[4 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip

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Question 14 Marks

In the given figure, $\text{AOC}$ is the diameter of the circle, with centre $O$. If arc $\text{AXB}$ is half of arc $\text{BYC}$, find $\angle BOC.$

Answer

Given :
$1. \text{AOC}$ is the diameter.
$2.$ Arc $\text{AXB} = \frac{1}{2}$ Arc $\text{BYC}$
From Arc $\text{AXB} = \frac{1}{2}$ Arc $\text{BYC}$
We can see that
Arc $\text{AXB}$ : Arc $\text{BYC} = 1 : 2$
$\Rightarrow \angle BOA : \angle BOC = 1 : 2$
Since $\text{AOC}$ is the diameter of the circle hence,
$\angle AOC = 180^\circ $
Now,
Assume that $\angle BOA = x^\circ $ and $\angle BOC = 2x^\circ $
$\angle AOC = \angle BOA + \angle BOC = 180^\circ $
$\Rightarrow x + 2x = 180^\circ $
$\Rightarrow 3x = 180^\circ $
$\Rightarrow x = 60^\circ $
Hence, $\angle BOA = 60^\circ $ and $\angle BOC = 120^\circ .$

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Question 24 Marks

In the given figure, $O$ is the center of the circle with radius $20 \ cm$ and $OD$ is perpendicular to $AB$. If $AB = 32 \ cm,$
find the length of $CD.$

Answer

To find: $CD$
Given : $AB = 32 \ cm$
$\Rightarrow AC = 16 \ cm($ Since Perpendicular is drawn from the centre to the chord, bisects the chord$ )$
In Right $\triangle OCA,$
$OA^2 = OC^2 + AC^2 ....($ By Pythagoras theorem $)$
$\Rightarrow OC^2 = OA^2 - AC^2$
$\Rightarrow OC^2 = 20^2 - 16^2$
$\Rightarrow OC^2= 144$
$\Rightarrow OC = 12\ cm$
Since $OD = 20 \ cm$ and $OC = 12 \ cm$
$\Rightarrow CD = OD - OC$
$= 20 - 12$
$= 8 \ cm.$

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Question 34 Marks

In a circle of radius $10 \ cm, AB$ and $CD$ are two parallel chords of lengths $16 \ cm$ and $12 \ cm$ respectively.
Calculate the distance between the chords, if they are on:$(i)$ the same side of the center.$(ii)$ the opposite sides of the center.

Answer

Given that $AB = 16 \ cm$ and $CD = 12 \ cm$
So, $AL = 8 \ cm$ and $CM = 6 \ cm ....( \perp $ from the center to the chord bisects the chord $)$
In right $\triangle OLA$ and $OMC,$
By Pythagoras theorem,
$OA^2 = OL^2 + AL^2$ and $OC^2 = OM^2 + CM^2$
$10^2 = OL^2 + 8^2$ and $10^2 = OM^2 + 6^2$
$OL^2 = 100 - 64$ and $OM^2 = 64$
$OL^2 = 6 \ cm$ and $OM^2 = 8 \ cm$
$(i)$ In the first case, distance between $AB$ and $CD$ is
$LM= OM - OL = 8 - 6 = 2 \ cm$
$(ii)$ In the second case , distance between $AB$ and $CD$ is
$LM = OM + OL = 8 + 6 = 14 \ cm$

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Question 44 Marks

Prove that equal chords of congruent circles subtend equal angles at their center.

Answer

Given: $AB$ and $CD$ are two equal chords of a congruent circle with centres $O$ and $O$ respectively.
To prove:$ \angle AOB = \angle CO'D$
Proof:
In $\triangle OAB$ and $\triangle O'CD$
$OA = O'C ...( \because $Radii of congruent circles $)$
$OB = O'D...( \because $ Radii of congruent circles $)$
$AB = CD ...($ Given $)$
$\triangle OAB \cong \triangle O'CD...($ By $\text{SSS}$ congruence criterion $)$
$\angle AOB = \angle CO'D...(\text{ c. p. c. t })$

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Question 54 Marks

The radius of a circle is $13 \ cm$ and the length of one of its chords is $24 \ cm$.Find the distance of the chord from the center.

