5 Marks Questions - MATHS STD 9 Questions - Vidyadip

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Question 15 Marks

In Figure (1), O is the centre of the circle. If $\angle\text{OAB}=40^\circ$ and $\angle\text{OCB}=30^\circ,$ find $\angle\text{AOC}.$
In Figure (2), A, B and C are three points on the circle with centre O such that $\angle\text{AOB}=90^\circ$ and $\angle\text{AOC}=110^\circ.$ Find $\angle\text{BAC}.$

Answer

Join BO.

In $\triangle\text{BOC},$ we have:
OC = OB [Radii of a circle]
$\Rightarrow\ \angle\text{OBC}=\angle\text{OCB}$
$\angle\text{OBC}=30^\circ\dots(\text{i})$
In $\triangle\text{BOA},$ we have:
OB = OA (Radii of a circle)
$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}$ $[\because\ \angle\text{OAB}=40^\circ]$
$\angle\text{OBA}=40^\circ\dots(\text{ii})$
Now, we have:
$\angle\text{ABC}=\angle\text{OBC}+\angle\text{OBA}$
$=30^\circ+40^\circ$ [From (i) and (ii)]
$\therefore\ \angle\text{ABC}=70^\circ$
The angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
i.e., $\angle\text{AOC}=2\angle\text{ABC}$
$=(2\times70^\circ)=140^\circ$

Here, $\angle\text{BOC}=[360^\circ-(90^\circ+110^\circ)]$
$=(360^\circ-200^\circ)=160^\circ$
We know that $\angle\text{BOC}=2\angle\text{BAC}.$
$\Rightarrow\ \angle\text{BAC}=\frac{\angle\text{BOC}}{2}=\Big(\frac{160^\circ}{2}\Big)=80^\circ$
Hence, $\angle\text{BAC}=80^\circ$

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Question 25 Marks

Two circles with centres O and O' intersect at two points A and B. A line PQ is drawn parallel to OO' through A or B, intersecting the circles at P and Q. Prove that PQ = 2OO'.

Answer

Given: Two circles with centres O and O' intersect at two points A and B.

Draw a line PQ parallel to OO' through B, OX perpendicular to PQ, O'Y perpendicular to PQ, join all.
We know that perpendicular drawn from the centre to the chord, bisects the chord.
$\therefore$ PX = XB and YQ = BY
$\therefore$ PX + YQ = XB + BY
On adding XB + BY on both sides, we get
PX + YQ + XB + BY = 2(XB + BY)
⇒ PQ = 2(XY)
⇒ PQ = 2(OO')
Hence, PQ = 2OO'

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Question 35 Marks

In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced. If $\angle\text{ABC}=92^\circ$ and $\angle\text{FAE}=20^\circ,$ find $\angle\text{BCD}.$

Answer

Given: ABCD is a cyclic quadrilateral.
Then $\angle\text{ABC}+\angle\text{ADC}=180^\circ$
$\Rightarrow\ 92^\circ+\angle\text{ADC}=180^\circ$
$\Rightarrow\ \angle\text{ADC}=(180^\circ-92^\circ)=88^\circ$
Again, AE parallel to CD.
Thus, $\angle\text{EAD}=\angle\text{ADC}=88^\circ$ [Alternate angles]
We know that the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.
$\therefore\ \angle\text{BCD}=\angle\text{DAF}$
$\Rightarrow\ \angle\text{BCD}=\angle\text{EAD}+\angle\text{EAF}$
$=88^\circ+20^\circ=108^\circ$
Hence, $\angle\text{BCD}=108^\circ$

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Question 45 Marks

Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB.

Answer

Given: Two equal circles intersect at point P and Q.
A straight line passes through P and meets the circle at points A and B.
To prove: QA = QB
Construction: Join PQ.

Proof:
Two circles will be congruent if and only if they have equal radii.
Here, PQ is the common chord to both the circles.
Thus, their corresponding arcs are equal (if two chords of a circle are equal,
then their corresponding arcs are congruent).
So, arc PCQ = arc PDQ
$\therefore\ \angle\text{QAP}=\angle\text{QBP}$ (Congruent arcs have the same degree in measure)
Hence, QA = QB (In isosceles triangle, base angles are equal)

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Question 55 Marks

In the adjoining figure, ABCD is a cyclic quadrilateral in which $\angle\text{BCD}=100^\circ$ and $\angle\text{ABD}=50^\circ.$ Find $\angle\text{ADB}.$

Answer

ABCD is a cyclic quadrilateral.
$\therefore\ \angle\text{A}+\angle\text{C}=180^\circ$ [Opposite of a cyclic quadrilateral are supplementary]
$\Rightarrow\ \angle\text{A}+100^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}=180^\circ-100^\circ=80^\circ$

Now, in $\triangle\text{ABD},$ we have:
$\angle\text{A}+\angle\text{ABD}+\angle\text{ADB}=180^\circ$
$\Rightarrow\ 80^\circ+50^\circ+\angle\text{ADB}=180^\circ$
$\Rightarrow\ \angle\text{ADB}=180^\circ-130^\circ=50^\circ$
$\therefore\ \angle\text{ADB}=50^\circ$

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Question 65 Marks

In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If $\angle\text{PSR}=150^\circ,$ find $\angle\text{RPQ}.$

Answer

In cyclic quadrilateral PQRS, we have:
$\angle\text{PSR}+\angle\text{PQR}=180^\circ$
$\Rightarrow\ 150^\circ+\angle\text{PQR}=180^\circ$
$\Rightarrow\ \angle\text{PQR}=(180^\circ-150^\circ)=30^\circ$
$\therefore\ \angle\text{PQR}=30^\circ\dots(\text{i})$
Also, $\angle\text{PQR}=90^\circ\dots(\text{ii})$ (Angle in a semicircle)
Now, in $\triangle\text{PRQ},$ we have:
$\angle\text{PQR}+\angle\text{PRQ}+\angle\text{RPQ}=180^\circ$
$\Rightarrow\ 30^\circ+90^\circ+\angle\text{RPQ}=180^\circ$ [From (i) and (ii)]
$\Rightarrow\ \angle\text{RPQ}=180^\circ-120^\circ=60^\circ$
$\therefore\ \angle\text{RPQ}=60^\circ$

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Question 75 Marks

A chord of length $30\ cm$ is drawn at a distance of $8\ cm$ from the centre of a circle. Find out the radius of the circle.

Answer

Let $AB$ be the chord of the given circle with centre $O.$ The perpendicular circle to the chord is $8\ cm.$
Join $OB.$
Then$ OM = 8\ cm$ and $AB = 30\ cm.$

We know that the perpendicular from the centre of a circle to a chord.
$\therefore\ \text{MB}=\Big(\frac{\text{AB}}{2}\Big)=\Big(\frac{30}{2}\Big)\text{ cm}=15\text{ cm}$
From the right $\triangle\text{OMB},$ we have:
$OB^2= OM^2 + MB^2$
$\Rightarrow OB^2 = 8^2 + 15^2$
$\Rightarrow OB^2 = 64 + 225$
$\Rightarrow OB^2 = 289$
$\Rightarrow\ \text{OB}=\sqrt{289}\text{ cm}=17\text{ cm}$
Hence, the required length of the radius is $17\ cm.$

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Question 85 Marks

In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line.

