NCERT Solutions for Class 9 Mathematics Chapter 5 - I’m Up and Down, and Round and Round

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NCERT Solutions for Class 9 Maths Chapter 5 I’m Up and Down, and Round and Round: Exercise Questions

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Think and Reflect (Page 94)

Question 1. What are the rotational symmetries of a square? How many lines of reflection symmetry does it have? What about a regular pentagon? A regular hexagon?

$\textbf{Answer:}$
We know that an object is symmetrical if it looks exactly the same after rotation.
Square:
A square can be rotated about its centre and still look exactly the same at
90°, 180°, 270°, and 360°
So, a square has 4 rotational symmetries (order 4).
It also has 4 lines of reflection symmetry: 2 diagonals and 2 lines joining the midpoints of opposite sides

Regular Pentagon:
A regular pentagon can be rotated about its centre and still look exactly the same at: 72°, 144°, 216°, 288°, and 360°
So, a regular pentagon has 5 rotational symmetries (order 5).
It also has 5 lines of reflection symmetry.

Regular Hexagon:
A regular hexagon can be rotated about its centre and still look exactly the same at: 60°, 120°, 180°, 240°, 300° and 360°
So, a regular pentagon has 6 rotational symmetries (order 6).
It also has 6 lines of reflection symmetry.

Question 2. What is the length of the longest chord in a circle of radius 5 units? Is there a smallest chord?

$\textbf{Answer:}$
The longest chord of a circle is the diameter.
We know that Diameter = 2 × Radius
Here radius = 5 units,
So, the longest chord of that circle, diameter = 2 × 5 = 10 units

A circle has infinitely many chords of different lengths, and we can draw chords shorter and shorter continuously. Hence, there is no smallest chord.

Question 3. The locus of points at a given distance from a given point is a circle. What can we say about the locus of points equidistant from two given points?
(Hint: We know that any point that is equidistant from two given points A and B lies on the perpendicular bisector of AB. Does this make the perpendicular bisector the locus? For this, we have to show that all the points on the perpendicular bisector are equidistant from A and B.)

$\textbf{Answer:}$
Let the two points of the circle be A and B.
We know that any point equidistant from A and B lies on the perpendicular bisector of AB.
Now we check the converse of the theorem:
Take any point P on the perpendicular bisector of AB.
So, by the property of perpendicular bisectors, PA = PB
So every point on the perpendicular bisector is equidistant from A and B.
Hence, the set (locus) of all points equidistant from two given points is:
The perpendicular bisector of the line segment joining the two points.

Question 1. Draw $\triangle \mathrm{ABC}$ with $\mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{~A}=70^{\circ}$ and $\angle \mathrm{B}=60^{\circ}$. Draw the circumcircle of $\triangle \mathrm{ABC}$. Is the centre inside or outside the triangle?

$\textbf{Answer:}$

Here, all the angles are less than 90°, so the triangle is acute-angled.

For an acute triangle, the circumcentre lies inside the triangle.

Hence, the centre is inside the triangle.

Question 2. Draw $\triangle \mathrm{ABC}$ with $\mathrm{AB}=5 \mathrm{~cm}, \angle \mathrm{~A}=100^{\circ}, \mathrm{AC}=4 \mathrm{~cm}$. Draw the circumcircle of $\triangle \mathrm{ABC}$. Is the centre inside or outside the triangle?

$\textbf{Answer:}$

Here, $\angle A$ is greater than 90°, so the triangle is obtuse angled.

For an obtuse triangle, the circumcentre lies outside the triangle.

Hence, the centre is outside the triangle.

Question 3. Draw $\triangle \mathrm{ABC}$, with $\mathrm{AB}=6 \mathrm{~cm}, \mathrm{BC}=7 \mathrm{~cm}$ and $\mathrm{CA}=7 \mathrm{~cm}$. Draw the circumcircle of $\triangle \mathrm{ABC}$. Let the circumcentre be O. Measure $\mathrm{OA}$, $\mathrm{OB}, \mathrm{OC}$.

$\textbf{Answer:}$

Since the circumcentre is equidistant from all three vertices of the triangle,
OA = OB = OC [O is the circum centre]

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

$\textbf{Answer:}$

The least possible radius of a circle passing through two points $A$ and $B$ occurs when $A B$ is the diameter of the circle.
If $AB=d$, then Radius $=\frac{d}{2}$
So, the least possible radius is: $\frac{A B}{2}$

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

$\textbf{Answer:}$

Here, PQ is the chord of the circle, and O is the centre.
We know that all radii of a circle are equal.
Here, OP and OQ are the Radii of the circle.
So, OP = OQ
Therefore, in $\triangle OPQ$, two sides are equal.
Hence, $\triangle OPQ$ is an isosceles triangle.

Question 2. Show that if two such isosceles triangles (occurring in the previous question) have equal base length, they are congruent to each other.