Answer

To find: $OM$
Given that $AB = 24 \ cm$
Since $OM \perp AB$
$\Rightarrow OM$ bisects $AB$
So, $AM = 12 \ cm$
In right
$\Rightarrow \text{OMA},$
$OA^2 = OM^2 + AM^2$
$OM^2 = OA^2 - AM^2$
$OM^2 = 13^2 - 12^2$
$OM^2 = 25$
$OM^2 = 5 \ cm$
Hence, the distance of the chord from the centre is $5 \ cm.$

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Question 64 Marks

In the given figure, $O$ is the center of the circle and the length of arc $AB$ is twice the length of arc $BC$. If $\angle AOB = 100^\circ ,$find:$ (i) \angle BOC (ii) \angle OAC$

Answer

We know that when two arcs are in ratio $2: 1$ then the subtended by them is also in ratio $2: 1$
As given arc $AB$ is twice the length of arc $BC.$
Therefore, arc $AB:$ arc $BC = 2: 1$
Hence, $\angle AOB: \angle BOC = 2: 1$
Now given that $\angle AOB = 100^\circ .$
So, $\angle BOC = \frac{1}{2} \angle AOB =\frac{1}{2} \times 100^{\circ}=50^{\circ}$
Now, $\angle AOC = \angle AOB + \angle BOC = 100^\circ + 50^\circ = 150^\circ .$
The triangle thus formed, $\angle AOC$ is an isosceles triangle with $OA = OC$ as they are radii of the same circle.
Thus,
$\angle OAC = \angle OCA$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ .$
So, $\angle COA + \angle OAC + \angle OCA = 180^\circ $
$2\angle OAC + 150^\circ = 180^\circ $
as,$ \angle OAC = \angle OCA$
$2\angle OAC = 180^\circ - 150^\circ $
$2\angle OAC = 30^\circ $
$\angle OAC = 15^\circ $
as $\angle OCA = \angle OAC$
So,
$\angle OCA = \angle OAC = 15^\circ .$

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Question 74 Marks

In the given figure, $AB$ is a side of a regular hexagon and $AC$ is a side of a regular eight$-$sided polygon.
Find$:(i) \angle AOB,(ii) \angle AOC,(iii) \angle BOC,(iv) \angle OBC$

Answer

As $AB$ is the side of a hexagon so the
$\angle AOB = \frac{360^{\circ}}{6} = 60^\circ $
$AC$ is the side of an eight$-$sided polygon so,
$\angle AOC =\frac{360^{\circ}}{8} = 45^\circ $
From the given figure we can see that:
$\angle BOC = \angle AOB + \angle AOC$
$\Rightarrow 60^\circ + 45^\circ = 105^\circ $
Again, from the figure, we can see that $\angle BOC$ is an isosceles triangle with sides $BO = OC$ as they are the radii of the same circle.
Angles $\angle OBC = \angle OCB$ as they are opposite angles to the equal sides of an isosceles triangle.
Sum of all the angles of a triangle is $180^\circ$
$\angle OBC + \angle OCB + \angle BOC = 180^\circ $
$2\angle OBC + 105^\circ = 180^\circ as, \angle OBC = \angle BOC$
$2\angle OBC = 180^\circ - 105^\circ $
$2\angle OBC = 75^\circ $
$\angle OBC = 37.5^\circ = 37^\circ 30'$
As,$ \angle OBC = \angle BOC$
$\angle OBC = \angle BOC = 37.5^\circ = 37^\circ 30'.$

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Question 84 Marks

In the following figure;$ P$ and $Q$ are the points of intersection of two circles with centers $O$ and $O'$. If straight lines $\text{APB}$ and $\text{CQD}$ are parallel to $OO';$prove that: $(i) OO' = \frac{1}{2} AB ; (ii) AB = CD$

Answer

Drop $OM$ and $O'N$ perpendicular on $AB$ and $OM'$ and $O'N'$ perpendicular on $CD.$

$\therefore OM, O'N, OM'$ and $O'N'$ bisect $AP, PB, CQ$ and $QD$ respectively.
$($ Perpendicular is drawn from the center of a circle to a chord bisects it.$ )$
$\therefore MP =\frac{1}{2} AP , PN =\frac{1}{2} BP , M ^{\prime} Q =\frac{1}{2} CQ , QN \prime=\frac{1}{2} QD$
Now ${ }^{\prime} OO ^{\prime}= MN = MP + PN =\frac{1}{2}( AP + BP )=\frac{1}{2} AB \ldots (i)$
and $OO ^{\prime}= M ^{\prime} N ^{\prime}= M ^{\prime} Q + QN ^{\prime}=\frac{1}{2}( CQ + QD )=\frac{1}{2} CD ...(ii)$
By $(i)$ and $(ii),$
$AB = CD.$

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Question 94 Marks

$AB$ and $CD$ are two equal chords of a circle with center $O$ which intersect each other at a right angle at point $P.$
If $OM \perp AB$ and $ON \perp CD;$
show that $\text{OMPN}$ is a square.

Answer

Clearly , all the angles of $\text{OMPN}$ are $90^\circ .$
$OM \perp AB$ and $ON \perp CD$
$\therefore BM =\frac{1}{2} AB =\frac{1}{2}CD = CN ....(i) ...[$ perpendicular drawn from the center of a circle to a chord bisects it $]$
As the two equal chords, $AB$ and $CD$ intersect at point $P$ inside the circle,
$\therefore AP = DP$ and $CP = BP .....(ii)$
Now, $CN - CP = BM - BP ...[$ by $(i)$ and $(ii) ]$
$\Rightarrow PN = MP$
$\therefore $ Quadrilateral $\text{OMPN}$ is $A$ square.