If $\angle\text{APB}=150^\circ$ and $\angle\text{BQD}=\text{x}^\circ,$ find the value of x.

Answer

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.
Here, arc AEB subtends $\angle\text{APB}$ at the centre and $\angle\text{ACB}$ at C on the circle.
$\therefore\ \angle\text{APB}=2\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACB}=\frac{150^\circ}{2}=75^\circ\dots(1)$
Since ACD is a straight line, $\angle\text{ACB}=2\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BCD}=180^\circ-75^\circ$
$\Rightarrow\ \angle\text{BCD}=105^\circ\dots(2)$
Also, arc BFD subtends reflex $\angle\text{BQD}$ at the centre and $\angle\text{BCD}$ at C on the circle.
$\therefore$ reflex $\angle\text{BQD}=2\angle\text{BCD}$
⇒ reflex $\angle\text{BQD}=2(105^\circ)=210^\circ\dots(3)$
Now,
reflex $\angle\text{BQD}+\angle\text{BQD}=360^\circ$
$\Rightarrow\ 210^\circ+\text{x}=360^\circ$
$\Rightarrow\ \text{x}=360^\circ-210^\circ$
$\Rightarrow\ \text{x}=150^\circ$
Hence, $\text{x}=150^\circ.$

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Question 95 Marks

In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of $\angle\text{BAC}.$

Answer

Given: AB and AC are two equal chords of a circle with centre O.

To prove: $\angle\text{OAB}=\angle\text{OAC}$
Construction: Join OA, OB and OC.
Proof:
In $\triangle\text{OAB}$ and $\triangle\text{OAC},$ we have:
AB = AC (Given)
OA = OA (Common)
OB = OC (Radii of a circle)
$\therefore\ \triangle\text{OAB}\cong\triangle\text{OAC}$ (By SSS congruency rule)
$\Rightarrow\ \angle\text{OAB}=\angle\text{OAC}$ (C.P.C.T.)
Hence, point O lies on the bisector of $\angle\text{BAC}.$

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Question 105 Marks

In the given figure, ABCD is a quadrilateral in which AD = BC and $\angle\text{ADC}=\angle\text{BCD}.$ Show that the points A, B, C, D lie on a circle.

Answer

ABCD is a quadrilateral in which AD = BC and $\angle\text{ADC}=\angle\text{BCD}.$
Draw $\text{DE}\perp\text{AB}$ and $\text{CF}\perp\text{AB}.$
In $\triangle\text{ADE}$ and $\triangle\text{BCF},$ we have:
$\angle\text{ADE}=\angle\text{ADC}-90^\circ=\angle\text{BCD}-90^\circ=\angle\text{BCF}$ $[$given: $\angle\text{ADC}=\angle\text{BCD}]$
AD = BC [given]
$\therefore\ \triangle\text{ADE}\cong\angle\text{BCF}$ [By AAS congruency]
$\Rightarrow\ \angle\text{A}=\angle\text{B}$
Now, $\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}=360^\circ$
$\Rightarrow\ 2\angle\text{B}=2\angle\text{D}=360^\circ$
$\Rightarrow\ \angle\text{B}+\angle\text{D}=180^\circ$
Hence, ABCD is a cyclic quadrilateral.

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Question 115 Marks

In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects $\angle\text{BPD}.$ Prove that AB = CD.

Answer

Given: O is the centre of a circle in which chords AB and CD intersect.
at P such that PO bisects $\angle\text{BPD}.$
To prove: AB = CD
Construction: Draw $\text{OE}\perp\text{AB}$ and $\text{OF}\perp\text{CD}.$
Proof: In $\triangle\text{OEP}$ and $\triangle\text{OFP},$ we have:
$\angle\text{OEP}=\angle\text{OFP}$ [90° each]
OP = OP [Common]
$\angle\text{OPE}=\angle\text{OPF}$ $[\because$ OP bisects $\angle\text{BPD}]$
Thus,
$\triangle\text{OEP}\cong\triangle\text{OFP}$ [AAS criterion]
⇒ OE = OF
Thus, chords AB and CD are equidistant from the centre O.
⇒ AB = CD $[\because$ Chords equidistant from the centre are equal$]$
$\therefore$ AB = CD

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Question 125 Marks

In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.

Answer

Given: BC is a diameter of a circle with centre O. AB and CD are two chords
such that AB || CD.
TO prove: AB = CD
Construction: Draw $\text{OL}\perp\text{AB}$ and $\text{OM}\perp\text{CD}.$

Proof:
In $\triangle\text{OLB}$ and $\triangle\text{OMC},$ we have:
$\angle\text{OLB}=\angle\text{OMC}$ [ 90° each]
$\angle\text{OBL}=\angle\text{OCD}$ [Alternate angles as AB || CD]
$\text{OB}=\text{OC}$ [Radii of a circle]
$\therefore\ \triangle\text{OLB}\cong\triangle\text{OMC}$ (AAS criterion)
Thus, OL = OM (C.P.C.T.)
We know that chords equidistant from the centre are equal.
Hence, AB = CD

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Question 135 Marks

Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

Answer

Given: AB and CD are two parallel chords of a circle with centre O.
POQ is a diameter which is perpendicular to AB.
To prove: $\text{PF}\perp\text{CD}$ and CF = FD
Proof:
AB || CD and POQ is a diameter.
$\angle\text{PEB}=90^\circ$ [Given]
$\angle\text{PFD}=\angle\text{PEB}$ $[\because$ AB || CD, Corresponding angles$]$
Thus, $\text{PF}\perp\text{CD}$
$\therefore\ \text{OF}\perp\text{CD}$
We know that the perpendicular from the centre to a chord bisects the chord.
i.e., CF = FD
Hence, POQ is perpendicular to CD and bisects it.

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Question 145 Marks

A chord of length $16\ cm$ is drawn in a circle of radius $10\ cm.$ Find the distance of the chord from the centre of the circle.

Answer

Let $AB$ be the chord of the given circle with centre $O$ and a radius of $10\ cm.$
Then $AB =16\ cm$ and $OB = 10\ cm$

From $O,$ draw $OM$ perpendicular to $AB.$
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore\ \text{BM}=\Big(\frac{16}{2}\Big)\text{ cm}=8\text{ cm}$
In the right $\triangle\text{OMB},$ we have:
$OB^2 = OM^2 + MB^2 [$Pythagoras theorem$]$
$\Rightarrow 10^2 = OM^2 + 8^2$
$\Rightarrow 100 = OM^2 + 64$
$\Rightarrow OM^2 = (100 - 64) = 36$
$\Rightarrow\ \text{OM}=\sqrt{36}\text{ cm}=6\text{ cm}$
Hence, the distance of the chord from the centre is $6\ cm.$

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Question 155 Marks

ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that $\angle\text{CBD}+\angle\text{CDB}=\frac{1}{2}\angle\text{BAD}.$

Answer

Construction: Take a point E on the circle. Join BE, DE and BD.
We know that the angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.
$\Rightarrow\ \angle\text{BAD}=2\angle\text{BED}$
$\Rightarrow\ \angle\text{BED}=\frac{1}{2}\angle\text{BAD}\dots\text(\text{i})$
Now, EBCD is a cyclic quadrilateral.
$\Rightarrow\ \angle\text{BED}+\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{BCD}=180^\circ-\angle\text{BED}$
$\Rightarrow\ \angle\text{BCD}=\frac{1}{2}\angle\text{BAD}\dots\text(\text{ii})$ [Using (ii)]
In $\triangle\text{BCD},$ by angle sum property
$\Rightarrow\ \angle\text{CBD}+\angle\text{CDB}\angle\text{BCD}=180^\circ$
$\Rightarrow\ \angle\text{CBD}+\angle\text{CDB}+180^\circ-\frac{1}{2}\angle\text{BAD}=180^\circ$ [Using (ii)]
$\Rightarrow\ \angle\text{CBD}+\angle\text{CDB}=\frac{1}{2}\angle\text{BAD}$

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Question 165 Marks

Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.