$\textbf{Answer:}$

Here, chords PQ and AB are of the same length. [Given]
We have two isosceles triangles $\triangle$OPQ and $\triangle$OAB,
Where
OP = OQ and OA = OB
So, all three corresponding sides are equal.
So, by Side-side-side (SSS) congruence, $\triangle$OPQ $\cong$ $\triangle$OAB

Hence, two isosceles triangles are congruent.

Question 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?
(Hint: Use Fig. 5.12. You are told that $\angle \mathrm{CMA}=\angle \mathrm{CMB}=90^{\circ}$. You need to show that $\mathrm{AM}=\mathrm{BM}$.)

$\textbf{Answer:}$

Let AB be the chord of the circle with centre C.
Here, CM $\perp$ AB
We have to prove AM = BM
In triangles $\triangle C M A$ and $\triangle C M B$,
We have,
$\angle$CMA = $\angle$CMB = 90
CA = CB = Radii of the same circle
CM is the common side of both triangles.
So, these two right triangles have equal hypotenuses and a common side.
Hence, by RHS congruence, $\triangle CMA \cong \triangle CMB$
So, the perpendicular from the centre to a chord bisects the chord.

Question 2. An isosceles triangle ABC is inscribed in a circle, with $\mathrm{AB}=\mathrm{AC}$. Show that the altitude from A to BC passes through the centre of the circle.

$\textbf{Answer:}$

Here, $\triangle$ABC is an isosceles triangle with AB = AC

D is the centre of the circle.

We know that in an isosceles triangle, the altitude from the vertex also bisects the base.

So, BM = CM and AM$\perp$BC

Therefore, AM is the perpendicular bisector of BC.

Now, the centre of the circumcircle lies on the perpendicular bisector of every chord.

Since $B C$ is a chord of the circle, the centre lies on the perpendicular bisector of BC.

Hence, the altitude AM passes through the centre of the circle.

Question 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.

$\textbf{Answer:}$

Here, AB = 8 cm and CD = 6 cm
Radius, OB = OD = 5 cm
Midpoint of AB is N, and the midpoint of CD is M.
We have to find the value of MN.

In $\triangle$OBN,
BN = $\frac{AB}2=\frac82=4$ cm
OB = 5 cm
Using Pythagoras's theorem, we get,
$OB^2=BN^2+ON^2$
$⇒5^2=4^2+ON^2$
$⇒ON^2=9$
$⇒ON =3$ cm

In $\triangle$ODM,
DM = $\frac{CD}2=\frac62=3$ cm
OD = 5 cm
Using Pythagoras's theorem, we get,
$OD^2=DM^2+OM^2$
$⇒5^2=3^2+OM^2$
$⇒OM^2=16$
$⇒OM =4$ cm

MN = ON + OM = 3 + 4 = 7 cm
Hence, the distance between the midpoints of the chords is 7 cm.

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

[Theorem 6: Chords of a circle having the same length are all at the same distance from the centre of the circle.]

$\textbf{Answer:}$

Let PQ and RS be two chords of the circle with centre O.
Here, PQ = RS
Also, OM $\perp$ RS and ON $\perp$ PQ
We have to prove OM = ON
Join the radii OS and OQ.
So, OS = OQ [$\because$ Radii of same circle]
⇒ $OS^2=OQ^2$
Since the perpendicular from the centre to a chord bisects the chord,
$MS=\frac{RS}2$ and $NQ=\frac{PQ}2=\frac{RS}2$ [$\because PQ=RS$]
So, MS = NQ
⇒ $MS^2=NQ^2$
Now applying the Baudhāyana-Pythagoras theorem in right triangles $\triangle OMS$ and $\triangle ONQ$, we get,
$OS^2= MS^2+OM^2$
$⇒OM^2=OS^2-MS^2----(1)$
and
$OQ^2=ON^2+NQ^2$
$⇒OS^2=ON^2+MS^2$
$⇒ON^2=OS^2-MS^2---(2)$
From equations 1 and 2, we get,
$OM^2=ON^2$
$⇒OM=ON$
Therefore, equal chords are at equal distances from the centre. [Proved]

Question 2. Consider Fig. 5.15. If CE is perpendicular to $\mathrm{AB}, \mathrm{CH}$ is perpendicular to GH, and CE $=\mathrm{CH}$, show that $\mathrm{AB}=\mathrm{GF}$.

If CE is perpendicular to $\mathrm{AB}$, that means BE = AE
Similarly, if $\mathrm{CH}$ is perpendicular to GH, GH = FH
We have to prove AB = GF
In triangles CEB and CHG,
CH = CE [Given]
CG = CB [Radii of the same circle]
$\angle$CHG = $\angle$CEG = 90
By RHS congruence, $\triangle$CEB $\cong \triangle$CHG
So, GH = BE [Corresponding Parts of Congruent Triangles]
⇒ 2GH = 2BE
⇒ AB = GF

Hence, proved.