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Question 104 Marks

The radius of a circle is $17.0 \ cm$ and the length of the perpendicular drawn from its center to a chord is $8.0 \ cm.$
Calculate the length of the chord.

Answer

Let $AB$ be the chord and $O$ be the center of the circle.
Let $OC$ be the perpendicular drawn from $O$ to $AB.$

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.
$\therefore AC = CB$
In $\triangle OCA,$
$OA^2 = OC^2 + AC^2 \dots...($ By Pythagoras theorem $)$
$\Rightarrow AC^2 = (17)^2 - (8)^2 = 225$
$\Rightarrow Ac = 15 \ cm$
$\therefore AB = 2 AC = 2 \times 15 = 30 \ cm.$

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Question 114 Marks

A chord of length $8 \ cm$ is drawn at a distance of $3 \ cm$ from the center of the circle.Calculate the radius of the circle.

Answer

Let $AB$ be the chord and $O$ be the center of the circle.
Let $OC$ be the perpendicular drawn from $O$ to $AB.$

We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.
$\therefore AB = 8 \ cm$
$\Rightarrow AC = CB = \frac{ AB }{2}$
$\Rightarrow AC = CB = \frac{ 8 }{2}$
$\Rightarrow AC = CB = 4 \ cm$
In $\text{OCA},$
$OA^2 = OC^2 + AC^2 ...($ By Pythagoras theorem $)$
$\Rightarrow OA^2 = ( 4 )^2 + ( 3 )^2 = 25$
$\Rightarrow OA = 5 \ cm$
Hence, radius oof the circle is $5 \ cm.$

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Question 124 Marks

In the given figure, $AB$ and $CD$ are two equal chords of a circle, with centre $O$. If $P$ is the mid$-$point of chord $AB, Q$ is the mid$-$point of chord $CD$ and $\angle POQ = 150^\circ ,$ find $\angle APQ.$

Answer

It is given in the question that point.
$P$ is the mid$-$point of the chord $AB$ and Point $Q$ is the mid$-$point of the $CD$.
$\Rightarrow \angle APO = 90^\circ \dots...($ as the straight line drawn from the center of a circle to bisect a chord, which is not a diameter, is at the right angle to the chord. $)$
As chords, $AB$ and $CD$ are equal therefore they are equidistant from the center i.e; $PO = OQ\dots ...( \because $ Equal chords of a circle are equidistant from the center$)$
Now, the $\triangle POQ$ is an isosceles triangle with $OP = OQ$ as its two equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ .$
$\Rightarrow \angle POQ + \angle OPQ + \angle PQO = 180^\circ $
$\Rightarrow \angle OPQ + \angle POQ + 150^\circ = 180^\circ \dots...($ Given: $\angle POQ = 150^\circ )$
$\Rightarrow 2\angle OPQ = 180^\circ - 150^\circ \dots ...$( As, $\angle OPQ = \angle PQO )$
$\Rightarrow 2\angle OPQ = 30^\circ $
$\Rightarrow \angle OPQ = 15^\circ As \angle APO = 90^\circ $
$\Rightarrow \angle APQ + \angle OPQ = 90^\circ $
$\Rightarrow \angle APQ = 90^\circ - 15^\circ \dots....($ As, $\angle OPQ = 15^\circ )$
$\Rightarrow \angle APQ = 75^\circ .$

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Question 134 Marks

Given two equal chords $AB$ and $CD$ of a circle with center $O$, intersecting each other at point $P.$
Prove that:
$(i) AP = CP;(ii) BP = DP$

Answer

In$ \triangle OMP$ and $\triangle ONP,$
$OP = OP \dots...($ common sides $)$
$\angle OMP = \angle ONP\dots ...($ both are right angles $)$
$OM = OM\dots ...($ side both the chords are equal, so the distance of the chords from the centre are also equal $)$
$\triangle OMP \cong \triangle ONP\dots ...(\text{RHS}$ congruence criterion $)$
$\Rightarrow MP = PN \dots...(\text{c.p.c.t} ) \dots....( a )$
$(i)$ Since $AB = CD\dots ...($ given $)$
$\Rightarrow AM = CN\dots ...($ drawn from the centre to the chord bisects the chord$ )$
$\Rightarrow AM + MP = CN + NP \dots.....($ from $a )$
$\Rightarrow AP = CP \dots....( b ) (ii)$ Since $AB = CD$
$\Rightarrow AP + BP = CP + DP$
$\Rightarrow BP = DP\dots ....($ from $b )$
Hence proved.

