NCERT Solutions · Class 9 Mathematics I’m Up and Down, and Round and Round

Perpendicular from the centre to a chord.Let \(\displaystyle O \) be the centre and \(\displaystyle AB \) the chord. "Distance from the centre" always means the perpendicular distance, so drop \(\displaystyle OM \perp AB \), with \(\displaystyle OM = 5 \) cm and \(\displaystyle OA = 13 \) cm (a radius).The reason this helps is a theorem from this chapter: the perpendicular from the centre to a chord bisects the chord. So \(\displaystyle M \) is the midpoint of \(\displaystyle AB \), and \[AB = 2\,AM. \]Triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \), so Pythagoras' theorem gives \[AM^{2} = OA^{2} - OM^{2} = 13^{2} - 5^{2} = 169 - 25 = 144, \] \[AM = 12 \text{ cm}. \]Therefore \(\displaystyle AB = 2 \times 12 = 24 \) cm.Check: \(\displaystyle 5,\,12,\,13 \) is a right triangle, since \(\displaystyle 25 + 144 = 169 \). And $\displaystyle 24$ cm is less than the diameter $\displaystyle 26$ cm, as any chord must be.The chord is $\displaystyle 24$ cm long.

Angle at the centre is twice the angle at the circle.Let the arc be \(\displaystyle AB \) and the centre \(\displaystyle O \), so that \(\displaystyle \angle AOB = 70^{\circ} \). Take a point \(\displaystyle P \) on the remaining part of the circle (that is, on the other arc, not on arc \(\displaystyle AB \) itself) and join \(\displaystyle PA \) and \(\displaystyle PB \).The theorem of this chapter says: the angle an arc subtends at the centre is twice the angle it subtends at any point on the remaining part of the circle. So \[\angle APB = \tfrac{1}{2}\,\angle AOB = \tfrac{1}{2}\times 70^{\circ} = 35^{\circ}. \]Two things worth noticing: The arc subtends \(\displaystyle 35^{\circ} \) at a point on the remaining part of the circle.

Perpendicular from the centre to a chord.The diameter is $\displaystyle 26$ cm, so the radius is \[r = \tfrac{26}{2} = 13 \text{ cm}. \]Let \(\displaystyle O \) be the centre, \(\displaystyle AB \) the chord of length $\displaystyle 24$ cm, and \(\displaystyle OM \perp AB \). Since the perpendicular from the centre bisects the chord, \[AM = \tfrac{24}{2} = 12 \text{ cm}. \]Triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \), so \[OM^{2} = OA^{2} - AM^{2} = 13^{2} - 12^{2} = 169 - 144 = 25, \] \[OM = 5 \text{ cm}. \]Check by working backwards: a chord $\displaystyle 5$ cm from the centre of a $\displaystyle 13$ cm circle has half-length \(\displaystyle \sqrt{169-25} = 12 \), so full length $\displaystyle 24$ cm. Correct.The chord is $\displaystyle 5$ cm from the centre.

Perpendicular from the centre to a chord.Let \(\displaystyle O \) be the centre, \(\displaystyle AB \) the chord, and \(\displaystyle OM \perp AB \), so \(\displaystyle OM = 9 \) cm and \(\displaystyle OA = 15 \) cm.The perpendicular from the centre bisects the chord, so \(\displaystyle M \) is the midpoint of \(\displaystyle AB \) and triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \): \[AM^{2} = OA^{2} - OM^{2} = 15^{2} - 9^{2} = 225 - 81 = 144, \] \[AM = 12 \text{ cm}. \]Hence \[AB = 2 \times 12 = 24 \text{ cm}. \]Check: \(\displaystyle 9, 12, 15 \) is just the \(\displaystyle 3,4,5 \) triangle scaled by $\displaystyle 3$, and \(\displaystyle 81 + 144 = 225 \). Also \(\displaystyle 24 < 30 \), the diameter, as it must be.The chord is $\displaystyle 24$ cm long.

