4 Marks Questions

Question 14 Marks

If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.

Answer

Given: $AB$ and $CD$ are two chords of a circle with centre $O$.
Diameter $POQ$ bisects them at points $L$ and $M$.
To prove: $AB || CD$ Proof: $AB$ and $CD$ are two chords of a circle with centre $O$.
Diameter $POQ$ bisects them at $L$ and $M$.

Then $\text{OL}\perp\text{AB}$ Also, $\text{OM}\perp\text{CD}$
$\therefore\ \angle\text{ALM}=\angle\text{LMD}=90^\circ$
Since alternate angles are equal, we have: $AB || CD$

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Question 24 Marks

In the given figure, $O$ is the centre of the circle. if $\angle\text{PBC}=25^\circ$ and $\angle\text{APB}=110^\circ,$find the value of $\angle\text{ADB}.$

Answer

From the given diagram, we have:

$\angle\text{ACB}=\angle\text{PCB}$
$\angle\text{BPC}=(180^\circ-110^\circ)=70^\circ$ [Linear pair]
Considering $\triangle\text{PCB},$ we have:
$\angle\text{PCB}+\angle\text{BPC}+\angle\text{PBC}=180^\circ$ [Angle sum property]
$\Rightarrow\ \angle\text{PCB}+70^\circ+25^\circ=180^\circ$
$\Rightarrow\ \angle\text{PCB}=(180^\circ-95^\circ)=85^\circ$
$\Rightarrow\ \angle\text{ACB}=\angle\text{PCB}=85^\circ$
We know that the angles in the same segment of a circle are equal.
$\therefore\ \angle\text{ADB}=\angle\text{ACB}=85^\circ$

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Question 34 Marks

Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Answer

Let $ABCD$ be a cyclic quadrilateral and let $O$ be the centre of the circle passing through $A, B, C, D$.
Then each of $AB, BC, CD,$ and $DA$ being a chord of the circle, its right bisector must pass through $O$.
$\therefore$ the right bisectors of $AB, BC, CD$ and $DA$ pass through and are concurrent.

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Question 44 Marks

Prove that the circles described with the four sides of a rhombus, as diameters, pass through the point of intersection of its diagonals.

Answer

$ABCD$ is a rhombus. Let the diagonal $AC$ and $BD$ of the rhombus $ABCD$ intersect at $O$.
But, we know that the diagonals of a rhombus bisect each other at right angles.
So, $\angle\text{BOC}=90^\circ$
$\therefore\ \angle\text{BOC}$ lies in a circle.

Thus the circle drawn with $BC$ as diameter will pass through $O$.
Similarly, all the circles described with $AB, AD,$ and $CD$ as diameter will pass through $O$.

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Question 54 Marks

In a cyclic quadrilateral $ABCD$, if $(\angle\text{B}-\angle\text{D})=60^\circ,$ show that the smaller of the two is $60^\circ$

Answer

$ABCD$ is a cyclic quadrilateral $\angle\text{B}-\angle\text{D}=60^\circ\dots(\text{i})$ And $\angle\text{B}+\angle\text{D}=180^\circ\dots(\text{ii})$
Adding $(i)$ and $(ii)$ we get, $2\angle\text{B}=240^\circ$
$\therefore\ \angle\text{B}=\frac{240}{2}=120^\circ$
Substituting the value of $\angle\text{B}=120^\circ$ in $(i)$ we get $120^\circ-\angle\text{D}=60^\circ$
$\Rightarrow\ \angle\text{D}=120^\circ-60^\circ=60^\circ$
The smaller of the two angles i.e. $\angle\text{D}=60^\circ.$

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Question 64 Marks

In the given figure, $\angle\text{BAD}=75^\circ,\angle\text{DCF}=\text{x}^\circ$ and $\angle\text{DEF}=\text{y}^\circ.$ Find the values of $x$ and $y$.