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Question 144 Marks

The circumference of a circle, with center $O,$ is divided into three arcs $\text{APB, BQC,}$ and $\text{CRA}$ such that:$\frac{\operatorname{arc} APB }{2}=\frac{\operatorname{arc} BQC }{3}=\frac{\operatorname{arc} CRA }{4}$Find $\angle BOC.$

Answer

From the given conditions given in the question
We can draw the circle with $arc\ APB, arc\ BQC$, and $arc\ CRA$

The given equation is
$\frac{\operatorname{arc} APB }{2}=\frac{\operatorname{arc} BQC }{3}=\frac{\operatorname{arc} CRA }{4}$
Let
$\frac{\operatorname{arc} APB }{2}=\frac{\operatorname{arc} BQC }{3}=\frac{\operatorname{arc} CRA }{4} = k ($ say $)$
then
$Arc\ APB = 2k, Arc\ BQC = 3k, Arc\ CRA = 4k$
or
$Arc\ APB : Arc\ BQC : Arc\ CRA = 2 : 3 : 4$
$\Rightarrow \angle AOB : \angle BOC : \angle AOC = 2 : 3 : 4$
and therefore,
and $\angle AOB = 2k^\circ , \angle BOC = 3k^\circ ,$ and $\angle AOC = 4k^\circ$
Now,
Angle in a circle is $360^\circ$
$So, 2k + 3k + 4k = 360^\circ$
$\Rightarrow 9k = 360^\circ$
$\Rightarrow k = 40^\circ$
Hence,
$\angle BOC = 3 \times 40^\circ = 120^\circ .$

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Question 154 Marks

In the given figure, $AB = BC = DC$ and $\angle AOB = 50^\circ .(i) \angle AOC;(ii) \angle AOD;(iii) \angle BOD;(iv) \angle OAC;(v) \angle ODA$

Answer

Since $arc\ AB$ and $BC$ are equal.
So, $\angle AOB = \angle BOC = 50^\circ $
Now,
$\angle AOC = \angle AOB + \angle BOC = 50^\circ + 50^\circ = 100^\circ $
As $arc\ AB, arc\ BC$ and $arc\ CD$ so,
$\angle AOB = \angle BOC = \angle COD = 50^\circ $
$\angle AOD = \angle AOB + \angle BOC + \angle COD = 50^\circ + 50^\circ + 50^\circ = 150^\circ $
Now, $\angle BOD = \angle BOC + \angle COD$
$\angle BOD = 50^\circ + 50^\circ $
$\angle BOD = 100^\circ $
The triangle thus formed, $\triangle AOC$ is an isosceles triangle with $OA = OC$ as they are radii of the same circle.
Thus $\angle OAC = \angle OCA$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ$
So, $\angle AOC + \angle OAC + \angle OCA = 180^\circ $
$2\angle OAC + 100^\circ = 180^\circ $ as,
$\angle OAC = \angle OCA$
$2\angle OAC = 180^\circ - 100^\circ $
$2\angle OAC = 80^\circ $
$\angle OAC = 40^\circ $
as $\angle OCA = \angle OAC$
So,
$\angle OCA = \angle OAC = 40^\circ $
The triangle thus formed, $\triangle AOD$ is an isosceles triangle with $OA = OD$ as they are radii of the same circle.
Thus, $\angle OAD = \angle ODA$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ$
So, $\angle AOD + \angle OAD + \angle ODA = 180^\circ $
$2\angle OAD + 150^\circ = 180^\circ $ as,
$ \angle OAD = \angle ODA$
$2\angle OAD = 180^\circ - 150^\circ $
$\angle OAD = 30^\circ $
as $\angle OAD = \angle ODA$ So,
$\angle OAD = \angle ODA = 15^\circ .$

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Question 164 Marks

In the given figure, the lengths of arcs $AB$ and $BC$ are in the ratio $3:2.$ If $\angle AOB = 96^\circ ,$find: $(i) \angle BOC (ii) \angle ABC$

Answer

We know that for two arcs are in ratio $3: 2$ then
$\angle AOB: \angle BOC = 3: 2$
As give $\angle AOC = 96^\circ $
So, $3x = 96$
$x = 32$
There $\angle BOC = 2 \times 32 = 64^\circ $
The triangle thus formed,$ \triangle AOB$ is an isosceles triangle with $OA = OB$ as they are radii of the same circle.
Thus $\angle OBA = \angle BAO$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ $
So, $\angle AOB + \angle OBA + \angle BAO = 180^\circ $
$2\angle OBA + 96^\circ = 180^\circ as, \angle OBA = \angle BAO$
$2\angle OBA = 180^\circ - 96^\circ $
$2\angle OBA = 84^\circ $
$\angle OBA = 42^\circ $
as, $\angle OBA = \angle BAO$ So,
$\angle OBA = \angle BAO = 42^\circ $
The triangle thus formed, $\triangle BOC$ is an isosceles triangle with $OB = OC$ as they are radii of the same circle.
Thus $\angle OBC = \angle OCB$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ $
So, $\angle BOC + \angle OBC + \angle OCB = 180^\circ $
$2\angle OBC + 64^\circ = 180^\circ $ as,
$\angle OBC = \angle OCB$
$2\angle OBC = 180^\circ - 64^\circ $
$2\angle OBC = 116^\circ $
$\angle OBC = 58^\circ $
As $\angle OBC = \angle OCB$ So,
$\angle OBC = \angle OCB = 58^\circ $
$\angle ABC = \angle BOA + \angle OBC = 42^\circ + 58^\circ = 100^\circ $