Answer

Let A, B and C be the given points.
With B as the centre and a radius equal to AC, draw an arc.
With C as the centre and AB as radius, draw another arc, which cuts the previous arcat D.

Then D is the required point BD and CD.
In $\triangle\text{ABC}$ and $\triangle\text{DCB}$
AB = DC
AC = DB
BC = CB [Common]
$\therefore\ \triangle\text{ABC}\cong\triangle\text{DCB}$ [By SSS]
$\Rightarrow\ \angle\text{BAC}=\angle\text{CDB}$ [C.P.C.T.]
Thus, BC subtends equal angles, $\angle\text{BAC}$ and $\angle\text{CDB}$ on the same side of it.
$\therefore$ Points A, B, C, D are concyclic.

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Question 175 Marks

The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

Answer

Given: Let ABCD be a cyclic quadrilateral whose diagonals AC and BD intersect at O at right angles.
Let $\text{OL}\perp\text{AB}$ such that LO produced meets CD at M.

To prove:
CM = MD
Proof:
$\angle\text{1}=\angle\text{2}$ [angles in the same segment]
$\angle\text{2}+\angle\text{3}=90^\circ$ $[\angle\text{OLB}=90^\circ]$
$\angle\text{3}+\angle\text{4}=90^\circ$ $[\because$ LOM is a straight line and $\angle\text{BOC}=90^\circ]$
$\therefore\ \angle\text{2}+\angle\text{3}=\angle\text{3}+\angle\text{4}$
$\Rightarrow\ \angle\text{2}=\angle\text{4}$
Thus,
$ \angle\text{1}=\angle\text{2}$
And
$ \angle\text{2}=\angle\text{4}$
$\Rightarrow\ \angle\text{1}=\angle\text{4}$
$\therefore\ \text{OM}=\text{CM}$
Similarly
$\text{OM}=\text{MD}$
Hence,
$\text{CM}=\text{MD}$

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Question 185 Marks

In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E. Prove that $\angle\text{AEC}=\frac{1}{2}$(angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre).

Answer

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.
Here, arc AXC subtends $\angle\text{AOC}$ at the centre and $\angle\text{ADC}$ at D on the circle.
$\therefore\ \angle\text{AOC}=2\angle\text{ADC}$
$\Rightarrow\ \angle\text{ADC}=\frac{1}{2}(\angle\text{AOC})\dots(\text{i})$
Also, arc DYB subtends $\angle\text{DOB}$ at the centre and $\angle\text{DAB}$ at A on the circle.
$\therefore\ \angle\text{DOB}=2\angle\text{DAB}$
$\Rightarrow\ \angle\text{DAB}=\frac{1}{2}(\angle\text{DOB})\dots(\text{ii})$
Now, in $\angle\text{ADE},$
$\angle\text{AEC}=\angle\text{ADC}+\angle\text{DAB}$ [Exterior angle]
$\Rightarrow\ \angle\text{AEC}=\frac{1}{2}(\angle\text{AOC}+\angle\text{DOB})$ [From (i) and (ii)]
Hence, $\angle\text{AEC}=\frac{1}{2}$(angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre).

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Question 195 Marks

In the given figure, O is the centre of a circle and $\angle\text{BOD}=150^\circ.$ Find the values of x and y.

Answer

O is the centre of the circle and $\angle\text{BOD}=150^\circ$
$\therefore$ reflex $\angle\text{BOD}=(360^\circ-\angle\text{BOD})$
$=(360^\circ-150^\circ)=210^\circ$

Now,
$\text{x}=\frac{1}{2}(\text{reflex}\angle\text{BOD})$
$=\frac{1}{2}\times210^\circ=105^\circ$
$\therefore\ \text{x}=105^\circ$
Again,
x + y = 180°
⇒ 105° + y = 180°
⇒ y = 180° - 105° = 75°
$\therefore$ y = 75°

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Question 205 Marks

In the given figure, O is the centre of a circle, $\angle\text{AOB}=40^\circ$ and $\angle\text{BDC}=100^\circ,$ find $\angle\text{OBC}.$

Answer

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
$\angle\text{AOB}=2\angle\text{ACB}$
$=2\angle\text{DCB}$ $[\because\ \angle\text{ACB}=\angle\text{DCB}]$
$\therefore\ \angle\text{DCB}=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\ \angle\text{DCB}=\Big(\frac{1}{2}\times40^\circ\Big)=20^\circ$
Considering $\triangle\text{DBC},$ we have:
$\angle\text{BDC}+\angle\text{DCB}+\angle\text{DBC}=180^\circ$
$\Rightarrow\ 100^\circ+20^\circ+\angle\text{DBC}=180^\circ$
$\Rightarrow\ \angle\text{DBC}=(180^\circ-120^\circ)=60^\circ$
$\Rightarrow\ \angle\text{OBC}=\angle\text{DBC}=60^\circ$
Hence, $\angle\text{OBC}=60^\circ$

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Question 215 Marks

In the given figure, O is the centre of the circle and $\angle\text{DAB}=50^\circ.$Calculate the values of x and y.

Answer

O is the centre of the circle and $\angle\text{DAB}=50^\circ.$
OA = OB [Radii of a circle]
$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=50^\circ$
In $\triangle\text{OAB},$ we have:
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$
$\Rightarrow\ 50^\circ+50^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\ \angle\text{AOB}=(180^\circ-100^\circ)=80^\circ$
Since AOD is a straight line, we have:
$\therefore\ \text{x}=180^\circ-\angle\text{AOB}$
$=(180^\circ-80^\circ)=100^\circ$
$\text{i.e},\text{x}=100^\circ$
The opposite angles of a cyclic quadrilateral are supplementary.
ABCD is a cyclic quadrilateral.
Thus, $\angle\text{DAB}+\angle\text{BCD}=180^\circ$
$\angle\text{BCD}=(180^\circ-50^\circ)=130^\circ$
$\therefore\ \text{y}=130^\circ$
Hence, $\text{x}=100^\circ$ and $\text{y}=130^\circ$

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Question 225 Marks

In the given figure, $\angle\text{BAC}=30^\circ.$ Show that BC is equal to the radius of the circumcircle of $\triangle\text{ABC}$ whose centre is O.

