I'm Up and Down, and Round and Round — NCERT Solutions

Let the perpendicular from the centre to the chord bisect the chord. Then half the chord, the distance from the centre, and the radius form a right triangle.

Radius $r = 13$ cm, distance from centre $d = 5$ cm.

Half-chord $= \sqrt{r^2 - d^2} = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12$ cm.

So chord length $= 2 \times 12 = 24$ cm.

Diameter $= 26$ cm, so radius $r = 13$ cm.

Let the perpendicular distance from the centre to the chord be $d$.
The chord length is 24 cm, so half-chord $= 12$ cm.

Using the right triangle formed by radius, half-chord, and perpendicular distance:

$$r^2 = d^2 + 12^2$$
$$13^2 = d^2 + 144$$
$$169 = d^2 + 144$$
$$d^2 = 25$$
$$d = 5$$

So the distance from the centre to the chord is 5 cm.

Radius $r = 15$ cm and distance from centre to chord $d = 9$ cm.

Half the chord length is

$$\sqrt{r^2 - d^2} = \sqrt{15^2 - 9^2} = \sqrt{225 - 81} = \sqrt{144} = 12\text{ cm}$$

So the full chord length is

$$2 \times 12 = 24\text{ cm}$$

Let the chord be $AB$ and let its perpendicular bisector meet the circle at some point(s). We know from the chapter that the centre of a circle is equidistant from the endpoints of any chord.

So if $O$ is the centre, then $OA = OB$. A point equidistant from $A$ and $B$ lies on the perpendicular bisector of $AB$. Therefore, the centre $O$ lies on the perpendicular bisector of the chord.

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

Let the radius be $r$ and the perpendicular distance from the centre to the chord be $6$ cm.

Half the chord is $16/2 = 8$ cm.

Using the right triangle formed:

$$r^2 = 6^2 + 8^2 = 36 + 64 = 100$$

So,

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

The sides are $5, 5, 12, 12$. Since it is cyclic, opposite angles are supplementary. A cyclic quadrilateral with sides in pairs like this is a kite-like configuration, and its area can be found by Brahmagupta’s formula:

$$\text{Area} = \sqrt{(s-a)(s-b)(s-c)(s-d)}$$
where
$$s = \frac{5+5+12+12}{2} = 17.$$

So,
$$\text{Area} = \sqrt{(17-5)(17-5)(17-12)(17-12)}$$
$$= \sqrt{12 \cdot 12 \cdot 5 \cdot 5}$$
$$= \sqrt{3600} = 60$$

Hence the area is 60 square units.

A cyclic quadrilateral’s circumcentre can be located by drawing the perpendicular bisectors of two sides (or of the diagonals’ sides). Their intersection is the circumcentre.

To tell whether it lies inside or outside the quadrilateral, use the chapter’s fact about the circumcentre of a triangle:

• If the quadrilateral is such that the circumcentre lies inside, all four vertices are around it.
• If one pair of opposite angles is obtuse, the centre may lie outside.

In practice, the best way is to examine the quadrilateral’s angles: if both opposite angles are less than $180^\circ$ and the figure is symmetric around the centre, the centre is inside; otherwise it may be outside. The chapter specifically notes that for triangles the circumcentre is inside for acute triangles, outside for obtuse triangles, and at the midpoint of the hypotenuse for right triangles.

Let the equal intersecting chords be divided into segments $x$ and $y$ on one chord, and $p$ and $q$ on the other. Since the two chords are equal,

$$x+y = p+q.$$

Also, for intersecting chords, the products of the segments are equal:

$$x\cdot y = p\cdot q.$$

If two pairs of numbers have the same sum and the same product, then they are the same pair up to order. Therefore, the corresponding segments are equal.

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

In a parallelogram, opposite angles are equal. If it is inscribed in a circle, then opposite angles of a cyclic quadrilateral are supplementary. So each pair of opposite angles is both equal and supplementary.

If two angles are equal and their sum is $180^\circ$, each must be $90^\circ$.

Therefore all angles of the parallelogram are right angles, so the parallelogram is a rectangle.

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

In a rectangle, the diagonals bisect each other. If the rectangle is inscribed in a circle, then each diagonal is a diameter of the circle, because the opposite angles are right angles and subtend semicircles.

The intersection point of the diagonals is the midpoint of both diagonals. Since each diagonal is a diameter, their midpoint is the centre of the circle.