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Question 174 Marks

In the given figure, $arc\ AB$ and $arc\ BC$ are equal in length. If $\angle AOB = 48^\circ ,$ find:$(i) \angle BOC;(ii) \angle OBC;(iii) \angle AOC;(iv) \angle OAC$

Answer

We know that the arc of equal lengths subtends equal angles at the center.

hence $\angle AOB = \angle BOC = 48^\circ $
Then $\angle AOC = \angle AOB + \angle BOC = 48^\circ + 48^\circ = 96^\circ $
The triangle thus formed, $\triangle BOC$ is an isosceles triangle with $OB = OC $as they are radii of the same circle.
Thus $\angle OBC = \angle OCB$as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ .$
So, $\angle BOC + \angle OBC + \angle OCB = 180^\circ $
$2\angle OBC + 48^\circ = 180^\circ $ as$ \angle OBC = \angle OCB$
$2\angle OBC = 180^\circ - 48^\circ $
$2\angle OBC = 132^\circ $
$\angle OBC = 66^\circ $
as $\angle OBC = \angle OCB$
So,$ \angle OBC = \angle OCB = 66^\circ $
The triangle thus formed, $\triangle AOC$ is an isosceles triangle with $OA = OC$ as they are radii of the same circle.
Thus $\angle OAC = \angle OCA$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ .$
So, $\angle COA + \angle OAC + \angle OCA = 180^\circ $
$2\angle OAC + 96^\circ = 180^\circ $ as, $\angle OAC = \angle OCA$
$2\angle OAC = 180^\circ - 96^\circ $
$2\angle OAC = 84^\circ $
$\angle OAC = 42^\circ $
as $ \angle OCA = \angle OAC$
So, $\angle OCA = \angle OAC = 42^\circ .$

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Question 184 Marks

In the given figure, $AB$ is a side of regular pentagon and $BC$ is a side of regular hexagon.$(i) \angle AOB;(ii) \angle BOC;(iii) \angle AOC;(iv) \angle OBA;(v) \angle OBC;(vi) \angle ABC$

Answer

As given that $AB$ is the side of a pentagon the angle subtended by each arm of the pentagon at the center of the circle is$ = \frac{360^{\circ}}{5} = 72^\circ $
Thus angle $\angle AOB = 72^\circ $
Similarly, as $BC$ is the side of a hexagon hence the angle subtended by $BC$ at the center is $= \frac{360^{\circ}}{6}$
i.e. $60^\circ $
$\angle BOC = 60^\circ $
Now $\angle AOC = \angle AOB + \angle BOC =72^\circ + 60^\circ = 132^\circ $
The triangle thus formed, $\triangle AOB$ is an isosceles triangle with $OA = OB$ as they are radii of the same circle.
Thus $\angle OBA = \angle BAO$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ $
so, $\angle AOB + \angle OBA + \angle BAO = 180^\circ $
$\Rightarrow 2\angle OBA + 72^\circ = 180^\circ $ as,
$\angle OBA = \angle BAO$
$\Rightarrow 2\angle OBA = 180^\circ - 72^\circ $
$\Rightarrow 2\angle OBA = 180^\circ $
$\Rightarrow 2\angle OBA =54^\circ $
as, $\angle OBA = \angle BAO$ So,
$\angle OBA = \angle BAO = 54^\circ $
The triangle thus formed, $\triangle BOC$ is an isosceles triangle with $OB = OC$ as they are radii of the same are.
Thus $\angle OBC = \angle OCB$ as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is $180^\circ $
so, $\angle BOC + \angle OBC + \angle OCB = 180^\circ $
$2\angle OBC + 60^\circ = 180^\circ $ as ,
$\angle OBC = \angle OCB$
$2\angle OBC = 180^\circ - 60^\circ $
$2\angle OBC = 120^\circ $
$\angle OBC = 60^\circ $
as $\angle OBC = \angle OCB$
So, $\angle OBC = \angle OCB = 60^\circ $
$\angle ABC = \angle OBA + \angle OBC = 54^\circ + 60^\circ = 114^\circ $

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Question 194 Marks

In the given figure, a square is inscribed in a circle with center $O.$ Find:

$\angle BOC$
$\angle OCB$
$\angle COD$
$\angle BOD$; Is $BD$ a diameter of the circle?

Answer

In the given figure we can extend the straight line $OB$ to $BD$ and $CO$ to $CA$ Then we get the diagonals of the square which intersect each other at $90$ by the property of Square.