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

$\textbf{Answer:}$

If CE is perpendicular to $\mathrm{AB}$, that means BE = $\frac12$AB
Similarly, if $\mathrm{CH}$ is perpendicular to GH, GH = $\frac12$GF
We have to prove AB = GF
Now, applying the Baudhāyana-Pythagoras theorem in right triangles $\triangle \mathrm{CHG}$ and $\triangle \mathrm{CEB}$, we get,
$CG^2=CH^2+GH^2$
And $BC^2=CE^2+BE^2$
As, CG = BC [Radii of the same circle], so $CG^2=BC^2$
Therefore, $CH^2+GH^2=CE^2+BE^2$
$⇒CE^2+GH^2=CE^2+BE^2$ [$\because$ CE = CH (Given)]
$⇒GH^2=BE^2$
$⇒GH=BE$
$⇒2GH=2BE$
$⇒GF=AB$

Hence, proved.

Question 1. Find the length of the chord of a circle where the radius is 7 cm and perpendicular distance is 6 cm.

$\textbf{Answer:}$

Here, PQ is the chord of the circle with centre O.
Here, OM = Perpendicular distance = 6 cm
Radius = OQ = 7 cm
We need to find PQ.
We know that the perpendicular from the centre bisects the chord.
So, PQ = 2 MQ
Using the Baudhāyana–Pythagoras theorem in $\triangle$OMQ, we get,
$OQ^2=OM^2+MQ^2$
$⇒7^2=6^2+MQ^2 $
$⇒49=36+MQ^2 $
$⇒MQ^2=13 $
$⇒ MQ=\sqrt{13}$
So, PQ $= 2 \times\sqrt{13}=2\sqrt{13}$ cm

Question 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 $2 \sqrt{r^2-d^2}$.

$\textbf{Answer:}$

Given:
The perpendicular distance of a chord from the centre is $d$, and the radius is $r$
PQ is a chord of the circle with centre O, and OM$\perp$PQ
Then, M is the midpoint of the chord.
So, MQ = $\frac{PQ}2$
We need to find PQ.
In right triangle $\triangle OMQ$,
$OQ^2=OM^2+MQ^2$
$⇒r^2=d^2+(\frac{PQ}2)^2$
$⇒r^2=d^2+\frac{PQ^2}4$
$⇒4r^2=4d^2+PQ^2$
$⇒PQ^2=4(r^2-d^2)$
$\therefore PQ = 2\sqrt{(r^2-d^2)}$

So, the chord length is $2\sqrt{(r^2-d^2)}$. [Proved]

*Question 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 $\mathrm{CD}=2 \mathrm{AB}$? Give reasons for your answer.

$\textbf{Answer:}$

No, we cannot conclude that CD = 2AB
The length of a chord does not change directly in proportion to its distance from the centre.
We know that in a circle:

• Chords nearer the centre are longer,
• Chords farther from the centre are shorter.

Also, Chord length $=2 \sqrt{r^2-d^2}$, where r = radius and d = perpendicular distance of the chord from the centre
It is not a linear relation.

So even if one chord is twice as far from the centre as another chord, its length will not necessarily be half.

Question 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?

$\textbf{Answer:}$

In $\triangle AOB$ :
$O A=O B=$ 12 cm [Radii of same circle.]
So, it is an isosceles triangle.
So, $\angle \mathrm{OAB}=\angle \mathrm{OBA}$ [Angles opposite to equal sides of triangle]
Let the equal angles be $x$.
In $\triangle$AOB,
$
\begin{aligned}
& \angle \mathrm{AOB}+\angle \mathrm{OAB}+\angle \mathrm{OBA}=180^{\circ}\\
& \Rightarrow 60^{\circ}+\mathrm{x}+\mathrm{x}=180^{\circ}\left[\text {Since, } \angle \mathrm{AOB}=60^{\circ}\right] \\
& \Rightarrow 2 \mathrm{x}=180^{\circ}-60^{\circ}=120^{\circ} \\
& \Rightarrow \mathrm{x}=\frac{120^{\circ} }2=60^{\circ} \\
& \Rightarrow \angle \mathrm{OAB}=\angle \mathrm{OBA}=60^{\circ}
\end{aligned}
$

Thus, triangle AOB is equilateral.
Therefore, $\mathrm{AB}=\mathrm{OA}=12 \mathrm{~cm}$
Hence, the length of chord AB is 12 cm.

Question 2. Let A and B be two points on a circle with centre O.
(i) Are there points $\mathrm{X}, \mathrm{Y}$ on the circle, on the same side of AB, such that $\angle \mathrm{AXB}$ is different from $\angle \mathrm{AYB}$?
(ii) Is it true that if $\angle \mathrm{AXB}=\angle \mathrm{AYB}$, then X and Y lie on the same side of the circle?
(iii) If $\angle \mathrm{AXB}=\angle \mathrm{AYB}$, and X and Y do not lie on the circle, does the circle through $\mathrm{A}, \mathrm{B}$ and X also pass through Y?

$\textbf{Answer:}$

(i)
No.
All angles subtended by the same chord AB at points on the same segment of the circle are equal.
So, if X and Y lie on the same side of AB, then $\angle A X B=\angle A Y B$
Therefore, the points cannot be different.

