NCERT Solutions for Class 9 Maths Chapter 5 — End-of-Chapter Exercises#

Here $r = 13$ cm and $d = 5$ cm. We find the length of the chord:

$$\text{Chord} = 2\sqrt{r^2 - d^2}$$

$$\text{Chord} = 2\sqrt{13^2 - 5^2}$$

$$\text{Chord} = 2\sqrt{169 - 25}$$

$$\text{Chord} = 2\sqrt{144}$$

$$\text{Chord} = 24 \text{ cm}$$

∴ The chord is 24 cm long.

By Theorem 9, the angle at a point on the circle (outside the arc) is half the angle at the centre:

$$\text{Angle at the circle} = \dfrac{1}{2} \times 70°$$

$$\text{Angle at the circle} = 35°$$

∴ The arc subtends 35° at a point on the circle (on the remaining part of the circle).

The radius is half the diameter:

$$r = \dfrac{26}{2} = 13 \text{ cm}$$

The perpendicular from the centre bisects the chord, so half the chord is 12 cm. In the right triangle formed by the radius, the half-chord and the distance $d$:

$$d^2 = r^2 - 12^2$$

$$d^2 = 169 - 144$$

$$d^2 = 25$$

$$\Rightarrow d = 5 \text{ cm}$$

∴ The chord is 5 cm from the centre.

Here $r = 15$ cm and $d = 9$ cm. We find the length of the chord:

$$\text{Chord} = 2\sqrt{r^2 - d^2}$$

$$\text{Chord} = 2\sqrt{225 - 81}$$

$$\text{Chord} = 2\sqrt{144}$$

$$\text{Chord} = 24 \text{ cm}$$

∴ The chord is 24 cm long.

Let $AB$ be a chord of a circle with centre $O$, and let $M$ be the midpoint of $AB$.

By Theorem 4, the line joining the centre $O$ to the midpoint $M$ is perpendicular to $AB$. So $OM$ is a line through $M$ perpendicular to $AB$.

At the point $M$, there is only one line perpendicular to $AB$ — the perpendicular bisector of $AB$. So the line $OM$ is the perpendicular bisector.

∴ The perpendicular bisector of $AB$ passes through the centre $O$.

Another way: $OA = OB$ (radii), so $O$ is equally far from $A$ and $B$. Every point equally far from $A$ and $B$ lies on the perpendicular bisector of $AB$, so $O$ lies on it.

$AB$ is a diameter, so the arc $AB$ that does not contain $C$ is a semicircle. It subtends a straight angle at the centre:

$$\angle AOB = 180°$$

By Theorem 9, the angle at $C$ on the circle is half of this:

$$\angle ACB = \dfrac{1}{2} \times 180°$$

$$\angle ACB = 90°$$

∴ $\angle ACB = 90°$. The angle in a semicircle is a right angle.

By Theorem 11, opposite angles of a cyclic quadrilateral add up to 180°.

$\angle A$ and $\angle C$ are opposite angles, so:

$$\angle A + \angle C = 180°$$

$$75° + \angle C = 180°$$

$$\Rightarrow \angle C = 105°$$

$\angle B$ and $\angle D$ are opposite angles, so:

$$\angle B + \angle D = 180°$$

$$110° + \angle D = 180°$$

$$\Rightarrow \angle D = 70°$$

∴ $\angle C = 105°$ and $\angle D = 70°$.

$\angle P$ and $\angle R$ are opposite angles of the cyclic quadrilateral $PQRS$, and opposite angles add up to 180°, so:

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

$$\Rightarrow 5x - 10 = 180$$

$$\Rightarrow 5x = 190$$

$$\Rightarrow x = 38$$

Substituting $x = 38$:

$$\angle P = 2(38) + 10 = 86°$$

$$\angle R = 3(38) - 20 = 94°$$

∴ $x = 38$, $\angle P = 86°$ and $\angle R = 94°$. (Check: $86° + 94° = 180°$ ✓)

Half the chord is 8 cm, and the distance is 6 cm. The radius is the hypotenuse of the right triangle they form:

$$r^2 = 8^2 + 6^2$$

$$r^2 = 64 + 36$$

$$r^2 = 100$$

$$\Rightarrow r = 10 \text{ cm}$$

∴ The radius is 10 cm.

There are two ways to arrange the sides in order. Both give the same area.

Case 1: sides in the order 5, 12, 5, 12. Opposite sides are equal, so the quadrilateral is a parallelogram. A parallelogram has equal opposite angles; in a cyclic quadrilateral they also add up to 180°, so each is 90°. So it is a rectangle of sides 5 and 12:

$$\text{Area} = 5 \times 12$$

$$\text{Area} = 60 \text{ square units}$$

Case 2: sides in the order 5, 5, 12, 12 — a kite $ABCD$ with $AB = AD = 5$ and $CB = CD = 12$.