From the above statement, we can see that
$\angle COD = 90^\circ $
The sum of the angle $\angle BOC$ and $\angle COD$ is $180^\circ $ as $BD$ is a straight line.
Hence$ \angle BOC + \angle OCD = \angle BOD = 180^\circ $
$\angle BOC + 90^\circ = 180^\circ $
$\angle BOC + 180^\circ - 90^\circ $
$\angle BOC = 90^\circ $
We can see that the $\text{OCB}$ is an isosceles triangle with sides $OB$ and $OC$ of EQual length as they are the radii of the same are.
In $\triangle OCB,$
$\angle OBC = \angle OCB$ as they are opposite angles to the two equal sides of an isosceles triangle.
Sum of all the angles of a triangle is $180^\circ $
so, $\angle OBC + \angle OCB + \angle BOC =180^\circ $
$\angle OBC + \angle OBC + 90^\circ = 180^\circ $ as,
$\angle OBC = \angle OCB$
$2\angle OBC = 180^\circ - 90^\circ $
$2\angle OBC = 90^\circ $
$2\angle OBC = 45^\circ $
as $\angle OBC = \angle OCB$ So,
$\angle OBC = \text{OCB} = 45^\circ $
Yes $BD$ is the diameter of the order.

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Question 204 Marks

In the given figure, an equilateral $\triangle ABC$ is inscribed in a circle with center $O.$Find:$ (i) \angle BOC;(ii) \angle OBC$

Answer

In the given figure, $\text{ABC}$ is an equilateral triangle.
Hence all the three angles of the triangle will be equal to $60^\circ $
i.e. $\angle A = \angle B = \angle C = 60^\circ $
As the triangle is an equilateral triangle, $BO$ and $CO$ will be the angle bisectors of $B$ and $C$ respectively.
Hence $\angle OBC = \frac{\angle ABC }{2}$
$= 30^\circ $
and as given in the figure we can see that $OB$ and $OC$ are the radii of the given circle.
Hence they are of equal length.
The $\triangle OBC$ is an isosceles triangle with $OB = OC$
In $\triangle OBC,$
$\angle OBC = \angle OCB$ as they are angles opposite to the two equal sides of an isosceles triangle.
Hence, $\angle OBC = 30^\circ $ and $\angle OCB = 30^\circ $
Since the sum of all angles of a triangle is $180^\circ $
Hence in triangle $OBC, \angle OCB + \angle OBC + \angle BOC + BOC = 180^\circ $
$30^\circ + 30^\circ + \angle BOC= 180^\circ $
$60^\circ + BOC = 180^\circ $
$\angle BOC = 180^\circ - 60^\circ $
$\angle BOC = 120^\circ $
Hence $\angle BOC =120^\circ $ and $\angle OBC =30^\circ $

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Question 214 Marks

The line joining the midpoints of two chords of a circle passes through its center.Prove that the chords are parallel.

Answer

Given: $AB$ and $CD$ are the two chords of a circle with center $O.$
L and M are the mid$-$points of $AB$ and $CD$ and $O$ lies in the line joining $ML$.
To prove:$ AB \| CD.$
Proof:
$AB$ and $CD$ are two chords of a circle with center $O.$
Line $LOM$ bisects them at $L$ and $M.$
Then,$OL \perp AB$
and, $OM \perp CD$
$\therefore \angle ALM = \angle LMD = 90^\circ $
But they are alternate angles
$\therefore AB \| CD.$

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Question 224 Marks

The length of the common chord of two intersecting circles is $30 \ cm.$ If the diameters of these two circles are $50 \ cm$ and $34 \ cm,$ calculate the distance between their centers.

Answer

$OA = 25 \ cm$ and $AB = 30 \ cm$
$\therefore AD =\frac{1}{2} \times AB =\left(\frac{1}{2} \times 30\right)cm = 15 \ cm$
Now in right angled $\text{ADO}$
$OA^2 + AD^2 + OD^2$
$\Rightarrow OD^2 = OA^2 - OD^2 = 25^2 - 15^2$
$= 625 - 225 = 400$
$\therefore OD =\sqrt{400} = 20 \ cm$
Again, we have $O\ 'A = 17 \ cm.$
In right$-$angle $\text{ADO}\ '$
$O\ 'A^2 = A\ 'D^2 + O\ 'D^2$
$\Rightarrow O\ 'D^2 = O\ 'A^2 - AD^2$
$= 17^2 - 15^2$
$= 289 - 225 = 64$
$\therefore O\ 'D = 8 \ cm$
$\therefore OO\ ' = ( OD + O\ 'D )$
$= ( 20 + 8 ) = 28 \ cm$
$\therefore $ the distance between their centres is $28 \ cm.$

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Question 234 Marks

In the following figure, $\text{OABC}$ is a square. A circle is drawn with $O$ as centre which meets $OC$ at $P$ and $OA$ at $Q$.Prove that$:( i ) \triangle OPA \cong \triangle OQC;( ii ) \triangle BPC \cong \triangle BQA$