(ii)
No. It is not true in all cases.
Equal angles can also occur when the points lie on opposite sides of AB.
For example:

• angles in the major segment are equal,
• angles in the minor segment are equal.

So, equality of angles alone does not guarantee that X and Y lie on the same side of the circle.

(iii)
Yes. It is true.
If $\angle \mathrm{AXB}=\angle \mathrm{AYB}$, then both points subtend the same chord AB at equal angles.
By the converse of the “angles in the same segment” theorem, the four points A, B, X, and Y lie on the same circle.
Hence, the circle through A, B, and X also passes through Y.

Question 3. Find $x$ in Fig. 5.26.

$\textbf{Answer:}$

Given: $\angle$ADC= 100°
We know that in a cyclic quadrilateral, opposite angles are supplementary.
That means, $\angle$ADB + $\angle$ABC = 180°
⇒ 100° + $\angle$ABC = 180°
$\therefore \angle$ABC = 180° - 100° = 80°

Hence, the value of $x$ is 80°.

Question 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?

$\textbf{Answer:}$

Let PQ be a chord of a circle with centre O.
OM $\perp$ PQ
OM = 5 cm [Given]
OQ = 13 cm = Radius of the circle [Given]
Applying the Baudhāyana–Pythagoras theorem to the right-angle triangle OMQ, we get,
$OQ^2=OM^2+MQ^2$
$⇒13^2=5^2+MQ^2$
$⇒MQ^2=169-25=144$
$⇒MQ=12$ cm
So, PQ = 2MQ = 2 × 12 = 24 cm

Hence, the length of the chord is 24 cm.

Question 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?

$\textbf{Answer:}$
The angle subtended by an arc at the centre is twice the angle subtended at a point on the circle.
Central angle = 70°
$\therefore$ Angle on the circle $=\frac{70^{\circ}}{2}=35^{\circ}$

Question 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.

$\textbf{Answer:}$

Let AB be the diameter, and PQ be the chord of a circle with centre O.
Given: AB = 26 cm and PQ = 24 cm
We know that the length of the radius is half of the diameter.
So, radius OQ = $\frac{26}2=13$ cm
Also, OM $\perp$ PQ
So, MQ = $\frac12$PQ = $\frac{24}2=12$ cm
Applying the Baudhāyana–Pythagoras theorem to the right-angle triangle OMQ, we get,
$OQ^2=OM^2+MQ^2$
$⇒13^2=OM^2+12^2$
$OM^2=169-144=25$
$OM = 5$ cm

Hence, the distance from the centre of the circle to the chord is 5 cm.

Question 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?

$\textbf{Answer:}$

Let PQ be the chord of a circle with centre O.
Let PQ be the chord of a circle with centre O.
Given: OQ = 15 cm = Radius of the circle
Also, OM $\perp$ PQ and OM = 9 cm
So, PQ = 2MQ
Applying the Baudhāyana–Pythagoras theorem to the right-angle triangle OMQ, we get,
$OQ^2=OM^2+MQ^2$
$⇒15^2=9^2+MQ^2$
$⇒MQ^2=225-81=144$
$⇒MQ=12$ cm
So, PQ = 2MQ = 2 × 12 = 24 cm

Hence, the length of the chord is 24 cm.

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

$\textbf{Answer:}$

Let PQ be a chord of a circle with centre O.
Suppose the perpendicular bisector of PQ meets at point M.
Since M lies on the perpendicular bisector of PQ,
So, MA = MB
The centre O is also equidistant from P and Q because OP = OQ
The locus of points equidistant from P and Q is the perpendicular bisector of PQ.
Hence, the centre O lies on the perpendicular bisector of the chord.

Hence, proved.

Question 6. The diameter of a circle is AB. Point C is on the circumference. What is the measure of the $\angle \mathrm{ACB}$? Explain your reasoning.

$\textbf{Answer:}$

Theorem: The angle subtended by a diameter at the circumference is a right angle.

Since AB is the diameter, the angle subtended by it at the circumference is a right angle.
Therefore, $\angle A C B=90^{\circ}$

Question 7. ABCD is a cyclic quadrilateral inscribed in a circle. If $\angle \mathrm{A}$ measures $75^{\circ}$, what is the measure of $\angle \mathrm{C}$ ? If $\angle \mathrm{B}$ measures $110^{\circ}$, what is the measure of $\angle \mathrm{D}$?

$\textbf{Answer:}$

In a cyclic quadrilateral, opposite angles are supplementary.

So, $\angle A+\angle C=180^{\circ}$

$⇒\angle C=180^{\circ}-75^{\circ}=105^{\circ}$ [$\angle A=75^{\circ}$(Given)]

Similarly,

$\angle B+\angle D=180^{\circ}$

$⇒\angle D=180^{\circ}-110^{\circ}=70^{\circ}$ [$\angle B=110^{\circ}$(Given)]

Question 8. Quadrilateral PQRS is inscribed in a circle. If $\angle \mathrm{P}=(2 x+10)^{\circ}$ and $\angle \mathrm{R}=(3 x-20)^{\circ}$, find the value of $x$ and the measures of $\angle \mathrm{P}$ and $\angle \mathrm{R}$.