Points equidistant from two points lie on their perpendicular bisector.Let \(\displaystyle AB \) be any chord of a circle with centre \(\displaystyle O \).Short proof. \(\displaystyle OA \) and \(\displaystyle OB \) are both radii, so \[OA = OB. \] So \(\displaystyle O \) is a point equidistant from \(\displaystyle A \) and \(\displaystyle B \). But the set of all points equidistant from \(\displaystyle A \) and \(\displaystyle B \) is exactly the perpendicular bisector of \(\displaystyle AB \). Hence \(\displaystyle O \) lies on the perpendicular bisector of \(\displaystyle AB \) — that is, the perpendicular bisector of the chord passes through the centre.The same thing with congruent triangles, in case you would rather see every step. Let \(\displaystyle M \) be the midpoint of \(\displaystyle AB \), and join \(\displaystyle OA \), \(\displaystyle OB \), \(\displaystyle OM \). In triangles \(\displaystyle OMA \) and \(\displaystyle OMB \): So \(\displaystyle \triangle OMA \cong \triangle OMB \) by SSS, and therefore \(\displaystyle \angle OMA = \angle OMB \). These two angles form a linear pair, so they add up to \(\displaystyle 180^{\circ} \): \[\angle OMA + \angle OMB = 180^{\circ} \quad\Longrightarrow\quad 2\,\angle OMA = 180^{\circ} \quad\Longrightarrow\quad \angle OMA = 90^{\circ}. \]So the line \(\displaystyle OM \) passes through the midpoint of \(\displaystyle AB \) and is perpendicular to \(\displaystyle AB \) — it is the perpendicular bisector of \(\displaystyle AB \). Since a segment has only one perpendicular bisector, the centre \(\displaystyle O \) lies on it.Why this is useful: if you are handed a circle with no marked centre, draw any two non-parallel chords and construct their perpendicular bisectors. Each must pass through the centre, so the point where they meet is the centre.Hence the perpendicular bisector of a chord always passes through the centre of the circle.

Angle in a semicircle.Let \(\displaystyle O \) be the centre. Since \(\displaystyle AB \) is a diameter, \(\displaystyle A \), \(\displaystyle O \), \(\displaystyle B \) lie on one straight line, so \[\angle AOB = 180^{\circ}. \]Reason $\displaystyle 1$ (using the central-angle theorem). The arc \(\displaystyle AB \) that does not contain \(\displaystyle C \) is a semicircle, and it subtends \(\displaystyle 180^{\circ} \) at the centre. An arc subtends at a point on the remaining part of the circle half the angle it subtends at the centre, so \[\angle ACB = \tfrac{1}{2} \times 180^{\circ} = 90^{\circ}. \]Reason $\displaystyle 2$ (from scratch, using only isosceles triangles). Join \(\displaystyle OC \). Then \(\displaystyle OA = OB = OC \), all radii. Now in triangle \(\displaystyle ABC \), the three angles are \(\displaystyle \angle A = x \), \(\displaystyle \angle B = y \), and \(\displaystyle \angle ACB = \angle OCA + \angle OCB = x + y \). Their sum is \(\displaystyle 180^{\circ} \): \[x + y + (x+y) = 180^{\circ} \quad\Longrightarrow\quad 2(x+y) = 180^{\circ} \quad\Longrightarrow\quad x + y = 90^{\circ}. \] So \(\displaystyle \angle ACB = 90^{\circ} \).This is true wherever \(\displaystyle C \) sits on the circle (as long as \(\displaystyle C \) is not \(\displaystyle A \) or \(\displaystyle B \)) — sliding \(\displaystyle C \) around changes \(\displaystyle x \) and \(\displaystyle y \), but never their sum.\(\displaystyle \angle ACB = 90^{\circ} \): the angle in a semicircle is a right angle.

Opposite angles of a cyclic quadrilateral are supplementary.Why the rule is true. \(\displaystyle \angle A \) stands on the arc \(\displaystyle BCD \) (the arc from \(\displaystyle B \) to \(\displaystyle D \) not containing \(\displaystyle A \)), and \(\displaystyle \angle C \) stands on the arc \(\displaystyle DAB \). Each inscribed angle is half its arc, and the two arcs together make the whole circle, \(\displaystyle 360^{\circ} \). So \[\angle A + \angle C = \tfrac{1}{2}\big(\text{arc } BCD + \text{arc } DAB\big) = \tfrac{1}{2}\times 360^{\circ} = 180^{\circ}. \] The same argument gives \(\displaystyle \angle B + \angle D = 180^{\circ} \).Applying it. \[\angle C = 180^{\circ} - \angle A = 180^{\circ} - 75^{\circ} = 105^{\circ}, \] \[\angle D = 180^{\circ} - \angle B = 180^{\circ} - 110^{\circ} = 70^{\circ}. \]Check: the four angles of any quadrilateral add to \(\displaystyle 360^{\circ} \), and indeed \[75^{\circ} + 110^{\circ} + 105^{\circ} + 70^{\circ} = 360^{\circ}. \]\(\displaystyle \angle C = 105^{\circ} \) and \(\displaystyle \angle D = 70^{\circ} \).