Answer

$(i)$ In $\triangle OPA$ and $\triangle OQC$,
$OP = OQ \dots....($ radii of same circle $)$
$\angle AOP = \angle COQ \dots... ($ both $90^\circ )$
$OA = OC \dots... ($ sides of the square $)$
By Side$- $Angle $-$ Side criterion of congruence.
$\therefore \triangle OPA \cong \triangle OQC ...($ by $\text{SAS})$
$(ii)$ Now, $OP = OQ\dots ...($ radii $)$
and $OC = OA\dots ...( $sides of the square$ )$
$\therefore OC - OP = OA - OQ$
$\Rightarrow CP = AQ\dots ....(i)$
In $\triangle BPC$ and $\triangle BQA,$
$BC = BA\dots ...($ sides of the square $)$
$\angle PCB = \angle QAB\dots ...($ both $90^\circ)$
$PC = QA ...($ by $( i ) )$
By Side$-$ Angle$-$Side criterion of congruence,
$\therefore \triangle BPC \cong \triangle BQA ...($ by $\text{SAS})$

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Question 244 Marks

Two equal chords $AB$ and $CD$ of a circle with center $O,$ intersect each other at point $P$ inside the circle.Prove that $: (i) \ AP = CP ; (ii)\ BP = DP$

Answer

Drop $OM$ and $ON$ perpendicular on $AB$ and $CD.$
Join $OP, OB,$ and $OD.$

$\therefore OM$ and $ON$ bisect $AB$ and $CD$ respectively$\dots.....($ Perpendicular drawn from the centre of a circle to a chord bisects it. $)$
$\therefore MB =\frac{1}{2} AB =\frac{1}{2} CD = ND \ldots (i)$
In right $\triangle \text{OMB},$
$OM^2 = OB^2 - MB^2\dots ....(ii)$
In right $\triangle \text{OND},$
$ON^2 = OD^2 - ND^2\dots ....(iii)$
From $(i), (ii),$ and $(iii),$
$OM = ON$
In $\triangle \text{OPM}$ and $\triangle \text{OPN},$
$\angle \text{OMP} = \angle \text{ONP}\dots....($ both $90^\circ )$
$OP = OP\dots....($ common $)$
$OM = ON\dots ....($ proved above $)$
By Right Angle$-$Hypotenuse$-$Side criterion of congruence,
$\therefore \triangle \text{OPM} \cong \triangle \text{OPN} ....($ by $\text{RHS} )$
The corresponding parts of the ongruent triangles are congruent.
$\therefore PM = PN\dots ....(\text{ c.p.c.t.} )$
Adding $(i)$ to both sides$,$
$MB + PM = ND + PN$
$\Rightarrow BP = DP$
Now$, AB = CD$
$\therefore AB - BP = CD - DP\dots...( \because BP = DP )$
$\Rightarrow AP = CP.$

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Question 254 Marks

A straight line is drawn cutting two equal circles and passing through the mid$-$point $M$ of the line joining their centers $O$ and $O'.$ Prove that the chords $AB$ and $CD,$ which are intercepted by the two circles, are equal.

Answer

Given: $A$ straight line $AD$ intersects two circles of equal radii at $A, B, C$ and $D.$
The line joining the centers $OO'$ intersect $AD$ at $M$ and $M$ is the midpoint of $OO'.$
To Prove: $AB = CD.$
Construction: From $O$, draw $OP \perp AB$ and from $O',$ draw
Proof:
In $\triangle OMP$ and $\triangle O'MQ,$
$\angle OMP = \angle O'MQ\dots ...($ Vertically Opposite angles$ )$
$\angle OPM = \angle O'QM\dots ...( $each $= 90^\circ )$
$OM = O'M\dots ...($ Given$ )$
By Angle$-$Angle$-$Side criterion of congruence,
$\therefore \triangle OMP \cong \triangle O'MQ, ...($ by $\text{AAS})$
The corresponding parts of the congruent triangles are congruent.
$\therefore OP = O'Q\dots ...( \text{c.p.c.t.})$
We know that two chords of a circle or equal circles which are equidistant from the center are equal.
$\therefore AB = CD.$

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Question 264 Marks

A chord $CD$ of a circle whose center is $O$ is bisected at $P$ by a diameter $AB.$ Given $OA = OB = 15 \ cm$ and $OP = 9 \ cm.$Calculate the lengths of $: (i) \ CD ; (ii)\ AD ; (iii) \ CB.$