$\textbf{Answer:}$

In a cyclic quadrilateral, opposite angles are supplementary.

So, $\angle P+\angle R=180^{\circ}$

$⇒(2 x+10)^{\circ}+(3 x-20)^{\circ}=180^{\circ}$

$⇒5x-10^{\circ}=180^{\circ}$

$⇒5x=190^{\circ}$

$\therefore x =38^{\circ}$

Now,

$\begin{aligned} & \angle P=2(38)+10=86^{\circ} \\ & \angle R=3(38)-20=94^{\circ}\end{aligned}$

Question 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.

$\textbf{Answer:}$

Let AB be the chord of the circle with centre O.
Radius of the circle is OB.
Given: AB = 16, OM = 6 cm
Also, OM $\perp$ AB
So, MB $=\frac 12 AB=\frac{16}2=8$ cm
Applying the Baudhāyana–Pythagoras theorem to the right-angle triangle OMB, we get,
$OB^2=OM^2+MB^2$
$⇒OB^2=6^2+8^2$
$⇒OB^2=100$
$⇒OB=10$ cm

Hence, the radius of the circle is 10 cm.

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

$\textbf{Answer:}$

The sides 5, 5, 12, and 12 units form an isosceles trapezium inscribed in a circle.
Using Brahmagupta's formula for a cyclic quadrilateral, we get,
$\text { Area }=\sqrt{(s-a)(s-b)(s-c)(s-d)}$, where $s$ = semi-perimeter, and a, b, c, and d are the sides
Semi perimeter, $s=\frac{5+5+12+12}2=\frac{34}2=17$
So,
$\text { Area }=\sqrt{(17-5)(17-5)(17-12)(17-12)}$
$=\sqrt{12 \times 12 \times 5 \times 5}$
$=\sqrt{3600}=60$

Hence, its area is 60 square units.

*Question 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? What is the best way of finding out?

$\textbf{Answer:}$

In a cyclic quadrilateral, the position of the circumcentre depends on the type of angles in the quadrilateral.

• If all angles are less than $90^{\circ}$, the circumcentre lies inside the quadrilateral.
• If one angle is greater than $90^{\circ}$, the circumcentre lies outside the quadrilateral.
• If one angle is exactly $90^{\circ}$, the circumcentre lies on the side opposite that angle (because that side becomes the diameter).

So, the best way is to:
1. Observe the angles of the cyclic quadrilateral.
2. Decide whether it is acute-angled, right-angled, or obtuse-angled.

*Question 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.

$\textbf{Answer:}$

Let chords AB and CD intersect at point P inside the circle, and O be the centre.
From the figure, $\mathrm{ON} \perp \mathrm{CD}$ and $\mathrm{OM} \perp \mathrm{AB}$.
Equal chords ⇒ equal distances from centre
Given: $\mathrm{AB}=\mathrm{CD} . . .(1)$
To Prove: $\mathrm{PB}=\mathrm{PD}$ and $\mathrm{AP}=\mathrm{CP}$.
So, the perpendicular distances from the centre are equal.
So, $\mathrm{OM}=\mathrm{ON}$

In $\triangle \mathrm{OPM}$ and $\triangle \mathrm{OPN}$, we have
OP = OP [common side]
OM = ON [Proved above]
$\angle \mathrm{OMP}=\angle \mathrm{ONP}=90^{\circ}$ [Perpendiculars]
So, by RHS congruency

$
\triangle \mathrm{OPM} \cong \triangle \mathrm{OPN} $
So, $\mathrm{PM}=\mathrm{PN}[\mathrm{CPCT}]$...

Now, $\mathrm{OM} \perp \mathrm{AB}$, so $\mathrm{AM}=\mathrm{MB}$ [Perpendicular from the centre bisect the chord]
Similarly, $\mathrm{ON} \perp \mathrm{CD}$, so $\mathrm{CN}=\mathrm{ND}$
Since, $\mathrm{AB}=\mathrm{CD}$ [From (1)]
Therefore, $\mathrm{BM}=\mathrm{DN}\left[\mathrm{BM}=\frac{1}{2} \mathrm{AB}\right.$ and $\left.\mathrm{DN}=\frac{1}{2} \mathrm{CD}\right]$.

Adding (2) and (3), we get

$ \mathrm{PM}+\mathrm{BM}=\mathrm{PN}+\mathrm{DN} $
$ \Rightarrow \mathrm{~PB}=\mathrm{PD} \ldots(4) \tag{4}$
Subtracting (4) from (1), we get

$
\begin{aligned}
& A B-P B=C D-P D \\
& \Rightarrow A P=C P
\end{aligned}
$
Hence proved.

*Question 13. 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?)