$\triangle ABC \cong \triangle ADC$ by SSS ($AB = AD$, $CB = CD$, $AC$ common), so $\angle B = \angle D$. In a cyclic quadrilateral $\angle B + \angle D = 180°$, so each is 90°. The kite is made of two right triangles with legs 5 and 12:

$$\text{Area} = 2 \times \dfrac{1}{2} \times 5 \times 12$$

$$\text{Area} = 60 \text{ square units}$$

∴ The area is 60 square units.

Look at the angle each side subtends at a vertex on the other side of it — for side $AB$ of $ABCD$, the angle $\angle ACB$ (or $\angle ADB$, which is equal, by Theorem 9).

• If a side subtends an obtuse angle, the arc on the other side of that side is more than a semicircle, so the centre lies on the far side of that side: the centre is outside the quadrilateral.
• If a side subtends a right angle, that side is a diameter, so the centre lies on that side.
• If every side subtends an acute angle, the centre is inside the quadrilateral.

The best way: for each side, measure the angle it subtends at the vertex across a diagonal: $\angle ACB$ for side $AB$, $\angle BDC$ for $BC$, $\angle CAD$ for $CD$ and $\angle DBA$ for $DA$. If all four are acute, the centre is inside; if one is 90°, the centre is on that side; if one is obtuse, the centre is outside, beyond that side. Only a protractor is needed — no circle has to be drawn.

Let the equal chords $AB$ and $CD$ of a circle with centre $O$ intersect at $P$, with $A$ and $C$ on the same side. Draw $OM \perp AB$ and $ON \perp CD$, and join $OP$.

Step 1. By Theorem 6, equal chords are equidistant from the centre:

$$OM = ON$$

Step 2. In right triangles $\triangle OMP$ and $\triangle ONP$: $OM = ON$, $OP$ is common, and $\angle OMP = \angle ONP = 90°$. By RHS congruence, $\triangle OMP \cong \triangle ONP$, so:

$$MP = NP$$

Step 3. By Theorem 5, $M$ and $N$ are the midpoints of the equal chords, so:

$$AM = \dfrac{AB}{2} = \dfrac{CD}{2} = CN$$

Step 4. Adding:

$$AP = AM + MP$$

$$CP = CN + NP$$

$$\Rightarrow AP = CP$$

And since $AB = CD$:

$$PB = AB - AP = CD - CP = PD$$

∴ $AP = CP$ and $PB = PD$ — the parts of one chord are equal to the corresponding parts of the other.

The radius must be:

$$r^2 = 3^2 + 3^2$$

$$r^2 = 18$$

$$\Rightarrow r = 3\sqrt{2} \approx 4.24 \text{ cm}$$

Construction (using the hint): the circle we want is the circumcircle of a right isosceles triangle with legs of 6 cm.

1. Draw $AB = 6$ cm. At $B$, draw $BC \perp AB$ with $BC = 6$ cm, and join $AC$.
2. $\angle ABC = 90°$, so the hypotenuse $AC$ is a diameter of the circumcircle (angle in a semicircle). Mark its midpoint $O$.
3. With centre $O$ and radius $OA$, draw the circle. It passes through $A$, $B$ and $C$.

Check:

$$AC^2 = 6^2 + 6^2 = 72$$

$$\Rightarrow AC = 6\sqrt{2}$$

So the radius is $OA = 3\sqrt{2}$ cm. The chord $AB = 6$ cm has half-length 3 cm, so its distance from $O$ is:

$$d^2 = (3\sqrt{2})^2 - 3^2 = 18 - 9 = 9$$

$$\Rightarrow d = 3 \text{ cm}$$

∴ In this circle, the chord $AB$ of length 6 cm is at a distance of 3 cm from the centre, as required.

Let $ABCD$ be a parallelogram inscribed in a circle.

In a parallelogram, opposite angles are equal:

$$\angle A = \angle C$$

In a cyclic quadrilateral, opposite angles add up to 180° (Theorem 11):

$$\angle A + \angle C = 180°$$

Substituting $\angle C = \angle A$:

$$2\angle A = 180°$$

$$\Rightarrow \angle A = 90°$$

A parallelogram with one right angle has all four angles 90°, so it is a rectangle.

∴ A parallelogram inscribed in a circle must be a rectangle. (And every rectangle can be inscribed, since its opposite angles add up to 180°, by Theorem 12.)

Let $ABCD$ be a rectangle inscribed in a circle with centre $O$.

