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Question 514 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, prove that 4q² = p² + 3r².

Answer

Given: In a circlr of radius r, there are two chords AB and AC such that AB = 2AC, Also, the distance of AB and AC from the centre are P and q, respectively.
To prove: $\text{4q}^2+\text{p}^2+3\text{r}^2,$
Proof: Let AC = a, then AB = 2a

From centre O, perpendicular is drawn to the chords AC and AB at M and N, respectively.
$\therefore\text{AM}=\text{MC}=\frac{\text{a}}{2}$
$\text{AN}=\text{NB}=\text{a}$
In $\triangle\text{OAM},\ \ \text{AO}^2=\text{AM}^2+\text{MO}^2$ [by pythagoras theorem]
$\Rightarrow\text{AO}^2=\Big(\frac{\text{a}}{2}\Big)^2+\text{q}^2\ \ ...\text{(i)}$
In $\triangle\text{OAN},$ use pythagoras theorem,
$\text{AO}^2=(\text{AN})^2+(\text{NO})^2$
$\Rightarrow\text{AO}^2=(\text{a})^2+\text{p}^2\ \ \ ...(\text{ii})$
From Eqs. (i) and (ii),
$\Big(\frac{\text{a}}{2}\Big)^2+\text{q}^2=\text{a}^2+\text{p}^2$
$\Rightarrow\frac{\text{a}^2}{4}+\text{q}^2=\text{a}^2+\text{p}^2$
$\Rightarrow\text{a}^2+4\text{q}^2=4\text{a}^2+4\text{p}^2$ [multiplying both sides by 4]
$\Rightarrow4\text{q}^2=3\text{a}^2+4\text{p}^2$
$\Rightarrow4\text{q}^2=\text{p}^2+3(\text{a}^2+\text{p}^2)$
$\Rightarrow4\text{q}^2=\text{p}^2+3\text{r}^2$ $\big[$In right angled $\triangle\text{OAN},\ \text{r}^2=\text{a}^2+\text{p}^2\big]$
Hence proved.

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Question 524 Marks

Prove that a diameter of a circle which bisects a chord of the circle also bisects the angle subtended by the chord at the centre of the circle.

Answer

Given:

PQ is a diameter of circle which bisects the chord AB at C.
To Prove: PQ bisects $\angle\text{AOB}$
Proof:
In $\angle\text{AOC}$ and $\angle\text{BOC}$
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [Given]
Then $\triangle\text{AOC}\cong\triangle\text{BOC}$ [By SSS condition]
$\angle\text{AOC}=\angle\text{BOC}$ [C.P.C.T]
Hence PQ bisects $\angle\text{AOB}.$

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Question 534 Marks

Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively Prove that $\angle\text{ACP}=\angle\text{QCD}.$

Answer

Join chords AP and DQ.
For chord AP,
$\angle\text{PBA}=\angle\text{ACP}$ (Angles in the same segment) ...(1)
For chord DQ,
$\angle\text{DBQ}=\angle\text{QCD}$ (Angles in the same segment) ...(2)
ABD and PBQ are line segments intersecting at B.
$\therefore\angle\text{PBA}=\angle\text{DBQ}$ (Vertically opposite angles) ...(3)
From equations (1), (2), and (3), we obtain
$\angle\text{ACP}=\angle\text{QCD}$

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Question 544 Marks

On a semi-circle with AB as diameter, a point C is taken, so that $\text{m}\big(\angle\text{CAB}\big)=30^\circ.$ Find $\text{m}\big(\angle\text{ACB}\big)$ and $\text{m}\big(\angle\text{ABC}\big).$

Answer

We have, $\angle\text{CAB}=30^\circ$
$\therefore\angle\text{ACB}=90^\circ$ [Angle in semicircle]
In $\triangle\text{ABC},$ by angle sum property
$\angle\text{CAB}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$
$\Rightarrow30^\circ+90^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-90^\circ-30^\circ=60^\circ$

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Question 554 Marks

If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD

Answer

Let us draw a perpendicular OM on line AD.