Answer

Join OB and OC.
$\angle\text{BOC}=2\angle\text{BAC}$ [As angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference]
$=2\times30^\circ$ $[\because\ \angle\text{BAC}=30^\circ]$
$=60^\circ\dots(\text{i})$
Consider $\triangle\text{BOC},$ we have:
OB = OC [Radii of a circle]
$\Rightarrow\ \angle\text{OBC}=\angle\text{OCB}\dots(\text{ii})$
In $\triangle\text{BOC},$ we have:
$\angle\text{BOC}+\angle\text{OBC}+\angle\text{OCB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 60^\circ+\angle\text{OCB}+\angle\text{OCB}=180^\circ$ [From (i) and (ii)]
$\Rightarrow\ 2\angle\text{OCB}=(180^\circ-60^\circ)=120^\circ$
$\Rightarrow\ \angle\text{OCB}=60^\circ\dots(\text{ii})$
Thus we have:
$\angle\text{OBC}=\angle\text{OCB}=\angle\text{BOC}=60^\circ$
Hence, $\triangle\text{BOC}$ is an equilateral triangle.
i.e., OB = OC = BC
$\therefore$ BC is the radius of the circumcircle.

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Question 235 Marks

In the given figure, O is the centre of a circle. If $\angle\text{AOC}=140^\circ$ and $\angle\text{CAB}=50^\circ,$ calculate

$\angle\text{EDB}$
$\angle\text{EBD}$

Answer

O is the centre of the circle where $\angle\text{AOC}=140^\circ$ and $\angle\text{CAB}=50^\circ.$

$\angle\text{BOD}=180^\circ-\angle\text{AOD}$

$=(180^\circ-140^\circ)=40^\circ$
We have the following:
OB = OD [Radii of a circle]
$\angle\text{OBD}=\angle\text{ODB}$
In $\triangle\text{OBD},$ we have:
$\angle\text{BOD}+\angle\text{OBD}+\angle\text{ODB}=180^\circ$
$\Rightarrow\ \angle\text{BOD}+\angle\text{OBD}+\angle\text{OBD}=180^\circ$ $[\because\ \angle\text{OBD}=\angle\text{ODB}]$
$\Rightarrow\ 40^\circ+2\angle\text{OBD}=180^\circ$
$\Rightarrow\ 2\angle\text{OBD}=(180^\circ-40^\circ)=140^\circ$
$\Rightarrow\ \angle\text{OBD}=70^\circ$
Since ABCD is a cyclic quadrilateral, we have:
$\angle\text{CAB}+\angle\text{BDC}=180^\circ$
$\Rightarrow\ \angle\text{CAB}+\angle\text{ODB}+\angle\text{ODC}=180^\circ$
$\Rightarrow\ 50^\circ+70^\circ+\angle\text{ODC}=180^\circ$
$\Rightarrow\ \angle\text{ODC}=(180^\circ- 120^\circ)= 60^\circ$
$\therefore\ \angle\text{ODC}= 60^\circ$
$\angle\text{EDB}=180^\circ-(\angle\text{ODC}+\angle\text{ODB})$
$=180^\circ-(60^\circ+70^\circ)$
$=180^\circ-130^\circ=50^\circ$
$\therefore\ \angle\text{EDB}= 50^\circ$

$\angle\text{EDB}=180^\circ-\angle\text{OBD}$

$=180^\circ-70^\circ$
$=\angle\text{110}^\circ$

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Question 245 Marks

In the given figure, $\angle\text{ABD}=54^\circ$ and $\angle\text{BCD}=43^\circ,$ calculate

$\angle\text{ACD}$
$\angle\text{BAD}$
$\angle\text{BDA}$

Answer

We know that the angles in the same segment of a circle are equal.

$\text{i.e.},\angle\text{ABD}=\angle\text{ACD}=54^\circ$

We know that the angles in the same segment of a circle are equal.

$\text{i.e.},\angle\text{BAD}=\angle\text{BCD}=43^\circ$

In $\triangle\text{ABD},$ we have:

$\angle\text{BAD}+\angle\text{ADB}+\angle\text{DBA}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 43^\circ+\angle\text{ADB}+54^\circ=180^\circ$
$\Rightarrow\ \angle\text{ADB}=(180^\circ-97^\circ)=83^\circ$
$\Rightarrow\ \angle\text{BDA}=83^\circ$

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Question 255 Marks

In the given figure, the diameter $CD$ of a circle with centre $O$ is perpendicular to chord $AB.$ If $AB = 12\ cm$ and $CE = 3\ cm,$ calculate the radius of the circle.

Answer

$CD$ is the diameter of the circle with centre $O$ and is perpendicular to chord $AB.$
Join $OA.$

Given:$ AB = 12\ cm$ and $CE = 3\ cm$
Let $OA = OC = r\ cm [$Radii of a circle$]$
Then $OE = (r - 3)cm$
Since the perpendicular from the centre of the circle to a chord bisects the chord,
we have:
$\text{AE}=\Big(\frac{\text{AB}}{2}\Big)=\Big(\frac{12}{2}\Big)\text{ cm}=6\text{ cm}$
Now, in right angled $\triangle\text{OEA},$ we have:
$\Rightarrow OA^2 = OE^2 + AE^2$
$\Rightarrow r^2 = (r - 3)^2 + 6^2$
$\Rightarrow r^2 = r^2 - 6r + 9 + 36$
$\Rightarrow r^2 - r^2 + 6r = 45$
$\Rightarrow 6r = 45$
$\Rightarrow\ \text{r}=\Big(\frac{45}{6}\Big)\text{ cm}=7.5\text{ cm}$
$\Rightarrow r = 7.5\ cm$
Hence, the required radius of the circle is $7.5 \ cm.$

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Question 265 Marks

In the adjoining figure, two circles with centres at $A$ and $B,$ and of radii $5\ cm$ and $3\ cm$ touch each other internally. If the perpendicular bisector of $AB$ meets the bigger circle in $P$ and $Q,$
find the length of $PQ.$

Answer

Two circles with centres $A$ and $B$ of respective radii $5\ cm$ and $3\ cm$ touch each other internally.
The perpendicular bisector of $AB$ meets the bigger circle at $P$ and $Q.$
Join $AP.$

Let $PQ$ intersect $AB$ at point $L.$
Here,$ AP = 5\ cm$
Then $AB = (5 - 3)cm = 2\ cm$
Since PQ is the perpendicular bisector of $AB,$ we have:
$\text{AL}=\Big(\frac{\text{AB}}{2}\Big)=\Big(\frac{2}{2}\Big)=1\text{cm}$
Now, in right angled $\triangle\text{PLA},$ we have:
$AP^2 = AL^2 + PL^2$
$\Rightarrow PL^2 = AP^2 - AL^2$
$= 5^2 - 1^2$
$= 25 - 1 = 24$
$\Rightarrow\ \text{PL}=\sqrt{24}=2\sqrt{6}\text{ cm}$
Thus $PQ = 2 \times PL$
$=\big(2\times2\sqrt{6}\big)=4\sqrt{6}\text{ cm}$
Hence, the required length of $PQ$ is $4\sqrt{6}\text{ cm}.$

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Question 275 Marks

Two parallel chords of lengths $30\ cm$ and $16\ cm$ are drawn on the opposite sides of the centre of a circle of radius $17\ cm.$ Find the distance between the chords.