So the diagonals intersect at the centre of the circumcircle.

For all chords of a fixed length, the chapter states that their midpoints are all at the same distance from the centre. The set of all such midpoints forms a locus, and that locus is a circle.

Since chords $AB$ and $AC$ are congruent, they subtend equal angles at the centre of the circle.

So if $O$ is the centre, then

$$\angle AOB = \angle AOC.$$

Thus $AO$ divides $\angle BAC$ into two equal angles, which means $AO$ is the angle bisector of $\angle BAC$.

Therefore, the centre of the circle lies on the angle bisector of $\angle BAC$.

Let the radius be $r$.

For the chord of length 10 cm, half-chord $= 5$ cm.
For the chord of length 24 cm, half-chord $= 12$ cm.

Since the chords are on the same side of the centre, the longer chord is closer to the centre. Let their distances from the centre be $d_1$ for 10 cm and $d_2$ for 24 cm, with $d_1-d_2=7$.

Using right triangles:
$$r^2 = 5^2 + d_1^2$$
$$r^2 = 12^2 + d_2^2$$

So,
$$25 + d_1^2 = 144 + d_2^2$$
$$d_1^2 - d_2^2 = 119$$
$$(d_1-d_2)(d_1+d_2)=119$$

Given $d_1-d_2=7$,
$$7(d_1+d_2)=119$$
$$d_1+d_2=17$$

Now solve:
$$d_1=12,\ d_2=5$$

Then
$$r^2 = 5^2 + 12^2 = 25 + 144 = 169$$
$$r=13\text{ cm}$$

In a regular hexagon inscribed in a circle, each side subtends an angle of $60^\circ$ at the centre. The chord of $60^\circ$ in a circle of radius $r$ has length

$$2r\sin 30^\circ = 2r\cdot \frac12 = r.$$

So the side length is $r$.

The perpendicular from the centre to a side bisects the side and forms a $30^\circ$-$60^\circ$-$90^\circ$ triangle. The distance from the centre to each side is

$$r\cos 30^\circ = r\cdot \frac{\sqrt3}{2}$$

But for a regular hexagon, the apothem is actually the perpendicular distance to the side, so from the geometry of the 60° central angle, the distance is

$$r\cos 30^\circ = \frac{\sqrt3}{2}r.$$

So the answers are: side $= r$, distance from centre $= \frac{\sqrt3}{2}r$.

Since $MN$ is a diameter, any angle subtended by $MN$ at a point on the circle is a right angle.

Therefore,

• $\angle MOP$ is the angle subtended at the centre by chord $MP$ and depends on the arc.
• $\angle MNP$ is an angle subtended by chord $MP$ at point $N$ on the circle.

If the question intended the angle subtended by the diameter $MN$, then the correct conclusion is:

$$\angle MPN = 90^\circ$$

because an angle in a semicircle is a right angle.

So the relevant statement from the chapter is that any angle standing on diameter $MN$ at the circumference is $90^\circ$.

Let $ABCD$ be a cyclic quadrilateral and let $E$ be a point on the extension of $CD$.

The exterior angle at $D$ is

$$\angle CDE = 180^\circ - \angle CDA.$$

But in a cyclic quadrilateral, opposite angles are supplementary:

$$\angle ABC + \angle CDA = 180^\circ.$$

So,

$$\angle ABC = 180^\circ - \angle CDA = \angle CDE.$$

Hence the exterior angle at any vertex of a cyclic quadrilateral equals the interior opposite angle.

The diameter is the longest chord of a circle. This follows from the chapter result that the chord nearest the centre is the longest one; the chord containing the centre has distance zero from the centre, so it is the diameter.

Any other chord lies at a positive distance from the centre, so it must be shorter than the diameter. Therefore, no chord can be longer than the diameter.

Let the circle have centre $O$, and let a chord through point $A$ be drawn. Among all chords through $A$, the shortest one is the chord whose distance from the centre is greatest.

Now the line through $A$ that is perpendicular to $OA$ is the chord farthest from the centre, because the perpendicular from the centre to a chord gives the shortest distance to that chord. Since greater distance from the centre means a shorter chord, this chord is the shortest.

Thus, the shortest chord through $A$ is the one perpendicular to $OA$.