It can be observed that BC is the chord of the smaller circle and AD is the chord of the bigger circle.
We know that perpendicular drawn from the centre of the circle bisects the chord.
$\therefore$ BM = MC ...(1)
And, AM = MD ...(2)
On subtracting equation (2) from (1), we obtain
AM − BM = MD − MC
⇒ AB = CD

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Question 564 Marks

Prove that the angle in a segment greater than a semi-circle is less than a right angle.

Answer

Given: $\angle\text{ACB}$ is an angle in minor segment.
To prove: $\angle\text{ACB}<90^\circ$
Proof: By degree measure theorem
$\angle\text{AOB}=2\angle\text{ACB}$
And $\angle\text{AOB}<180^\circ$
Then, $2\angle\text{ACB}<180^\circ$
$\Rightarrow\angle\text{ACB}<\frac{180^\circ}{2}$
$\Rightarrow\angle\text{ACB}<90^\circ$

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Question 574 Marks

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

Answer

Given: In a figure, two chords AB and CD intersecting each other at point E.
To prove: $\angle\text{AEC}=\frac{1}{2}$ [angle subtended by arc C × A at centre + angle subtended by arc DYB at the centre]

Construction: Extend the line DO and BO at the points l and H on the circle. Also, join AC.
Proof: We know that, in a circle, the angle subtended by an arc at the centre is twice the angle subtended by it at the remaining part of the circle.
$\therefore\angle1=2\angle6\ \ ...(\text{i})$ and $\angle3=2\angle7\ \ ...(\text{ii)}$
In $\triangle\text{AOC},\ \ \text{OC}=\text{OA}$ [both are the radius of circle]
$\angle\text{OCA}=\angle4$ [angles opposite to equal sides are equal]
Also, $\angle\text{AOC}+\angle\text{OCA}+\angle4=180^\circ$ [by angle sum property of triangle]
$\Rightarrow\angle\text{AOC}+\angle4+\angle4=180^\circ$
$\Rightarrow\angle\text{AOC}=180^\circ-2\angle4\ \ \ ...(\text{iii})$
Now, in $\triangle\text{AEC},\ \ \ \angle\text{AEC}+\angle\text{ECA}+\angle\text{CAE}=180^\circ$ [by angle property sum of a triangle] $\Rightarrow\angle\text{AEC}=180^\circ-(\angle\text{ECA}+\angle\text{CAE})$
$\Rightarrow\angle\text{AEC}=180^\circ-[(\angle\text{ECO}+\angle\text{OCA})+\angle\text{CAO}+\angle\text{OAE}]$
$=180^\circ-(\angle6+\angle4+\angle4+\angle5)$ $\big[$In $\triangle\text{OCD},\angle6=\angle\text{ECO}$ angles opposite to equal sides are equal$\big]$
$=180^\circ-(2\angle4+\angle5+\angle6)$
$=180^\circ-(180^\circ-\angle\text{AOC}+\angle7+\angle6)$
[From Eq. (iii) and in $\triangle\text{AOB}.\angle5=\angle7,$ as (angles opposite to equal sides are equal)]
$=\angle\text{AOC}-\frac{\angle3}{2}-\frac{\angle1}{2}$ [from Eqs. (i) and (ii)]
$=\angle\text{AOC}-\frac{\angle1}{2}-\frac{\angle2}{2}-\frac{\angle3}{2}+\frac{\angle2}{2}$ [adding and subtracting $\frac{\angle2}{2}$]
$=\angle\text{AOC}-\frac{1}{2}(\angle1+\angle2+\angle3)+\frac{\angle8}{2}$ [$\because\angle2=\angle8$ (vertically opposite angles)]
$=\angle\text{AOC}=\frac{\angle\text{AOC}}{2}+\frac{\angle\text{DOB}}{2}$
$\Rightarrow\angle\text{AEC}=\frac{1}{2}(\angle\text{AOC}+\angle\text{DOB})$
$=\frac{1}{2}$ [angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre]