$\textbf{Answer:}$

1. Draw a line segment $A B=6 \mathrm{~cm}$.
2. Find the midpoint $M$ of $A B$.
3. Draw a perpendicular line to $AB$ through $M$.
4. On this perpendicular, mark a point $O$ such that: $OM=3 \mathrm{~cm}$
5. Join $O A$ and $O B$.
6. With centre $O$ and radius $O A$, draw a circle.
The circle obtained is the required circle because:
$A B=6 \mathrm{~cm}$ is a chord and the perpendicular distance from the centre $O$ to the chord $A B$ is 3 cm.

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

$\textbf{Answer:}$

Let PQRS be a parallelogram inscribed in a circle.
We know that in a cyclic quadrilateral, opposite angles are supplementary.
So, $\angle$P + $\angle$R = 180°
But in a parallelogram, opposite angles are equal.
So, $\angle$P = $\angle$Q
Now, $\angle$P + $\angle$R = 180°
⇒ $\angle$P +$\angle$P = 180°
⇒ 2$\angle$P = 180°
⇒ $\angle$P = 90°
Similarly, all angles are 90°.

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

*Question 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.

$\textbf{Answer:}$

Let PQRS be a parallelogram inscribed in a circle, and diagonals PR and QS intersect at point O.

We know that in a rectangle, the diagonals are equal in length and bisect each other.

So, OP = OR and OS = OQ

Since PR = QS, OP = OR = OS = OQ

Thus, O is equidistant from all four vertices.

But the centre of a circle is the unique point equidistant from all points on the circle. Therefore, O is the centre of the circle.

*Question 16. Consider all chords of a circle of a fixed length. What is the shape formed by the midpoints of all these chords?

$\textbf{Answer:}$

Equal chords are equidistant from the centre of a circle.

So every chord of the fixed length lies at the same perpendicular distance from the centre O.

The midpoint of a chord lies on the perpendicular from the centre to the chord. Hence, all such midpoints are at the same fixed distance from O.

Therefore, the set of all such midpoints forms a circle concentric with the original circle.

*Question 17. 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 $\angle \mathrm{BAC}^{\prime \prime}$.

$\textbf{Answer:}$

Given: AB = AC

We have joined OA, OB, and OC.

In $\triangle \mathrm{AOB}$ and $\triangle \mathrm{AOC}$, we have

Now, $\mathrm{OB}=\mathrm{OC}$ [Radii of same circle]

OA = OA [Common Side]

$\mathrm{AB}=\mathrm{AC}$ [Given]

Therefore, by Side-Side-Side (SSS) congruence

$\triangle \mathrm{AOB} \cong \triangle \mathrm{AOC}$,

So, $\angle \mathrm{BAO}=\angle \mathrm{OAC}$

Hence, AO bisects $\angle \mathrm{BAC}$.

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

Question 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.

$\textbf{Answer:}$

Let the radius be $r$.
We know that the distance $(d)$ from the centre to a chord of length $x$ is:
$\sqrt{r^2-(\frac x2)^2}$
Distance from centre to chord 1 of 24 cm
= $\sqrt{r^2-(\frac {24}2)^2}=\sqrt{r^2-12^2}=\sqrt{r^2-144}$
Distance from centre to chord 1 of 10 cm
= $\sqrt{r^2-(\frac {10}2)^2}=\sqrt{r^2-5^2}=\sqrt{r^2-25}$
Both chords are on the same side of the centre and are 7 cm apart.
So, $\sqrt{r^2-25}-\sqrt{r^2-144}=7$
$⇒\sqrt{r^2-25}=7+\sqrt{r^2-144}$
$⇒(\sqrt{r^2-25})^2=(7+\sqrt{r^2-144})^2$
$⇒r^2-25=49+r^2-144+14 \sqrt{r^2-144}$
$⇒70=14 \sqrt{r^2-144}$
$⇒ \sqrt{r^2-144}=5$
$⇒( \sqrt{r^2-144})^2=5^2$
$⇒r^2-144=25$
$⇒r^2=169$
$\therefore r=13$ cm

Therefore, the radius of the circle is 13 cm.

*Question19. 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.

$\textbf{Answer:}$

If we join the centre to adjacent vertices, it divides the hexagon into 6 equilateral triangles.
Hence, each side of the hexagon equals the radius.
Let AB be one side of the hexagon with centre O.
Let OA = OB = AB = $r$
So, each side of the hexagon is equal to $r$.
Let OM $\perp$ AB, and OM is the altitude of the triangle, which is also the distance of a side from the centre.
We know that the altitude of an equilateral triangle
$=\frac{\sqrt{3}}{2} \times$ Side length $=\frac{\sqrt{3}}{2}r$

Therefore, the side length of the hexagon is $r$, and the distance from the centre is $\frac{\sqrt{3}}{2}r$.

Question 20. A quadrilateral MNOP is inscribed in a circle. If MN is a diameter, what can you say about $\angle \mathrm{MOP}$ and $\angle \mathrm{MNP}$? Explain your reasoning.

$\textbf{Answer:}$

We know that angles subtended by the same chord are equal.

Both $\angle \mathrm{MOP}$ and $\angle \mathrm{MNP}$ are inscribed angles that subtend the exact same arc, which is arc $M P$.

According to the Angles Subtended by the Same Arc Theorem, angles subtended by the same arc at the remaining part of the circle are equal.