Answer

Let $AB$ and $CD$ be two chords of a circle such that $AB$ is parallel to $CD$ and they are on the opposite sides of the centre.
Given:$ AB = 30\ cm$ and $CD = 16\ cm$
Draw $\text{OL}\perp\text{AB}$ and $\text{OM}\perp\text{CD}.$

Join $OA$ and $OC.$
$OA = OC = 17\ cm [$Radii of a circle$]$
The perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore\ \text{AL}=\Big(\frac{\text{AB}}{2}\Big)=\Big(\frac{30}{2}\Big)=15\text{ cm}$
Now, in right angled $\triangle\text{OLA},$ we have:
$OA^2 = AL^2 + LO^2$
$\Rightarrow LO^2= OA^2 - AL^2$
$\Rightarrow LO^2 = 17^2 - 15^2$
$\Rightarrow LO^2 = 289 - 225 = 64$
$\therefore LO = 8\ cm$
Similarly, $\therefore\ \text{CM}=\Big(\frac{\text{CD}}{2}\Big)=\Big(\frac{16}{2}\Big)=8\text{ cm}$
In right angled $\triangle\text{CMO},$ we have:
$\Rightarrow OC^2 = CM^2 + MO^2$
$\Rightarrow MO^2 = OC^2 - CM^2$
$\Rightarrow MO^2 = 17^2 - 8^2$
$\Rightarrow MO^2 = 289 - 64 = 225$
$\therefore MO = 15\ cm$
Hence, distance between the chords $= (LO + MO) = (8 + 15)\ cm = 23\ cm$

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Question 285 Marks

In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, $\text{OP}\perp\text{AB}$ and $\text{OQ}\perp\text{AC},$
prove that PB = QC.

Answer

Given: AB and AC are chords of the circle with centre O. AB = AC, $\text{OP}\perp\text{AB}$ and $\text{OQ}\perp\text{AC}.$

To prove: PB = QC
Proof: AB = AC (Given)
$\Rightarrow\ \frac{1}{2}\text{AB}=\frac{1}{2}\text{AC}$
The perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore$ MB = NC ...(1)
Also, OM = ON (Equal chords of a circle are equidistant from the centre)
and OP = OQ (Radii)
⇒ OP - OM = OQ - ON
$\therefore$ PM = QN ...(2)
Now, in $\triangle\text{MPB}$ and $\triangle\text{NQC},$ we have:
MB = NC [From (i)]
$\angle\text{PMB}=\angle\text{QNC}$ [90° each]
PM = QN [From (ii)]
i.e., $\triangle\text{MPB}\cong\triangle\text{NQC}$ (SAS criterion)
$\therefore$ PB = QC (C.P.C.T)

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Question 295 Marks

In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If $\angle\text{ABC}=25^\circ,$ calculate $\angle\text{CED}.$

Answer

$\angle\text{BCD}=\angle\text{ABC}=25^\circ$ [Alternate angles]
Join CO and DO.
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by an arc at any point on the circumference.
Thus, $\angle\text{BOD}=2\angle\text{BCD}$
$\Rightarrow\ \angle\text{BOD}=2\times25^\circ=50^\circ$
Similarly, $\angle\text{AOC}=2\angle\text{ABC}$
$\Rightarrow\ \angle\text{AOC}=2\times25^\circ=50^\circ$
AB is a straight line passing through the centre.
$\text{i.e.},\angle\text{AOC}+\angle\text{COD}+\angle\text{BOD}=180^\circ$
$\Rightarrow\ 50^\circ+\angle\text{COD}+50^\circ=180^\circ$
$\Rightarrow\ \angle\text{COD}=(180^\circ-100^\circ)=80^\circ$
$\Rightarrow\ \angle\text{CED}=\frac{1}{2}\angle\text{COD}$
$\Rightarrow\ \angle\text{CED}=\Big(\frac{1}{2}\times80^\circ\Big)=40^\circ$
$\therefore\ \angle\text{CED}=40^\circ$

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Question 305 Marks

In the given figure, O is the centre of the given circle and measure of arc ABC is 100°. Determine $\angle\text{ADC}$ and $\angle\text{ABC}.$

Answer

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.
Thus, $\angle\text{AOC}=2\angle\text{ADC}$
$\Rightarrow\ 100^\circ=2\angle\text{ADC}$
$\therefore\ \angle\text{ADC}=50^\circ$
The opposite angles of a cyclic quadrilateral are supplementary and ABCD is a cyclic quadrilateral.
Thus, $\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 50^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=(180^\circ-50^\circ)=130^\circ$
$\therefore\ \angle\text{ADC}=50^\circ$ and $ \angle\text{ABC}=130^\circ$

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Question 315 Marks

Find the length of a chord which is at a distance of $3\ cm$ from the centre of a circle of radius $5\ cm.$

Answer

Let $AB$ be the chord of the given circle with centre $O$ and a radius of $5\ cm.$
Then $OM = 3\ cm$ and $OB = 5\ cm$

From the right $\triangle\text{OMB},$ we have:
$OB^2 = OM^2 + MB^{2} [$ Pythagoras theorem$]$
$\Rightarrow 5^2 = 3^2 + MB^2$
$\Rightarrow 25 = 9 + MB^2$
$\Rightarrow MB^2 = (25 - 9) = 16$
$\Rightarrow\ \text{MB}=\sqrt{16}\text{ cm}=4\text{ cm}$
Since the perpendicular from the centre of a circle to a chord bisects the chord, we have:
$AB = 2 \times MB = (2 \times 4)cm = 8\ cm$
Hence, the required length of the chord is $8\ cm.$

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Question 325 Marks

In the given figure, O is the centre of a circle in which $\angle\text{OAB}=20^\circ$and $\angle\text{OCB}=55^\circ.$ Find

$\angle\text{BOC}$
$\angle\text{AOC}$

Answer

OB = OC [Radii of a circle]

$\Rightarrow\ \angle\text{OBC}=\angle\text{OCB}=55^\circ$
Considering $\triangle\text{BOC},$ we have:
$\angle\text{BOC}+\angle\text{OCB}+\angle\text{OBC}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{BOC}+55^\circ+55^\circ=180^\circ$
$\Rightarrow\ \angle\text{BOC}=(180^\circ-110^\circ)=70^\circ$

OA = OB [Radii of a circle]

$\Rightarrow\ \angle\text{OBA}=\angle\text{OAB}=20^\circ$
Considering $\triangle\text{AOB},$ we have:
$\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{AOB}+20^\circ+20^\circ=180^\circ$
$\Rightarrow\ \angle\text{AOB}=(180^\circ-40^\circ)=140^\circ$
$\therefore\ \angle\text{AOC}=\angle\text{AOB}-\angle\text{BOC}$
$=(140^\circ-70^\circ)$
$=70^\circ$
Hence, $\angle\text{AOC}=70^\circ$

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Question 335 Marks

In the given figure, AB and CD are straight lines through the centre O of a circle. If $\angle\text{AOC}=80^\circ$ and $\angle\text{CDE}=40^\circ,$ find

$\angle\text{DCE}$
$\angle\text{ABC}$

Answer

$\angle\text{CED}=90^\circ$ [Angle in a semi circle]