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Question 584 Marks

A circular park of radius 40m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Answer

Given that AB = BC = CA

So, ABC is an equilateral triangle
OA (radius) = 40m
Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC. We also know that median intersect each other at the ratio 2 : 1 As AD is the median of equilateral triangle ABC, we can write:
$\frac{\text{OA}}{\text{OD}}=\frac{2}{1}$
$\Rightarrow\frac{\text{4OM}}{\text{OD}}=\frac{2}{1}$
$\Rightarrow\text{OD}=20\text{m}$
Therefore, AD = OA + OD = (40 + 20)m = 60m In $\triangle\text{ADC}$ By using Pythagoras theorem
$\text{AC}^2 = \text{AD}^2 + \text{DC}^2$
$\text{AC}^2=60^2+\big(\text{AC}^2\big)^2$
$\text{AC}^2=3600+\frac{\text{AC}^2}{4}$
$\Rightarrow\frac{3}{4}\text{AC}^2=3600$
$\Rightarrow\text{AC}^2=4800$
$\Rightarrow\text{AC}=40\sqrt{3}\text{m}$
So, length of string of each phone will be $40\sqrt{3}\text{m}.$

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Question 594 Marks

In the given figure, two congruent circles with centres O and O' intersect at A and B. If $\angle\text{AOB} = 50^\circ,$ then find $\angle\text{APB.}$

Answer

Since both the circles are congruent, they will have equal radii. Let their radii be ‘r’
So, from the given figure we have,
OA = OB = O'A = O'B = r

Now, since all the sides of the quadrilateral OBO’A are equal it has to be a rhombus.
One of the properties of a rhombus is that the opposite angles are equal to each other.
So, since it is given that $\angle\text{AO}'\text{B}=50^\circ$ we can say that the angle opposite it, that is to say that
$\angle\text{AO}\text{B}$ should also have the same value.
Hence we get $\angle\text{AO}\text{B}=50^\circ$
Now, consider the first circle with the centre ‘O’ alone. ‘AB’ forms a chord and it subtends an angle of 50° with its centre, that is $\angle\text{AO}\text{B}=50^\circ.$
A property of a circle is that the angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.
This means that,
$\angle\text{APB}=\frac{\angle\text{AOB}}{2}$
$=\frac{50^\circ}{2}$
$=25^\circ$
Hence the measure of $\angle\text{APB}$ is 25°.

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Question 604 Marks

In the given figure, two circles intersect at A and B. The centre of the smaller circle is O and it lies on the circumference of the larger circle. If $\angle\text{APB} = 70^\circ,$ find $\angle\text{ACB.}$

Answer

Consider the smaller circle whose centre is given as ‘O’.
The angle subtended by an arc at the centre of the circle is double the angle subtended by the arc in the remaining part of the circle.

So, here we have
$\angle\text{AOB}=2\angle\text{APB}$
$=2(70^\circ)$
$\angle\text{AOB}=140^\circ$
Now consider the larger circle and the points ‘A’, ‘C’, ‘B’ and ‘O’ along its circumference. ‘ACBO’ form a cyclic quadrilateral.
In a cyclic quadrilateral it is known that the opposite angles are supplementary, meaning that the opposite angles add up to 180°.
$\angle\text{AOB}+\angle\text{ACB}=180^\circ$
$\angle\text{ACB}=180^\circ-\angle\text{AOB}$
$=180^\circ-140^\circ$ $\angle\text{ACB}=40^\circ$
Hence ,the measure of $\angle\text{ACB}$ is $40^\circ.$

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Question 614 Marks

Find the length of a chord which is at a distance of 5cm from the centre of a circle of radius 10cm.