So, $\angle \mathrm{MOP}=\angle \mathrm{MNP}$

Therefore, we can say that $\angle \mathrm{MOP}$ and $\angle \mathrm{MNP}$ are equal, because they are angles subtended by the same arc (or chord) in the same segment of the circle. Both vertices $O$ and $P$ lie on the circumference on the same side of the chord $M P$.

Question 21. Let ABCD be a cyclic quadrilateral. Explain why the exterior angle at any vertex is equal to the interior opposite angle (e.g., $\angle \mathrm{CDE}=\angle \mathrm{ABC}$, where E is a point on the extension of side CD).

$\textbf{Answer:}$

In a cyclic quadrilateral,

$\angle C D A+\angle A B C=180^{\circ}$------(1)

Also, exterior angle and interior angle form a linear pair:

$\angle C D E+\angle C D A=180^{\circ}$------(2)

From equations 1 and 2, we get,

$\angle C D A+\angle A B C=\angle C D E+\angle C D A$

$⇒\angle A B C=\angle C D E$

Hence, proved.

Hence, the exterior angle of a cyclic quadrilateral equals the interior opposite angle.

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

$\textbf{Answer:}$

The diameter is the longest distance between any two points on a circle.

For any chord AB, the perpendicular from the centre to the chord bisects it.

If $d$ is the distance from the centre and $r$ the radius, then:

$A B=2 \sqrt{r^2-d^2}$

Since $d^2 \geq 0$,

$⇒A B \leq 2 r$

We know that $2r$ is the diameter.

Hence, no chord can be longer than the diameter.

Note that equality occurs only when $d=0$, i.e. when the chord passes through the centre and becomes the diameter.

*Question 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.

$\textbf{Answer:}$

Let the circle have centre $O$ and let $A$ be a point inside it.

Any chord passing through A will have some perpendicular distance from the centre 0.

We know that the farther a chord is from the centre, the shorter it is.

So, the shortest chord through A will be the chord whose distance from O is maximum.

Among all lines passing through A, the perpendicular distance from O to the line is greatest when the line is perpendicular to OA.

Therefore, the chord through A that is perpendicular to OA is farthest from the centre.

Hence, it is the shortest chord passing through A.

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

Hence proved.

Question 24. How would you use the following figure to justify the statement that the angle in a semicircle is $90^{\circ}$?

$\textbf{Answer:}$

Let the endpoints of the diameter be B and C, and let A be any point on the semicircle with centre at O.
So, OA = OB = OC [Radii of the same circle]
So, triangles AOB and AOC are isosceles.
Let:$\angle A B O=a \text { and } \angle A C O=b$
Then, because of the isosceles triangles, $\angle B A O=a \text { and } \angle O A C=b$
So, $\angle B A C=\angle B A O+\angle O A C=a+b$
In $\triangle$ABC,
$\angle BAC +\angle ABC +\angle ACA=180°$
$⇒a+b+a+b=180°$
$⇒2(a+b)=180°$
$⇒a+b=90°$
Therefore, $\angle \mathrm{BAC}=90^{\circ}$

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

*Question 25. In a circle, two chords $\mathrm{CC}^{\prime}$ and $\mathrm{DD}^{\prime}$ are drawn perpendicular to a diameter AB. Prove that the segment $\mathrm{MM}^{\prime}$ joining the midpoints of the chords CD and $\mathrm{C}^{\prime} \mathrm{D}^{\prime}$ is perpendicular to AB.

$\textbf{Answer:}$

Let AB be the diameter of the circle.

Since chords $\mathrm{CC}^{\prime}$ and $\mathrm{DD}^{\prime}$ are perpendicular to AB , the points C and $\mathrm{C}^{\prime}$ are symmetric about AB.

Similarly, D and $\mathrm{D}^{\prime}$ are also symmetric about AB.

Now, let $M$ be the midpoint of $C D$ and $M^{\prime}$ be the midpoint of $C^{\prime} D^{\prime}$.

Since $C$ corresponds to $C^{\prime}$ and $D$ corresponds to $D^{\prime}$ by reflection in $AB$, the midpoint of CD corresponds to the midpoint of C'D'.

Therefore, $M$ and $M^{\prime}$ are symmetric about $A B$.

The line joining two points that are symmetric about a line is perpendicular to that line.

Hence, $\mathrm{MM}^{\prime} \perp \mathrm{AB}$

Therefore, the segment joining the midpoints of $C D$ and $C^{\prime} D^{\prime}$ is perpendicular to $AB$.

Hence proved.

*Question 26. How would you use the following figure to justify the statement that the sum of the opposite angles of a cyclic quadrilateral is $180^{\circ}$?

$\textbf{Answer:}$

Let ABCD be a cyclic quadrilateral with centre O.

Join $\mathrm{OA}, \mathrm{OB}, \mathrm{OC}$ and OD .

Since $\mathrm{OA}, \mathrm{OB}, \mathrm{OC}$ and OD are radii of the same circle:

$O A=O B=O C=O D$

So the triangles formed with the centre are isosceles.