In $\triangle\text{CED},$ we have:
$\angle\text{CED}+\angle\text{EDC}+\angle\text{DCE}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 90^\circ+40^\circ+\angle\text{DCE}=180^\circ$
$\Rightarrow\ \angle\text{DCE}=(180^\circ-130^\circ)=50^\circ\dots(\text{i})$
$\therefore\ \angle\text{DCE}=50^\circ$

As $\angle\text{AOC}$ and $\angle\text{BOC}$ are linear pair, we have:

$\angle\text{BOC}=(180^\circ-80^\circ)=100^\circ\dots(\text{ii})$
In $\triangle\text{BOC},$ we have:
$\angle\text{ABC}+\angle\text{DCB}+\angle\text{BOC}=180^\circ$ $[\because\ \angle\text{OBC}=\angle\text{ABC}$ and $\angle\text{OCB}=\angle\text{DCE}]$
$\Rightarrow\ \angle\text{ABC}=180^\circ-(\angle\text{BOC}+\angle\text{DCE})$
$\Rightarrow\ \angle\text{ABC}=180^\circ-(100^\circ+50^\circ)$ [From (i) and (ii)]
$\Rightarrow\ \angle\text{ABC}=(180^\circ-150^\circ)=30^\circ$

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Question 345 Marks

In the adjoining figure, chords AC and BD of a circle with centre O, intersect at right angles at E. If $\angle\text{OAB}=25^\circ,$ calculate $\angle\text{EBC}.$

Answer

OA = OB [Radii of a circle]
Thus, $\angle\text{OBA}=\angle\text{OAB}=25^\circ$
Join OB.

Now in $\triangle\text{OAB},$ we have:
$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 25^\circ+25^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\ 50^\circ+\angle\text{AOB}=180^\circ$
$\Rightarrow\ \angle\text{AOB}=(180^\circ-50^\circ)=130^\circ$
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.
$\text{i.e},\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\ \angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=\Big(\frac{1}{2}\times130^\circ\Big)=65^\circ $
Here, $\angle\text{ACB}=2\angle\text{ECB}$
$\therefore\ \angle\text{ECB}=65^\circ\dots(\text{i})$
Considering the right angled $\triangle\text{BEC},$ we have:
$\angle\text{EBC}+\angle\text{BEC}+\angle\text{ECB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{EBC}+90^\circ+65^\circ=180^\circ$ [From (i)]
$\Rightarrow\ \angle\text{EBC}=(180^\circ-155^\circ)=25^\circ$
Hence, $ \angle\text{EBC}=25^\circ$

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Question 355 Marks

In the given figure, PQ is a diameter of a circle with centre O. If $\angle\text{PQR}=65^\circ,\angle\text{SPR}=40^\circ$ and $\angle\text{PQM}=65^\circ,$ find $\angle\text{QPR},\angle\text{QPM}$ and $\angle\text{PRS}.$

Answer

Here, PQ is the diameter and the angle in a semicircle is a right angle.
$\text{i.e},\angle\text{PRQ}=90^\circ$
In $\triangle\text{PRQ},$ we have:
$\angle\text{QPR}+\angle\text{PRQ}+\angle\text{PQR}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{QPR}+90^\circ+65^\circ=180^\circ$
$\Rightarrow\ \angle\text{QPR}=(180^\circ-155^\circ)=25^\circ$
In $\triangle\text{PQM}$ PQ is the diameter.
$\angle\text{QPM}+\angle\text{PMQ}+\angle\text{PQM}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{QPM}+90^\circ+50^\circ=180^\circ$
$\Rightarrow\ \angle\text{QPM}=(180^\circ-145^\circ)=40^\circ$
Now, in quadrilateral PQRS, we have:
$\angle\text{QPS}+\angle\text{SRQ}=180^\circ$ [Opposite angles of a cyclic quadrilateral]
$\Rightarrow\ \angle\text{QPR}+\angle\text{RPS}+\angle\text{PRQ}+\angle\text{PRS}=180^\circ$
$\Rightarrow\ 25^\circ+40^\circ+90^\circ+\angle\text{PRS}=180^\circ$
$\Rightarrow\ \angle\text{PRS}=180^\circ-155^\circ=25^\circ$
$\therefore\ \angle\text{PRS}=25^\circ$
Thus, $ \angle\text{QPR}=25^\circ, \angle\text{QPM}=25^\circ,\angle\text{PRS}=25^\circ$

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Question 365 Marks

In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC || CD and AC = 2 × OD.

Answer

Given: BC is a diameter of a circle with centre O and $\text{OD}\perp\text{AB}.$
To prove: AC parallel to OD and AC = 2 × OD
Construction: Join AC.
Proof:
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
Here, $\text{OD}\perp\text{AB}.$
D is the mid point of AB.
i.e., AD = BD
Also, O is the midpoint of BC.
i.e., OC = OB
Now, in $\triangle\text{ABC},$ we have:
D is the midpoint of AB and O is the mid point of BC.
According to the mid point theorem, the line segment joining the mid points of any two sides of a triangle is parallel to the third side and equal to half of it.
i.e., OD || AC and $\text{OD}=\frac{1}{2}\text{AC}$
$\therefore$ AC = 2 × OD
Hence, proved.

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Question 375 Marks

$AB$ and $AC$ are two chords of a circle of radius r such that $AB = 2AC.$ If p and q are the distances of $AB$ and $AC$ from the centre then prove that $4q^2 = p^2 + 3r^2$.

Answer

Let $AC = a.$
Since$, AB = 2AC, \therefore AB = 2a.$

From centre $O,$ perpendicular is drawn to the chords $AB$ and $AC$ at points $M$ and $N,$ respectively.
It is given that $OM = p$ and $ON = q.$
We know that perpendicular drawn from the centre to the chord, bisects the chord.
$\therefore AM = MB = a ...(1)$
and $\text{AN}=\text{NC}=\frac{\text{a}}{2}\dots(2)$
In $\triangle\text{OAN},$
$(AN)^2 + (NO)^2 = (OA)^2 [$Pythagoras theorem$]$
$\Rightarrow\ \Big(\frac{\text{a}}{2}\Big)^2+(\text{q})^2=(\text{r})^2$
$\Rightarrow\ \frac{\text{a}^2}{4}+\text{q}^2=\text{r}^2$
$\Rightarrow \frac{\text{a}^2+4\text{q}^2}{4}=\text{r}^2$
$\Rightarrow\ \text{a}^2+4\text{q}^2=4\text{r}^2$
$\Rightarrow\ \text{a}^2=4\text{r}^2-4\text{q}^2\dots(3)$
In $\triangle\text{OAM},$
$(\text{AM})^2+(\text{MO})^2=(\text{OA})^2$ [Pythagoras theorem]
$\Rightarrow\ (\text{a})^2+\text{p}^2=(\text{r)}^2$
$\Rightarrow\ (\text{a})^2=(\text{r)}^2-\text{p}^2\dots(4)$
From equation $(3)$ and $(4),$
$4\text{r}^2-4\text{q}^2=\text{r}^2-\text{p}^2$
$\Rightarrow\ 4\text{r}^2-\text{r}^2+\text{p}^2=4\text{q}^2$
$\Rightarrow\ 3\text{r}^2+\text{p}^2=4\text{q}^2$
Hence, $4\text{q}^2=\text{p}^2+3\text{r}^2$

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Question 385 Marks

In the given figure, $\triangle\text{ABC}$ is equilateral. Find

$\angle\text{BDC}$
$\angle\text{BEC}$

Answer

Given: $\triangle\text{ABC}$ is an equilateral triangle

i.e., each of its angle = 60°
$\Rightarrow\ \angle\text{BAC}=\angle\text{ABC}=\angle\text{ACB}=60^\circ$
Angles in the same segment of a circle are equal.
$\text{i.e.},\angle\text{BDC}=\angle\text{BAC}=60^\circ$
$\therefore\ \angle\text{BDC}=60^\circ$

The opposite angles of a cyclic quadrilateral are supplementary.