Answer

Given that,

Distance (OC) = 5cm
Radius of the circle (OA) = 10cm
In $\triangle\text{OCA,}$ by Pythagoras theorem
OC²+ AC²= OA²
⇒ 5² + AC² = 10²
⇒ 25 + AC² = 100
⇒ AC² = 100 - 25
⇒ AC² = 75
$\Rightarrow\text{AC}=\sqrt{75}$
⇒ AC = 8.66cm We know that, the perpendicular from the centre to chord bisects the chord
Therefore, AC = BC = 8.66cm
Then the chord AB = 8.66 + 8.66
= 17.32cm

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Question 624 Marks

O is the circumference of the triangle ABC and OD is perpendicular on BC. Prove that $\angle\text{BOD}=\angle\text{A}.$

Answer

Given O is the circum centre of triangle ABC and $\text{OD}\perp\text{BC}$

To prove $\angle\text{BOD}=2\angle\text{A}$ Proof: In $\triangle\text{OBD}$ and $\triangle\text{OCD}$
$\angle\text{ODB}=\angle\text{ODC}$ [Each 90°] OB = OC [Radius of circle] OD = OD [Common] Then
$\triangle\text{OBD}\cong\triangle\text{OCD}$ [By RHS Condition].
$\therefore\angle\text{BOD}=\angle\text{COD}\dots(\text{i})$ [PCT].
By degree measure theorem
$\angle\text{BOC}=2\angle\text{BAC}$
$\Rightarrow2\angle\text{BOD}=2\angle\text{BAC}$ [By using (i)]
$\Rightarrow\angle\text{BOD}=\angle\text{BAC}$

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Question 634 Marks

In figure, O is the centre of the circle, BO is the bisector of $\angle\text{ABC}.$ Show that AB = AC.

Answer

Given, BO is the bisector of $\angle\text{ABC}$
To prove AB = BC
Proof:
Since, BO is the bisector of $\angle\text{ABC}$
Then, $\angle\text{ABO}=\angle\text{CBO}\dots(\text{i})$
Since, OB = OA [Radius of circle]
Then, $\angle\text{ABO}=\angle\text{DAB}\dots(\text{ii})$ [opposite angles to equal sides]
Since OB = OC [Radius of circle]
Then, $\angle\text{OAB}=\angle\text{OCB}\dots(\text{iii})$ [opposite angles to equal sides]
Compare equations (i), (ii) and (iii)
$\angle\text{OAB}=\angle\text{OCB}\dots(\text{iii})$
In $\triangle\text{OAB}$ and $\triangle\text{OCB}$
$\angle\text{OAB}=\angle\text{OCB}$ [From (iv)]
$\angle\text{OBA}=\angle\text{OBC}$ [Given]
OB = OB [Common]
Then, $\triangle\text{OAB}\cong\triangle\text{OCB}$ [By AAS condition]
$\therefore\text{AB}=\text{BC}$ [C.P.C.T.]

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Question 644 Marks

A circle has radius $\sqrt{2}\text{cm}.$ It is divided into two segments by a chord of length 2cm. Prove that the angle subtended by the chord at a point in major segment is 45°.

Answer

Draw a circle having centre O. Let AB = 2cm be a chord of a circle. A chord AB is divided by the line OM in two equal segments.

To prove, $\angle\text{APB}=45^\circ$
Here, AN = NB = 1cm and $\text{OB}=\sqrt{2}\text{cm}$
In $\triangle\text{ONB},\ \ \text{OB}^2=\text{ON}^2+\text{NB}^2$ [use pythagoras theorem]
$\Rightarrow(\sqrt{2})^2=\text{NO}^2+(1)^2$
$\Rightarrow\text{ON}^2=2-1=1$
$\Rightarrow\text{ON}=1\text{cm}$ [taking positive square root, because distance is always positive]
Also, $\angle\text{ONB}=90^\circ$ [ON is the perpendicular bisector of the chord AB]
$\therefore\angle\text{NOB}=\angle\text{NBO}=45^\circ$
Similarly, $\angle\text{AON}=45^\circ$
Now, $\angle\text{AOB}=\angle\text{AON}+\angle\text{NOB}$
$=45^\circ+45^\circ=90^\circ$
We know that, chord subtends an angle to the circle is half the angle subtended by it to the centre.
$\therefore\angle\text{APB}=\frac{1}{2}\angle\text{AOB}$
$=\frac{90^\circ}{2}=45^\circ$
Hence, proved.

