). Each formed triangle with the centre is equilateral with sides \(r\). Thus, Side length \(= r\).
2. Distance from centre: Height of the equilateral triangle of side \(r\) is \(\frac{\sqrt{3}}{2}r\).

20. 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: \(\angle MPN = 90^\circ\) (angle in a semicircle). In cyclic quadrilateral \(MNOP\), \(\angle MOP + \angle MNP = 180^\circ\) because they are opposite angles.

21. Let \(ABCD\) be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., \(\angle CDE = \angle ABC\)).

Proof: Since \(ABCD\) is cyclic, \(\angle ABC + \angle ADC = 180^\circ\). Also, \(\angle ADC + \angle CDE = 180^\circ\) (linear pair along straight line \(ADE\)). Equating the two: \(\angle ABC + \angle ADC = \angle ADC + \angle CDE \implies \angle CDE = \angle ABC\).

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

Justification: Length of any chord at distance \(d \ge 0\) from centre is \(2\sqrt{r^2 - d^2}\). Since \(d^2 \ge 0\), \(2\sqrt{r^2 - d^2} \le 2\sqrt{r^2} = 2r = \text{diameter}\). The maximum value occurs when \(d = 0\), which gives the length equal to the diameter.

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 \(OA\).

Proof: Let \(PQ\) be a chord through \(A\) perpendicular to \(OA\). The distance of \(PQ\) from centre \(O\) is \(OA\). For any other chord \(RS\) passing through \(A\) not perpendicular to \(OA\), drop perpendicular \(OM \perp RS\). In right triangle \(\Delta OMA\), hypotenuse \(OA > OM\). Since chord length \(L = 2\sqrt{r^2 - d^2}\), a greater distance \(d\) gives a smaller chord length. Since distance \(OA > OM\), chord \(PQ\) has greater distance from centre than \(RS\), making \(PQ\) shorter than \(RS\).

24. How would you use Fig. 5.30 to justify the statement that the angle in a semicircle is \(90^\circ\)?

Justification: In \(\Delta ABC\), join centre \(O\) to vertex \(C\). \(OA = OB = OC = r\). In \(\Delta AOC\), \(\angle OAC = \angle OCA = a\). In \(\Delta BOC\), \(\angle OBC = \angle OCB = b\). Total angle \(\angle ACB = a + b\). Sum of angles in \(\Delta ABC = a + b + (a + b) = 2(a + b) = 180^\circ \implies a + b = 90^\circ\). Thus, \(\angle ACB = 90^\circ\).

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 \(AB\).

Proof: Since \(CC' \perp AB\) and \(DD' \perp AB\), chords \(CC'\) and \(DD'\) are parallel. By reflection symmetry of the circle across diameter \(AB\), \(AB\) bisects both chords \(CC'\) and \(DD'\) perpendicularly. The figure \(CD D'C'\) forms an isosceles trapezium symmetric about \(AB\). The line segment connecting the midpoints \(M\) and \(M'\) lies along the line of symmetry, which is perpendicular to \(AB\).

26. How would you use Fig. 5.31 to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is \(180^\circ\)?

Justification: Let central angles subtended by arcs be \(u\) and \(v\). Complete angle at centre \(u + v = 360^\circ\). Inscribed opposite angle \(\angle A = \frac{1}{2}u\) and \(\angle C = \frac{1}{2}v\). Sum of opposite angles \(\angle A + \angle C = \frac{1}{2}u + \frac{1}{2}v = \frac{1}{2}(u + v) = \frac{1}{2}(360^\circ) = 180^\circ\).