Then in cyclic quadrilateral ABEC, we have:
$\angle\text{BAC}+\angle\text{BEC}=180^\circ$
$\Rightarrow\ 60^\circ+\angle\text{BEC}=180^\circ$
$\Rightarrow\ \angle\text{BEC}=(180^\circ-60^\circ)=120^\circ$
$\therefore\ \angle\text{BDC}=60^\circ$ and $\angle\text{BEC}=120^\circ$

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Question 395 Marks

In the given figure, O is the centre of the circle and $\angle\text{BCO}=30^\circ.$ Find x and y.

Answer

In the given figure, OD is parallel to BC.
$\therefore\ \angle\text{BCO}=\angle\text{COD}$ [Alternate interior angles]
$\Rightarrow \angle\text{COD}=30^\circ\dots(\text{i})$
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.
Here, arc CD subtends $\angle\text{COD}$ at the centre and $\angle\text{CBD}$ at B on the circle.
$\therefore\ \angle\text{COD}=2\angle\text{CBD}$
$\Rightarrow \angle\text{CBD}=\frac{30^\circ}{2}=15^\circ$ [From (i)]
$\therefore\ \text{y}=15^\circ\dots(\text{ii})$
Also, arc AD subtends $\angle\text{AOD}$ at the centre and $\angle\text{ABD}$ at B on the circle.
$\therefore\ \angle\text{AOD}=2\angle\text{ABD}$
$\Rightarrow \angle\text{ABD}=\frac{90^\circ}{2}=45^\circ$
In $\triangle\text{ABE},$
$\text{x}+\text{y}+\angle\text{ABD}+\angle\text{AEB}=180^\circ$ [Sum of the angles of a triangle]
$\Rightarrow\ \text{x}+15^\circ+45^\circ+90^\circ=180^\circ$ [From (ii) and (iii)]
$\Rightarrow\ \text{x}=180^\circ-(15^\circ+45^\circ+90^\circ)$
$\Rightarrow\ \text{x}=180^\circ-150^\circ$
$\Rightarrow\ \text{x}=30^\circ$
Hence, $\text{x}=30^\circ$ and $\text{y}=15^\circ.$

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Question 405 Marks

In the given figure, a circle with centre $O$ is given in which a diameter $AB$ bisects the chord $CD$ at a point $E$ such that $CE = ED = 8\ cm$ and $EB = 4\ cm$. Find the radius of the circle.

Answer

$AB$ is the diameter of the circle with centre $O,$ which bisects the chord $CD$ at point $E.$
Given: $CE = ED = 8\ cm$ and $EB = 4\ cm$
Join $OC.$

Let $OC = OB = r\ cm [$Radii of a circle$]$
Then $OE = (r - 4)\ cm$
Now, in right angled $\triangle\text{OEC},$ we have:
$OC^2 = OE^2 + EC^{2 }[$Pythagoras theorem$]$
$\Rightarrow r^2 = (r - 4)^2 + 8^2$
$\Rightarrow r^2 = r^2 - 8r + 16 + 64$
$\Rightarrow r^2 = r^2 + 8r = 80$
$\Rightarrow 8r = 80$
$\Rightarrow\ \text{r}=\Big(\frac{80}{8}\Big)\text{cm}=10\text{cm}$
$\Rightarrow r = 10\ cm$
Hence, the required radius of the circle is $10\ cm.$

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Question 415 Marks

In the adjoining figure, OPQR is a square. A circle drawn with centre O cuts the square at X and Y. Prove that QX = QY.

Answer

Given: OPQR is a square. A circle with centre O cuts the square at X and Y.
To prove: QX = QY
Construction: Join OX and OY.

Proof:
In $\triangle\text{OXP}$ and $\triangle\text{OYP},$ we have:
$\angle\text{OPX}=\angle\text{ORY}$ (90° each)
OX = OY (Radii of a circle)
OP = OR (Sides of a square)
$\therefore\ \triangle\text{OXP}\cong\triangle\text{OYP}$ (BY R.H.S. congruency rule)
⇒ PX = RY (By C.P.C.T.)
⇒ PQ - PX = QR - RY (PQ and QR are sides of a square)
⇒ QX = QY
Hence, proved.

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Question 425 Marks

Two circles of radii $10\ cm$ and $8\ cm$ intersect each other, and the length of the common chord is $12\ cm.$ Find the distance between their centres.

Answer

Given: $OA = 10\ cm, O'A = 8\ cm$ and $AB = 12\ cm$
$\text{AD}=\Big(\frac{\text{AB}}{2}\Big)=\Big(\frac{12}{2}\Big)=6\text{ cm}$
Now, in right angled $\triangle\text{ADO},$ we have:
$OA^2 = AD^2 + OD^2$
$\Rightarrow OD^2 = OA^2 - AD^2$
$= 10^2 - 6^2$
$= 100 - 36 = 64$
$\therefore OD = 8\ cm$
Similarly, in right angled $\triangle\text{ADO}$ we have:
$O'A^2 = AD^2 + O'D^2$
$\Rightarrow O'D^2= O'A^2 - AD^2$
$= 8^2 - 6^2$
$= 64 - 36$
$= 28$
$\Rightarrow\ \text{O}'\text{D}=\sqrt{28}=2\sqrt{7}\text{ cm}$
Thus, $OO' = (OD + O'D)$
$=\big(8+2\sqrt{7}\big)\text{ cm}$
Hence, the distance between their centres is $\big(8+2\sqrt{7}\big)\text{ cm}.$

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Question 435 Marks

In the given figure, O is the centre of the circle, BD = OD and $\text{CD}\perp\text{AB}.$ Find $\angle\text{CAB}.$

Answer

In the given figure, BD = OD and $\text{CD}\perp\text{AB}.$

Join AC and OC.
In $\triangle\text{ODE}$ and $\triangle\text{DBE},$
$\angle\text{DOE}=\angle\text{DBE}$ [given]
$\angle\text{DEO}=\angle\text{DEB}=90^\circ$
OD = DB [given]
$\therefore$ By AAS conguence rule, $\triangle\text{ODE}\cong\triangle\text{BDE},$
Thus, OE = EB ...(1)
Now, in $\triangle\text{COE}$ and $\triangle\text{CBE},$
CE = CE [common]
$\angle\text{CEO}=\angle\text{CEB}=90^\circ$
OE = EB [From (1)]
$\therefore$ By AAS conguence rule, $\triangle\text{COE}\cong\triangle\text{CBE},$
Thus, CO = CB ...(2)
Also, CO = OB = OA [radius of the circle] ...(3)
From (2) and (3),
CO = CB = OB
$\therefore\ \triangle\text{COB}$ is equilateral triangle.
$\therefore\ \triangle\text{COB}=60^\circ\dots(4)$
We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by it on the remaining part of the circle.
Here, arc CB subtends $\angle\text{COB}$ at the centre and $\angle\text{CAB}$ at A on the circle.
$\therefore\ \angle\text{COB}=2\angle\text{CAB}$
$\Rightarrow\ \angle\text{CAB}=\frac{60^\circ}{2}=30^\circ$ [From (4)]
Hence, $\angle\text{CAB}=30^\circ.$

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Question 445 Marks

In a circle of radius $5\ cm, AB$ and $CD$ are two parallel chords of lengths $8\ cm$ and $6\ cm$ respectively. Calculate the distance between the chords if they are:

On the same side of the centre.
On the opposite sides of the centre.