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Question 654 Marks

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.

Answer

Given: AB and CD are two chords of a circle whose centre of O. The mid-points of AB and CD are L and M respectively.

To prove: AB || CD
Proof: $\because$ L is the mid-point of chord AB $\therefore\text{OL}\bot\text{AB},\ \text{or}\ \angle\text{ALO}=90^\circ$
[$\because$ The line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord]
Similarly, $\angle\text{CMO}=90^\circ$
$\therefore\angle\text{ALO}=\angle\text{CMO}$
But, these are corresponding angles.
So, AB || CD.
Hence, proved.

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Question 664 Marks

In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

Answer

$\angle\text{EDC}+\angle\text{CDA}=180^\circ$ [Linear pair of angles]
$\Rightarrow80^\circ+\angle\text{CDA}=180^\circ$
$\Rightarrow\angle\text{CDA}=180^\circ-80^\circ=100^\circ$
Since, ABCD is a cyclic quadrilateral.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow100^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\angle\text{ABC}=180^\circ-100^\circ=80^\circ$
Now, $\angle\text{ABC}+\angle\text{ABF}=180^\circ$ [Linear pair of angles]
$\Rightarrow80^\circ+\text{x}^\circ=180^\circ$
$\Rightarrow\text{x}=180^\circ-80^\circ=100^\circ$

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Question 674 Marks

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment.

Answer

Since chord of a circle is equal radius, so we have AB = OA = OB. Therefore, ABC is an equilateral triangle.

Since each angle of an equilateral triangle is 60°, so we have $\angle\text{AOB}=60^\circ$ Since angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle, so we have $\angle\text{AOB}=2\angle\text{ACB}$.
Hence, $\angle\text{ACB}=\frac{1}{2}\angle\text{AOB}=\frac{1}{2}\times60^\circ=30^\circ$

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Question 684 Marks

In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and Frespectively. If $\angle\text{CBF}=130^\circ$ and $\angle\text{CDE}=\text{x}^\circ,$ find the value of x.

Answer

ABCD is a cyclic quadrilateral.
We know that in a cyclic quadrilateral, the exterior angle = interior opposite angle.
$\therefore\ \angle\text{CBF}=\angle\text{CDA}=(180^\circ-\text{x})$
$\Rightarrow\ 130^\circ=180^\circ-\text{x}$
$\Rightarrow\ \text{x}=180^\circ-130^\circ=50^\circ$
$\Rightarrow\ \text{x}=50^\circ$

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Question 694 Marks

In figure, O is the centre of a circle and PQ is a diameter. If $\angle\text{ROS} = 40^\circ, $ find $\angle\text{RTS}.$

Answer

Since PQ is diameter
Then,
$\angle\text{PRQ} = 90^\circ$ [Angle in semicircle]
$\therefore\angle\text{PRQ}+\angle\text{TRQ}=180^\circ$ [Linear pair of angle]
$900+\angle\text{TRQ}=180^\circ$
$\angle\text{TRQ}=180^\circ-90^\circ=90^\circ$
By degree measure theorem
$\angle\text{ROS}=2\angle\text{RQS}$
$\Rightarrow40^\circ=2\angle\text{RQS}$
$\Rightarrow\angle\text{RQS}=\frac{40^\circ}{2}=20^\circ$
In $\triangle\text{RQT},$ by Angle sum property
$\angle\text{RQT}+\angle\text{QRT}+\angle\text{RTS}=180^\circ$
$\Rightarrow20^\circ+90^\circ+\angle\text{RTS}=180^\circ$
$\Rightarrow\angle\text{RTS}=180^\circ-20^\circ-90^\circ=70^\circ$

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Question 704 Marks

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

Answer

Let ABC be an equilateral triangle of side 9cm and let AD is one of its median.