Opposite angles of a cyclic quadrilateral are supplementary.In the cyclic quadrilateral \(\displaystyle PQRS \), the angles \(\displaystyle \angle P \) and \(\displaystyle \angle R \) are opposite, so \[\angle P + \angle R = 180^{\circ}. \]Substituting the given expressions: \[(2x + 10)^{\circ} + (3x - 20)^{\circ} = 180^{\circ}, \] \[5x - 10 = 180, \] \[5x = 190, \] \[x = 38. \]Now put \(\displaystyle x = 38 \) back into each expression: \[\angle P = (2 \times 38 + 10)^{\circ} = (76 + 10)^{\circ} = 86^{\circ}, \] \[\angle R = (3 \times 38 - 20)^{\circ} = (114 - 20)^{\circ} = 94^{\circ}. \]Check: \(\displaystyle 86^{\circ} + 94^{\circ} = 180^{\circ} \). Both angles are positive and less than \(\displaystyle 180^{\circ} \), so this is a genuine quadrilateral.\(\displaystyle x = 38 \), \(\displaystyle \angle P = 86^{\circ} \), \(\displaystyle \angle R = 94^{\circ} \).

Perpendicular from the centre to a chord.Let \(\displaystyle O \) be the centre, \(\displaystyle AB \) the chord of length $\displaystyle 16$ cm, and \(\displaystyle OM \perp AB \), with \(\displaystyle OM = 6 \) cm.The perpendicular from the centre bisects the chord, so \[AM = \tfrac{16}{2} = 8 \text{ cm}. \]Triangle \(\displaystyle OMA \) is right angled at \(\displaystyle M \), and \(\displaystyle OA \) is the radius \(\displaystyle r \): \[r^{2} = OM^{2} + AM^{2} = 6^{2} + 8^{2} = 36 + 64 = 100, \] \[r = 10 \text{ cm}. \]Check: \(\displaystyle 6, 8, 10 \) is the \(\displaystyle 3,4,5 \) triangle doubled. And the radius $\displaystyle 10$ cm must be more than half the chord ($\displaystyle 8$ cm) — it is.The radius of the circle is $\displaystyle 10$ cm.

Equal chords cut off equal arcs, then split the quadrilateral by a diagonal.Call the quadrilateral \(\displaystyle ABCD \), taken round the circle with \[AB = 5,\quad BC = 5,\quad CD = 12,\quad DA = 12. \]Step $\displaystyle 1$: the arcs. In one circle, equal chords cut off equal arcs. Since \(\displaystyle AB = BC \), arc \(\displaystyle AB = \) arc \(\displaystyle BC \); call each of them \(\displaystyle p \). Since \(\displaystyle CD = DA \), arc \(\displaystyle CD = \) arc \(\displaystyle DA \); call each of them \(\displaystyle q \). Going once round the circle uses up all four arcs: \[p + p + q + q = 360^{\circ} \quad\Longrightarrow\quad 2(p+q) = 360^{\circ} \quad\Longrightarrow\quad p + q = 180^{\circ}. \]Step $\displaystyle 2$: a right angle appears. The angle \(\displaystyle \angle DAB \) is the inscribed angle at \(\displaystyle A \). It stands on the arc going from \(\displaystyle B \) to \(\displaystyle D \) the long way round through \(\displaystyle C \), which is arc \(\displaystyle BC \) + arc \(\displaystyle CD \) \(\displaystyle = p + q = 180^{\circ} \). An inscribed angle is half its arc, so \[\angle DAB = \tfrac{1}{2}\times 180^{\circ} = 90^{\circ}. \] And since \(\displaystyle ABCD \) is cyclic, \(\displaystyle \angle BCD = 180^{\circ} - 90^{\circ} = 90^{\circ} \) as well.Step $\displaystyle 3$: two right triangles. Draw the diagonal \(\displaystyle BD \). It cuts \(\displaystyle ABCD \) into \(\displaystyle \triangle ABD \) and \(\displaystyle \triangle CBD \), and both have a right angle: (As a bonus, \(\displaystyle \angle DAB = 90^{\circ} \) means \(\displaystyle BD \) is a diameter, so the circle has radius \(\displaystyle 6.5 \) units.)Step $\displaystyle 4$: add the areas. \[\text{Area} = \tfrac{1}{2}(5)(12) + \tfrac{1}{2}(5)(12) = 30 + 30 = 60. \]What if the sides come in the order \(\displaystyle 5, 12, 5, 12 \) instead? Then opposite sides are equal, so the quadrilateral is a parallelogram; and a parallelogram inscribed in a circle must be a rectangle (see the question on that). A \(\displaystyle 5 \times 12 \) rectangle has area \(\displaystyle 5 \times 12 = 60 \) — the same answer. That is not a coincidence: for a cyclic quadrilateral the area depends only on the four side lengths, not on the order they are arranged in.The area is $\displaystyle 60$ square units.