Answer

We have:

Let $AB$ and $CD$ be two chords of a circle such that $AB$ is parallel to $CD$ on the same side of the circle.

Given: $AB = 8\ cm, CD = 6\ cm$ and $OB = OD = 5\ cm$
Join $OL$ and $OM.$

The perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore\ \text{LB}=\frac{\text{AB}}{2}=\Big(\frac{8}{2}\Big)=4\text{ cm}$
Now, in right angled $\triangle\text{BLO},$ we have:
$OB^2 = LB^2 + LO^2$
$\Rightarrow LO^2= OB^2 - LB^2$
$\Rightarrow LO^2= 5^2 - 4^2$
$\Rightarrow LO^{2}= 25 - 16 = 9$
Similarly,
$\Rightarrow MO = 4\ cm$
$\therefore$ Distance between the chords $= (MO - LO) = (4 - 3)\ cm = 1\ cm.$

Let $AB$ and $CD$ be two chords of a circle such that $AB$ is parallel to $CD$ and they are on the opposite sides of the centre.

Given: $AB = 8\ cm$ and $CD = 6\ cm$
Draw $\text{OL}\perp\text{AB}$ and $\text{OM}\perp\text{CD}.$

Join $OA$ and $OC.$
$OA = OC = 5\ cm [$Radii of a circle$]$
The perpendicular from the centre of a circle to a chord bisects the chord.
$\therefore\ \text{AL}=\frac{\text{AB}}{2}=\Big(\frac{8}{2}\Big)=4\text{ cm}$
Now, in right angled $\triangle\text{OLA},$ we have:
$OA^2 = AL^2 + LO^2$
$\Rightarrow LO^2= OA^2 - AL^2$
$\Rightarrow LO^2 = 5^2 - 4^2$
$\Rightarrow LO^2 = 25 - 16 = 9$
$\therefore LO = 3\ cm$
Similarly, $\text{CM}=\frac{\text{CD}}{2}=\Big(\frac{6}{2}\Big)=3\text{ cm}$
In right angled $\triangle\text{CMO},$ we have:
$OC^2 = CM^2 + MO^2$
$\Rightarrow MO^2 = OC^2 - CM^2$
$\Rightarrow MO^{2}= 5^2 - 3^2$
$\Rightarrow MO^{2}= 25 - 9 = 16$
$\therefore MO = 4\ cm$
Hence, distance between the chords $= (MO + LO) = (4 + 3) \ cm = 7\ cm.$

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Question 455 Marks

In the given figure, AB is a diameter of a circle with centre O and DO || CB. If $\angle\text{BCD}=120^\circ,$ calculate

$\angle\text{BAD}$
$\angle\text{ABD}$
$\angle\text{CBD}$
$\angle\text{ADC}$

Also, show that $\triangle\text{AOD}$ is an equilateral triangle.

Answer

We have,
AB is a diameter of the circle where O is the centre, DO || BC and $\angle\text{BCD}=120^\circ.$
Since ABCD is a cyclic quadrilateral, we have:
$\angle\text{BCD}+\angle\text{BAD}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{BAD}=180^\circ$
$\Rightarrow\ \angle\text{BAD}=(180^\circ-120^\circ)=60^\circ$
$\therefore\ \angle\text{BAD}=60^\circ$
$ \angle\text{BAD}=90^\circ$ [Angle in a semicircle]
In $\triangle\text{ABD},$ we have:
$\angle\text{BDA}+\angle\text{BAD}+\angle\text{ABD}=180^\circ$
$\Rightarrow\ 90^\circ+60^\circ+\angle\text{ABD}=180^\circ$
$\Rightarrow\ \angle\text{ABD}=(180^\circ-150^\circ)=30^\circ$
$\therefore\ \angle\text{ABD}=30^\circ$
OD = OA [Radii of a circle]
$\angle\text{ODA}=\angle\text{OAD}$
$=\angle\text{BAD}=60^\circ$
$\angle\text{ODB}=90^\circ-\angle\text{ODA}=(90^\circ-60^\circ)=30^\circ$
Here, DO || BC [Given; alternate angles]
$\angle\text{CBD}=\angle\text{ODB}=30^\circ$
$\therefore\ \angle\text{CBD}=30^\circ$
$\angle\text{ADC}=\angle\text{ADB}+\angle\text{CDB}$
$=90^\circ+30^\circ=120^\circ$
In $\triangle\text{AOD},$ we have:
$\angle\text{ODA}+\angle\text{OAD}+\angle\text{AOD}=180^\circ$
$\Rightarrow\ 60^\circ+60^\circ+\angle\text{AOD}=180^\circ$
$\Rightarrow\ \angle\text{AOD}=(180^\circ-120^\circ)=60^\circ$
Since all the angles of $\triangle\text{AOD}$ are of 60° each, $\triangle\text{AOD}$ is an equilateral triangle.

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Question 465 Marks

An equilateral triangle of side 9cm is inscribed in a circle. Find the radius of the circle.

Answer

Let $\triangle\text{ABC}$ be an equilateral triangle of side 9cm.
Let AD be one of its median.

Then, $\text{AD}\perp\text{BC}$ $[\triangle\text{ABC}$ is an equilateral triangle$]$
Also,
$\text{BD}=\Big(\frac{\text{BC}}{2}\Big)=\Big(\frac{9}{2}\Big)=4.5\text{cm}$
In right angled $\triangle\text{ADB},$ we have:
$\text{AB}^2=\text{AD}^2+\text{BD}^2$
$\Rightarrow\ \text{AD}^2=\text{AB}^2-\text{BD}^2$
$\Rightarrow\ \text{AD}=\sqrt{\text{AB}^2-\text{BD}^2}$
$=\sqrt{(9)^2-\Big(\frac{9}{2}\Big)^2}\text{cm}$
$=\frac{9\sqrt{3}}{2}\text{cm}$
In the equilateral triangle, the centroid and circumcentre coincide and AG : GD = 2 : 1.
Now, radius $=\text{AG}=\frac{2}{3}\text{AD}$
$\Rightarrow\ \text{AG}=\Big(\frac{2}{3}\times\frac{9\sqrt{3}}{2}\Big)=3\sqrt{3}\text{cm}$
$\therefore$ The radius of the circle is $3\sqrt{3}\text{cm}.$

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