Let G be the centroid of $\triangle\text{ABC}$ Then AG : GD = 2 : 1
We know that in an equilateral triangle, centroid coincides with the circum centre.
Therefore, G is the centre of the circumference with circum radius GA.
Also G is the centre and GD is perpendicular to BC.
Therefore, In right triangle ADB, we have
AB² = AD² + DB²
⇒ 9² = AD² + DB²$\Rightarrow\text{AD}=\sqrt{81-\frac{81}{4}}=\frac{9\sqrt3}{2}\text{cm}$\
$\therefore\text{Radius}=\text{AG}=\frac{2}{3}\text{AD}$
$=3\sqrt3\text{cm}.$

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Question 714 Marks

In a cyclic quadrilateral ABCD if AB || CD and $\angle\text{B}=70^\circ,$ find the remaining angles.

Answer

We have, $\angle\text{B}=70^\circ$
Since, ABCD is a cyclic quadrilateral
Then, $\angle\text{B}+\angle\text{D}=180^\circ$
$\Rightarrow70^\circ+\angle\text{D}=180^\circ$
$\Rightarrow\angle\text{D}=180^\circ-70^\circ=110^\circ$
Since, AB || DC
Then, $\angle\text{B}+\angle\text{C}=180^\circ$ [Co-interior angles]
$\Rightarrow70^\circ+\angle\text{C}=180^\circ$
$\Rightarrow\angle\text{C}=180^\circ-70^\circ=110^\circ$
Now, $\angle\text{A} +\angle\text{C}=180^\circ$ [Opposite angles of cyclic quad.]
$\Rightarrow\angle\text{A}+110^\circ=180^\circ$
$\Rightarrow\angle\text{A}=180^\circ-110^\circ=70^\circ$

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Question 724 Marks

In figure, O is the centre of the circle. If $\angle\text{APB}=50^\circ,$ find $\angle\text{AOB}$ and $\angle\text{OAB}.$

Answer

$\angle\text{APB}=50^\circ$

By degree measure theorem

$\angle\text{AOB}=2\angle\text{APB}$

$\Rightarrow\angle\text{APB}=2\times50^\circ=100^\circ$ since OA = OB [Radius of circle]

Then $\angle\text{OAB}=\angle\text{OBA}$ [Angles opposite toequalsides]

Let $\angle\text{OAB}=\text{x}$

In $\triangle\text{OAB},$ by angle sum property

$\angle\text{OAB}+\angle\text{OBA}+\angle\text{AOB}=180^\circ$

$\Rightarrow\text{x}+\text{x}+100^\circ=180^\circ$

$\Rightarrow\text{2x}=180^\circ-100^\circ$

$\Rightarrow\text{2x}=80^\circ$

$\Rightarrow\text{x}=40^\circ$

$\angle\text{OAB}=\angle\text{OBA}=40^\circ$

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Question 734 Marks

In the given figure, O is the centre of the circle. If $\angle\text{ABD}=35^\circ$ and $\angle\text{BAC}=70^\circ,$find $\angle\text{ACB}.$

Answer

It is clear that BD is the diameter of the circle. Also, we know that the angle in a semicircle is a right angle. i.e.,
$\angle\text{BAD}=90^\circ$ Now, considering the $\triangle\text{BAD},$ we have:
$\angle\text{ADB}+\angle\text{BAD}+\angle\text{ABD}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{ADB}+90^\circ+35^\circ=180^\circ$
$\Rightarrow\ \angle\text{ADB}=(180^\circ-125^\circ)=55^\circ$
Angles in the same segment of a circle are equal.
Hence, $\angle\text{ACB}=\angle\text{ADB}=55^\circ$