Answer

$(i) OP \perp CD$
$\therefore OP$ bisects $CD\dots. ....($ Perpendicular drawn from the centre of a circle to a chord bisects it. $)$
$\Rightarrow CP = \frac{C D}{2}$
In right $\triangle \text{OPC},$
$OC^2 = OP^2 + CP^2$
$\Rightarrow CP^2 = OC^2 - OP^2$
$\Rightarrow 15^2 - 9^2 = 144$
$\therefore CP = 12 \ cm$
$\therefore CD = 12 \times 2 = 24 \ cm$
$(ii)$ Join $BD,$
$\therefore BP = OB - OP = 15 - 9 = 6 \ cm.$
In right $\triangle \text{BPD},$
$BD^2 = BP^2 + PD^2$
$= 6^2 + 12^2 = 180$
In $\triangle \text{ADB},$
$\angle \text{ADB} = 90^\circ\dots ...($ Angle in a semi$-$circle is a right angle $)$
$\therefore AB^2 = AD^2 + BD^2$
$\Rightarrow AD^2 = AB^2 - BD^2$
$= 30^2 - 180 = 720$
$\therefore AD = \sqrt{720} = 26.83 \ cm$
$(iii)$ Also$, BC = BD = \sqrt{180} = 13.42 \ cm.$

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Question 274 Marks

In a circle of radius $17 \ cm,$ two parallel chords of lengths $30 \ cm$ and $16 \ cm$ are drawn. Find the distance between the chords,if both the chords are :$(i)$ on the opposite sides of the centre;$(ii)$ on the same side of the centre.

Answer

Let $O$ be the center of the circle and $AB$ and $CD$ be the two parallel chords of length $30 \ cm$ and $16 \ cm$ respectively.
Drop $OE$ and $OF$ perpendicular on $AB$ and $CD$ from the center $O.$

$OE \perp AB$ and $OF \perp CD.$
$\therefore OE$ bisects $AB$ and $OF$ bisects $CD\dots. ...($ Perpendicular is drawn from the centre of a circle to a chord bisects it. $)$
$\Rightarrow AE =\frac{30}{2} = 15 \ cm;$
$CF = \frac{16}{2}= 8 \ cm$
In right $\triangle \text{OAE},$
$OA^2 = OE^2 + AE^2$
$\Rightarrow OE^2 = OA^2 - AE^2 = 17^2 - 15^2 = 64$
$\therefore OE = 8 \ cm$
In right $\triangle \text{OCF},$
$OC^2 = OF^2 + CF^2$
$\Rightarrow OF^2 = OC^2 - CF^2 = 17^2 - 8^2 = 225$
$\therefore OF = 15 \ cm$
$(i)$ The chord are on the opposite sides of the centre :
$\therefore EF = EO + OF = 8 + 15 = 23\ cm$
$(ii)$ The chord are on the same side of the centre :
$\therefore EF = OF - OE = 15 - 8 = 7 \ cm.$

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Question 284 Marks

In the following figure$, AD$ is a straight line$, OP \perp AD$ and $O$ is the centre of both circles. If $OA = 34\ cm, OB = 20 \ cm$ and $OP = 16 \ cm;$find the length of $AB.$

Answer

For the inner circle$, BC$ is a chord and $OP \perp BC.$
We know that the perpendicular to a chord, from the center of a circle, bisects the chord.
$\therefore BP = PC$
By Pythagoras theorem,
$OB^2 = OP^2 + BP^2$
$\Rightarrow BP^2 = 20^2 - 16^2 = 144$
$\therefore BP = 12 \ cm$
For the outer circle$, AD$ is the chord and $OP \perp AD.$
We know that the perpendicular to a chord, from the center of a circle, bisects the chord.
$\therefore AP = PD$
By Pythagoras Theorem,
$OA^2 = OP^2 + AP^2$
$\Rightarrow AP^2 = (34)^2 - (16)^2= 900$
$\Rightarrow AP = 30 \ cm$
$AB = AP - BP = 30 - 12 = 18 \ cm$

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Question 294 Marks

A chord of length $24 \ cm$ is at a distance of $5 \ cm$ from the center of the circle. Find the length of the chord of the same circle which is at a distance of $12 \ cm$ from the center.

Answer

Let $AB$ be the chord of length $24 \ cm$ and $O$ be the center of the circle.
Let $OC$ be the perpendicular drawn from $O$ to $AB.$
We know, that the perpendicular to a chord, from the center of a circle, bisects the chord.
$\therefore AC = CB = 12 \ cm$
In $\text{OCA},$
$OA^2 = OC^2 + AC^2\dots ....($ By Pythagoras theorem $)$
$=(5)^2 + ( 12 )^2 = 169$
$\Rightarrow OA = 13 \ cm$
$\therefore $ radius of the circle $= 13 \ cm.$
Let $A\ 'B\ '$ be the new chord at a distance of $12 \ cm$ from the center.
$\therefore ( OA\ ' )^2 = ( OC\ ' )^2 + ( A\ 'C\ ' )^2$
$\Rightarrow ( A\ 'C\ ' )^2 = ( 13 )^2 - ( 12 )^2 = 25$
$\therefore A\ 'C\ ' = 5 \ cm$
Hence$,$ length of the new chord $= 2 \times 5 = 10 \ cm.$

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[4 marks sum] - MATHEMATICS STD 9 Questions - Vidyadip