Let the diameter be $AB$ and let the point on the semicircle be $C$. The arc $AB$ is a semicircle, so the angle subtended by the arc at the centre is a straight angle, $180^\circ$.

By the theorem in the chapter, the angle subtended by an arc at any point on the circle outside the arc is half the angle subtended at the centre.

Therefore,

$$\angle ACB = \frac{1}{2} \times 180^\circ = 90^\circ.$$

So the angle in a semicircle is a right angle.

Let the diameter $AB$ be the $x$-axis with centre at the origin. Since the chords $CC'$ and $DD'$ are perpendicular to the diameter, each chord is perpendicular to $AB$, so both chords are vertical lines.

Because a line through the centre perpendicular to a chord bisects the chord, the diameter through the centre bisects each of these chords. So the midpoint of $CC'$ lies on $AB$, and the midpoint of $DD'$ lies on $AB$.

Now let the midpoint of chord $CD$ be $M$, and the midpoint of chord $C'D'$ be $M'$. Since $C$ and $D$ lie on one vertical chord and $C'$ and $D'$ lie on the other, the points $M$ and $M'$ have the same horizontal coordinate as the corresponding chords. Hence the segment joining their midpoints is also vertical, while $AB$ is horizontal.

Therefore, $MM'$ is perpendicular to $AB$.

So the required segment is perpendicular to the diameter.

Let $ABCD$ be the cyclic quadrilateral, and let $O$ be the centre of the circle.

Consider the arc $BCD$. Point $A$ lies on the circle outside this arc, so by the theorem on angles subtended by an arc,
$$\angle BAD = \tfrac12(\text{angle subtended by arc }BCD \text{ at } O).$$
This is the reflex angle $\angle BOD$.

Similarly, the opposite angle $\angle BCD$ is half the angle subtended by the opposite arc $BAD$ at the centre.

So,
$$\angle BAD + \angle BCD = \tfrac12(360^\circ)=180^\circ.$$

Thus, the sum of opposite angles of a cyclic quadrilateral is $180^\circ$.

Since $AB$ is a diameter, the arc from $A$ to $B$ subtends a straight angle at the centre, i.e. $180^\circ$.

Point $C$ lies on the circle on the remaining part of the circle, so the angle subtended by the diameter at $C$ is half of $180^\circ$:
$$\angle ACB=\frac12\times 180^\circ=90^\circ.$$

So, the angle in a semicircle is a right angle.

In a cyclic quadrilateral, opposite angles are supplementary.

So,
$$\angle A+\angle C=180^\circ$$
$$75^\circ+\angle C=180^\circ$$
$$\angle C=180^\circ-75^\circ=105^\circ$$

Also,
$$\angle B+\angle D=180^\circ$$
$$110^\circ+\angle D=180^\circ$$
$$\angle D=180^\circ-110^\circ=70^\circ$$

So the required angles are $105^\circ$ and $70^\circ$.

In a cyclic quadrilateral, opposite angles are supplementary:
$$\angle P+\angle R=180^\circ$$
Given
$$\angle P=(2x+10)^\circ,\qquad \angle R=(3x-20)^\circ$$
So,
$$(2x+10)+(3x-20)=180$$
$$5x-10=180$$
$$5x=190$$
$$x=38$$
Now substitute:
$$\angle P=2(38)+10=86^\circ$$
$$\angle R=3(38)-20=94^\circ$$

So the correct values are $x=38$, $\angle P=86^\circ$, and $\angle R=94^\circ$.

If the printed expression was meant exactly as written, then these are the computed values.

Use the angles on the same segment idea.

For a cyclic quadrilateral, if we want to know whether the centre of the circumcircle lies inside or outside the quadrilateral, we can examine the type of triangle/angles formed by its vertices:

• If the quadrilateral has a pair of opposite angles summing to $180^\circ$, then it is cyclic.
• To locate the centre, draw the perpendicular bisectors of two sides. Their intersection is the circumcentre.
• Then observe whether this point lies inside or outside the quadrilateral.

A useful school-level rule from the chapter is:

• for an acute-angled triangle, the circumcentre lies inside;
• for an obtuse-angled triangle, the circumcentre lies outside;
• for a right-angled triangle, it lies at the midpoint of the hypotenuse.

So the best way is to construct the perpendicular bisectors of two sides and see where they meet. That point is the circumcentre, and its position tells us whether it lies inside or outside the quadrilateral.