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Question 744 Marks

In figure, O is the centre of the circle. Find $\angle\text{BAC}.$

Answer

We have $\angle\text{AOB}=80^\circ$

And $\angle\text{AOC}=110^\circ$

Therefore, $\angle\text{AOB}+\angle\text{AOC}+\angle\text{BOC}=360^\circ$ [Complete angle]

$\Rightarrow80^\circ+100^\circ+\angle\text{BOC}=360^\circ$

$\Rightarrow\angle\text{BOC}=360^\circ-80^\circ-110^\circ$

$\Rightarrow\angle\text{BOC}=170^\circ$

By degree measure theorem

$\angle\text{BOC}=2\angle\text{BAC}$

$\Rightarrow170^\circ=2\angle\text{BAC}$

$\Rightarrow\angle\text{BAC}=\frac{170^\circ}{2}=85^\circ$

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Question 754 Marks

In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that $\triangle\text{AEB}$ is isosceles.

Answer

AB and CD are two parallel chords of a circle. BDE and ACE are two straight lines that intersect at E.
If one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle.
$\therefore$ Exteriore $\angle\text{EDC}=\angle\text{A}\dots(\text{i})$
Exteriore $\angle\text{DCE}=\angle\text{B}\dots(\text{ii})$
Also, AB parallel to CD. Then, $\angle\text{EDC}=\angle\text{B}$ [Corresponding angles]
$\angle\text{DCE}=\angle\text{A}$ [Corresponding angles]
$\therefore\ \angle\text{A}=\angle\text{B}$ [From (i) and (ii)]
Hence, $\triangle\text{AEB}$ is isosceles.

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Question 764 Marks

The circumcentre of the triangle ABC is O. Prove that $\angle\text{OBC} + \angle\text{BAC} = 90^\circ.$

Answer

ABC is a triangle and O is the circumcentre.

Draw $\text{OD}\bot\text{BC}.$ Join OB and OC.
In right $\triangle\text{OBD}$ and right $\triangle\text{OCD}$, we have hyp. OB = hyp. OC [Raddi of the same circle]
OD = OD [Common side]
$\therefore\triangle\text{OBD}\cong\triangle\text{OCD}$ [By RHS cong. Rule]
$\therefore\angle1=\angle2$ and $\angle1=\angle4$ [By C.P.C.T.]
Now, $\angle\text{BOC}=2\angle1$ and $\angle\text{BOC}=2\angle\text{A}$
$\therefore2\angle1=2\angle\text{A}\Rightarrow\angle1=\angle\text{A}$
$\therefore\angle\text{A}=\angle2\ ....(1)\ \ \ [\because\angle1=\angle2]$
$\Rightarrow\angle\text{A}+\angle4=\angle2+\angle4$ [Adding $\angle4$ to both sides]
$\Rightarrow\angle\text{A}+\angle3=90^\circ\ \ [\because\angle2+\angle4=90^\circ\ \text{and}\ \angle4=\angle3]$
$\Rightarrow\angle\text{BOC}=\angle\text{A}=90^\circ$
Hence, proved.

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Question 774 Marks

In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If $\angle\text{CBD}=60^\circ,$calculate $\angle\text{CDE}.$

Answer

Angles in the same segment of a circle are equal.
$\text{i.e},\angle\text{CAD}=\angle\text{CBD}=60^\circ$
We know that an angle in a semicircle is a right angle.
$\text{i.e},\angle\text{ACD}=90^\circ$ In $\triangle\text{ADC},$ we have:
$\angle\text{ACD}+\angle\text{ADC}+\angle\text{CAD}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ \angle\text{ACD}+90^\circ+60^\circ=180^\circ$
$\Rightarrow\ \angle\text{ACD}=180^\circ-(90^\circ+60^\circ)$
$=(180^\circ-150^\circ)=30^\circ$
$\Rightarrow\ \angle\text{CDE}=\angle\text{ACD}=30^\circ$ [Alternate angles as AC parallel to DE]
Hence, $\angle\text{CDE}=30^\circ$

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Question 784 Marks

In the given figure, O and O' are centers of two circles intersecting at B and C. ACD is a straight line, find x.