$\angle ABC = 90°$. By Theorem 9, the arc $AC$ (not containing $B$) subtends double this at the centre:

$$\angle AOC = 2 \times 90° = 180°$$

So $A$, $O$, $C$ lie on a straight line: $AC$ is a diameter and passes through $O$. In the same way, $\angle BCD = 90°$ shows that $BD$ is a diameter and passes through $O$.

Two different diameters meet only at the centre.

∴ The diagonals $AC$ and $BD$ intersect at the centre $O$ of the circle.

Let the circle have centre $O$ and radius $r$, and let every chord have length $l$. By Theorem 6, all these chords are at the same distance from the centre:

$$d = \sqrt{r^2 - \dfrac{l^2}{4}}$$

The distance from $O$ to a chord is measured along the perpendicular, which meets the chord at its midpoint (Theorem 5). So every midpoint is at the same distance $d$ from $O$.

∴ The midpoints form a circle with the same centre $O$ and radius $\sqrt{r^2 - \dfrac{l^2}{4}}$ — a smaller concentric circle. (If the chords are diameters, the “circle” shrinks to the single point $O$.)

Join $OA$, $OB$ and $OC$. Consider $\triangle OAB$ and $\triangle OAC$:

Part	Reason
$AB = AC$	given (congruent chords)
$OB = OC$	radii
$OA = OA$	common side

By SSS congruence, $\triangle OAB \cong \triangle OAC$, so the corresponding angles at $A$ are equal:

$$\angle OAB = \angle OAC$$

So $AO$ divides $\angle BAC$ into two equal parts.

∴ $AO$ is the bisector of $\angle BAC$, and the centre $O$ lies on it.

Let the radius be $r$, and let the 24 cm chord be at distance $x$ from the centre. The shorter chord is farther from the centre (Theorem 8), and the chords are on the same side, 7 cm apart, so the 10 cm chord is at distance $x + 7$.

Half the 24 cm chord is 12 cm, and its distance is $x$, so:

$$r^2 = 12^2 + x^2 \quad \cdots (1)$$

Half the 10 cm chord is 5 cm, and its distance is $x + 7$, so:

$$r^2 = 5^2 + (x + 7)^2 \quad \cdots (2)$$

Both are equal to $r^2$, so:

$$144 + x^2 = 25 + (x + 7)^2$$

$$\Rightarrow 144 + x^2 = 25 + x^2 + 14x + 49$$

$$\Rightarrow 144 = 74 + 14x$$

$$\Rightarrow 14x = 70$$

$$\Rightarrow x = 5$$

Substituting $x = 5$ in equation (1):

$$r^2 = 12^2 + x^2$$

$$r^2 = 144 + 25$$

$$r^2 = 169$$

$$\Rightarrow r = 13 \text{ cm}$$

∴ The radius is 13 cm. (The chords are 5 cm and 12 cm from the centre.)

Join the centre $O$ to the six vertices. This makes six congruent triangles, so each central angle is:

$$\dfrac{360°}{6} = 60°$$

Each triangle has two sides equal to $r$ (radii) and a 60° angle between them, so its base angles are also 60°: each triangle is equilateral.

∴ Each side of the hexagon is $r$.

The distance of a side from the centre is the perpendicular to that side. With half-side $\dfrac{r}{2}$ and radius $r$:

$$d^2 = r^2 - \left(\dfrac{r}{2}\right)^2$$

$$d^2 = \dfrac{3r^2}{4}$$

$$\Rightarrow d = \dfrac{\sqrt{3}}{2}r$$

∴ Each side is at a distance of $\dfrac{\sqrt{3}}{2}r$ (about $0.87r$) from the centre.

∠MOP and ∠MNP are equal. Both are angles subtended by the chord $MP$ at points $O$ and $N$ of the circle. In the quadrilateral $MNOP$, the vertices $N$ and $O$ come between $M$ and $P$, so they lie on the same arc, on the same side of $MP$. Angles in the same segment are equal (Theorem 9):

$$\angle MOP = \angle MNP$$

Using that MN is a diameter: the angles subtended by the diameter $MN$ at the other two vertices are angles in a semicircle:

$$\angle MON = 90°$$

$$\angle MPN = 90°$$

So the quadrilateral has right angles at $O$ and $P$.

$E$ is on $CD$ extended beyond $D$, so $\angle ADE$ and $\angle ADC$ form a straight angle. Taking the exterior angle at $D$ as $\angle ADE$:

$$\angle ADE + \angle ADC = 180° \quad \cdots (1)$$

$ABCD$ is cyclic, so by Theorem 11 its opposite angles add up to 180°:

$$\angle ABC + \angle ADC = 180° \quad \cdots (2)$$

Comparing equation (1) and equation (2):

$$\Rightarrow \angle ADE = 180° - \angle ADC = \angle ABC$$

∴ The exterior angle at $D$ equals the interior opposite angle $\angle ABC$. The same argument works at every vertex.