From the figure:

Let $\angle \mathrm{OAB}=\mathrm{p}$ and $\angle \mathrm{OBA}=\mathrm{q}$.

Let $\angle \mathrm{ODC}=\mathrm{v}$ and $\angle \mathrm{OCD}=\mathrm{u}$.

The angle at the centre corresponding to the arcs gives twice the angle at the circumference.

Thus: $\angle \mathrm{A}+\angle \mathrm{C}=180^{\circ}$

Similarly: $\angle \mathrm{B}+\angle \mathrm{D}=180^{\circ}$

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

Hence proved.

I’m Up and Down, and Round and Round Class 9 Chapter 5: Topics

Topics you will learn in NCERT Class 9 Maths Chapter 5 Pair of I’m Up and Down, and Round and Round include:

• 5.1 Definitions
• 5.2 Symmetries of a Circle
• 5.3 How Many Circles?
• 5.4 Chords and the Angles They Subtend
• 5.5 Midpoints and Perpendicular Bisectors of Chords
• 5.6 Distance of Chords from the Centre
• 5.7 Angles Subtended by an Arc
• 5.8 Concyclicity of Points

Class 9 Maths NCERT Chapter 5 Solutions: Extra Questions

Question 1: The radius of a circle is 5 cm, and the length of the chord is 6 cm. Find the distance from the centre to its chord.

$\textbf{Answer:}$

Let AB = 6 cm be the chord of the circle with radius AO = 5 cm. Draw OP perpendicular to AB.

Now, apply Pythagoras' theorem

(AO)² = (OP)² + (AP)²

Here, AP = $\frac{1}{2}$ AB

So, AP = 3 cm

Now, putting values, we get:

5² = (OP)² + 3²

25 = (OP)² + 9

(OP)² = 16

Therefore, OP = 4 cm

Thus, the distance from the centre to its chord is 4 cm.

Question 2:

ABCD is a cyclic quadrilateral in which angle B is opposite to angle D. If $\angle \mathrm{B}=(\mathrm{x}+10)^{\circ}$ and $\angle \mathrm{D}=(2 \mathrm{x}+35)^{\circ}$, then what is the value of $\mathrm{x}$?

$\textbf{Answer:}$

Given that ABCD is a cyclic quadrilateral, in which angle B is opposite to angle D.
$\angle \mathrm{B}=(\mathrm{x}+10)^{\circ}$ and $\angle \mathrm{D}=(2 \mathrm{x}+35)^{\circ}$
We know the opposite angles of a cyclic quadrilateral are supplementary.
So, $\angle \mathrm{B}+\angle \mathrm{D}=180^\circ$
⇒ $\mathrm{x}+10^\circ+2 \mathrm{x}+35^\circ=180^\circ$
⇒ $3\mathrm{x}+45^\circ=180^\circ$
⇒ $3\mathrm{x}=135^\circ$
⇒ $\mathrm{x}=45^\circ$
Hence, the correct answer is 45$^{\circ}$.

Question 3:

In a cyclic quadrilateral ${EFGH}, \angle {E}$ is opposite to $\angle{G}$. If $\angle{E}=95°$, then what is the value of $\angle {G}$?

$\textbf{Answer:}$

Given,
$\angle E = 95°$
EFGH is a cyclic quadrilateral
In a cyclic quadrilateral, the opposite angles are supplementary.
⇒ $\angle E + \angle G = 180°$
⇒ $95° + \angle G = 180°$
⇒ $\angle G = 180° - 95° = 85°$
Hence, the correct answer is 85°.

Question 4:

The angle subtended by a chord on the centre of a circle is 180°. What will be the angle subtended by the same chord on the circumference of this circle?

$\textbf{Answer:}$

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.
Angle subtended on the circumference = $\frac{180°}{2}$ = 90°
Therefore, the angle subtended by the same chord on the circumference is 90°.
Hence, the correct answer is 90°.

Question 5:

A is a point outside of a circle with centre O. AP and AQ are two tangents of the circle. If AP = $a^2+14$ and AQ = 239, then what is the value of a?

$\textbf{Answer:}$

Given,
Length of AP = $a^2+14$
Length of AQ = 239
We know that tangents drawn from an external point to a circle are equal
⇒ AP = AQ
⇒ $a^2$ + 14 = 239
⇒ $a^2$ = 225
⇒ a = $\small\sqrt{225}$
⇒ a = 15
Hence, the correct answer is 15.

NCERT Solutions for Class 9 Maths Chapter Wise

We at Careers360 compiled all the NCERT Class 9 Maths solutions in one place for easy student reference. The following links will allow you to access them.

NCERT Class 9 Books & Syllabus

As students step into a new class, they must first explore the latest syllabus to identify the chapters included. Below are the links to the most recent syllabus and essential reference books.

• NCERT Books Class 9 Maths
• NCERT Books for Class 9 Science
• NCERT Syllabus Class 9 Maths
• NCERT Books Class 9
• NCERT Syllabus Class 9