Answer

By degree measure theorem

$\angle\text{AOB} = 2\angle\text{ACB}$

$\Rightarrow130^\circ=2\angle\text{ACB}$

$\Rightarrow\angle\text{ACB}=\frac{130^\circ}{2}=65^\circ$ [Liner a pair of angles]

$\Rightarrow65^\circ+\angle\text{BCD}=180^\circ$

$\Rightarrow\angle\text{BCD}=180^\circ-65^\circ=115^\circ$ By degree measure theorem reflex

$\angle\text{BOD}=2\angle\text{BCD}$

$\Rightarrow\text{reflex }\angle\text{BOD}=2\times115^\circ=230^\circ$

Now, reflex $\angle\text{BOD}+\angle\text{BO}'\text{D}=360^\circ$ [Complex angle]

$\Rightarrow230^\circ+\text{x}=360^\circ$

$\Rightarrow\text{x}=360^\circ-230^\circ$

$\therefore\text{x}=130^\circ$

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Question 794 Marks

Given an arc of a circle, complete the circle.

Answer

Steps of Construction:

Take three points A, B and C on the given arc.
Join AB and BC.
Draw the perpendicular bisectors of chords AB and BC which intersect each other at point O. Then O will be the required centre of the required circle.
Join OA.
With centre O and radius OA, complete the circle.

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Question 804 Marks

Circles are described on the sides of a triangle as diameters. Proved that the circle on any two sides intersect each other on the third side (or third side produced).

Answer

Since, AB is a diameter
Then, $\angle\text{ADB}=90^\circ\dots(1)$ [Angles in semicircle]
Since, AC is a diameter
Then, $\angle\text{ADC}=90^\circ\dots(2)$ [Angles in semicircle]
Add equations (1) and (2)
$\angle\text{ADB}+\angle\text{ADC}=90^\circ+90^\circ$
$\Rightarrow\angle\text{BDC}=180^\circ$
Then, BDC is a line
Hence, the circles on any two sides intersect each other on the third side.

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Question 814 Marks

In Fig. $\angle\text{ACB}=40^\circ$. Find $\angle\text{OAB}.$

Answer

Given, $\angle\text{ACB}=40^\circ$
We know that, a segment subtends an angle to the circle is half the angle subtends to the centre.
$\therefore\angle\text{AOB}=2\angle\text{ACB}$
$\Rightarrow\angle\text{ACB}=\frac{\angle\text{AOB}}{2}$
$\Rightarrow40^\circ=\frac{1}{2}\angle\text{AOB}$
$\Rightarrow\angle\text{AOB}=80^\circ\ \ \ ...(\text{i})$
In $\triangle\text{AOB},\ \text{AO}=\text{BO}$ [both are the radius of a circle]
$\Rightarrow\angle\text{OBA}=\angle\text{OAB}\ \ ...(\text{ii})$ [angle opposite to the equal sides are equal]
We know that, the sum of all three angles in a triangle AOB is 180°.
$\therefore\angle\text{AOB}+\angle\text{OBA}+\angle\text{OAB}=180^\circ$
$\Rightarrow80^\circ+\angle\text{OAB}+\angle\text{OAB}=180^\circ$ [from Eqs. (i) and (ii)]
$\Rightarrow2\angle\text{OAB}=180^\circ-80^\circ$
$\Rightarrow2\angle\text{OAB}=100^\circ$
$\therefore\angle\text{OAB}=\frac{100^\circ}{2}=50^\circ$

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Question 824 Marks

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

Answer

Given: Consider AB and CD are two equal chords of a circle, which meet at point E.
To prove: AE = CE and BE = DE
Construction: Draw $\text{OM}\bot\text{AB}$ and $\text{ON}\bot\text{CD}$ and join OE where O is the centre of circle.