Let $AB$ be any chord of a circle with centre $O$ and radius $r$.

If $AB$ passes through $O$, it is a diameter, of length $2r$.

If $AB$ does not pass through $O$, then $O$, $A$, $B$ form a triangle. In a triangle, any side is shorter than the sum of the other two:

$$AB < OA + OB$$

$$AB < r + r$$

$$AB < 2r$$

∴ Every chord is at most $2r$, the length of a diameter. (Also, from Theorem 8 and the formula $2\sqrt{r^2 - d^2}$: the chord is longest when $d = 0$, that is, when it passes through the centre.)

Take any chord through $A$, and let $M$ be the foot of the perpendicular from $O$ to it. Its distance from the centre is $OM$, and its length is $2\sqrt{r^2 - OM^2}$.

The chord passes through $A$. If $M \neq A$, then $\triangle OMA$ is right-angled at $M$ with hypotenuse $OA$, so:

$$OM < OA$$

If $M = A$ (that is, the chord is perpendicular to $OA$), then:

$$OM = OA$$

So $OM$ is largest, equal to $OA$, exactly when the chord is perpendicular to $OA$. A chord farther from the centre is shorter (Theorem 8), so this chord is the shortest.

∴ The shortest chord through $A$ is the one perpendicular to $OA$, and its length is $2\sqrt{r^2 - OA^2}$.

In Fig. 5.30, $PQ$ is a diameter with centre $O$, and $A$ is on the semicircle. The base angles are $a$ at $P$ and $b$ at $Q$, and $OA$ is joined.

$OP = OA = OQ$ (all radii), so the figure splits into two isosceles triangles.

In $\triangle OPA$, $OP = OA$, so the angles opposite them are equal:

$$\angle OAP = \angle OPA = a$$

In $\triangle OQA$, $OQ = OA$, so:

$$\angle OAQ = \angle OQA = b$$

So the angle at $A$ is:

$$\angle PAQ = a + b$$

The angles of $\triangle PAQ$ add up to 180°:

$$a + b + (a + b) = 180°$$

$$\Rightarrow 2(a + b) = 180°$$

$$\Rightarrow a + b = 90°$$

∴ $\angle PAQ = 90°$: the angle in a semicircle is a right angle.

Place the circle on a coordinate plane with its centre at the origin and the diameter $AB$ along the x-axis.

The chord $CC'$ is perpendicular to the diameter $AB$, so the diameter — which passes through the centre — is the perpendicular from the centre to $CC'$. By Theorem 5, it bisects $CC'$. So $C$ and $C'$ are mirror images in the x-axis:

$$C = (c, h), \quad C' = (c, -h)$$

In the same way, $D$ and $D'$ are mirror images:

$$D = (e, k), \quad D' = (e, -k)$$

By the midpoint formula:

$$M = \left(\dfrac{c + e}{2}, \dfrac{h + k}{2}\right)$$

$$M' = \left(\dfrac{c + e}{2}, \dfrac{-h - k}{2}\right)$$

$M$ and $M'$ have the same $x$-coordinate, so the segment $MM'$ is vertical, that is, parallel to the y-axis.

∴ $MM'$ is perpendicular to the x-axis, which is the diameter $AB$. (In fact $M'$ is the mirror image of $M$ in $AB$.)

In Fig. 5.31, $ABCD$ is inscribed in a circle with centre $O$, and $O$ is joined to all four vertices.

$OA = OB = OC = OD$ (radii), so the four triangles $OAB$, $OBC$, $OCD$ and $ODA$ are all isosceles, and each has equal base angles. Name these base angles:

Triangle	Equal base angles
$\triangle OAB$	$p$ at $A$ and $p$ at $B$
$\triangle OBC$	$q$ at $B$ and $q$ at $C$
$\triangle OCD$	$u$ at $C$ and $u$ at $D$
$\triangle ODA$	$v$ at $D$ and $v$ at $A$

Each angle of the quadrilateral is made of two of these:

$$\angle A = p + v, \quad \angle B = p + q, \quad \angle C = q + u, \quad \angle D = u + v$$

The angles of the quadrilateral add up to 360°:

$$(p + v) + (p + q) + (q + u) + (u + v) = 360°$$

$$\Rightarrow 2(p + q + u + v) = 360°$$

$$\Rightarrow p + q + u + v = 180°$$

So, for a pair of opposite angles:

$$\angle A + \angle C = (p + v) + (q + u)$$

$$\angle A + \angle C = p + q + u + v$$

$$\angle A + \angle C = 180°$$

In the same way, $\angle B + \angle D = 180°$.

∴ The opposite angles of a cyclic quadrilateral add up to 180°. (This argument uses the case where the centre is inside the quadrilateral, as in the figure.)