Answer

We know that if one side of a cyclic quadrilateral is produced, then the exterior angle is equal to the interior opposite angle $\text{i.e.},\angle\text{BAD}=\angle\text{DCF}=75^\circ$
$\Rightarrow\ \angle\text{DCF}=\text{x}=75^\circ$ Again, the sum of opposite angles in a cyclic quadrilateral is $180^\circ$.
Thus, $\angle\text{DCF}=\angle\text{DEF}=180^\circ$
$\Rightarrow\ 75^\circ+\text{y}=180^\circ$
$\Rightarrow\ \text{y}=(180^\circ-75^\circ)=105^\circ$
Hence, $\text{x}=75^\circ$ and $\text{y}=105^\circ.$

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Question 74 Marks

In the given figure, $BD = DC$ and $\angle\text{CBD}=30^\circ,$ find $\angle\text{BAC}.$

Answer

$BD = DC$
$\Rightarrow\ \angle\text{BCD}=\angle\text{CBD}=30^\circ$ In $\triangle\text{BCD},$
we have: $\angle\text{BCD}+\angle\text{CBD}+\angle\text{CDB}=180^\circ$ [Angle sum property of a triangle]
$\Rightarrow\ 30^\circ+30^\circ+\angle\text{CDB}=180^\circ$
$\Rightarrow\ \angle\text{CDB}=(180^\circ-60^\circ)=120^\circ$
The opposite angles of a cyclic quadrilateral are supplementary.
Thus, $\angle\text{CDB}+\angle\text{BAC}=180^\circ$
$\Rightarrow\ 120^\circ+\angle\text{BAC}=180^\circ$
$\Rightarrow\ \angle\text{BAC}=(180^\circ-120^\circ)=60^\circ$
$\therefore\ \angle\text{BAC}=60^\circ$

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Question 84 Marks

Two chords $AB$ and $CD$ of a circle intersect each other at $P$ outside the circle. If $AB = 6\ cm, BP = 2\ cm$ and $PD = 25\ cm$, find $CD$.

Answer

$AB$ and $CD$ are two chords of a circle which intersect each other at $P$ outside the circle.
$AB = 6\ cm, BP = 2\ cm$ and $PD = 2.5\ cm$
$\therefore$ $AP \times BP = CP \times DP$
$ \Rightarrow 8 \times 2 = (CD + 2.5) \times 2.5$
$[\because$ $CP = CD + DP$$]$
Let $CD = x\ cm$ Thus, $8 \times 2 = (CD + 2.5) \times 2.5 $
$\Rightarrow 16 = 2.5x + 6.25 $
$\Rightarrow 2.5x = (16 - 6.25) = 9.75$
$\Rightarrow\ \text{x}=\frac{9.75}{2.5}=3.9$
Hence, $CD = 3.9\ cm$

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Question 94 Marks

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

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. $\angle\text{AOB}=2\angle\text{ACB}$
$=2\times50^\circ$ [Given] $\angle\text{AOB}=100^\circ\dots(\text{i})$
Let us consider the triangle $\triangle\text{OAB}.$
$\text{OA}=\text{OB}$ (Radii of a circle)
Thus, $\angle\text{OAB}=\angle\text{OBA}$ In $\triangle\text{OAB},$
we have: $\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$
$\Rightarrow\ 100^\circ+\angle\text{OAB}+\angle\text{OBA}=180^\circ$
$\Rightarrow\ 100^\circ+2\angle\text{OAB}=180^\circ$
$\Rightarrow\ 2\angle\text{OAB}=180^\circ-100^\circ=80^\circ$
$\Rightarrow\ \angle\text{OAB}=40^\circ$
Hence, $\angle\text{OAB}=40^\circ$

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Question 104 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 114 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 124 Marks

Prove that two different circles cannot intersect each other at more than two points.

Answer

Given: Two distinct circles To prove: Two distinct circles cannot intersect each other in more than two points.
Proof: Suppose that two distinct circles intersect each other in more than two points.
$\therefore$ These points are non-collinear points. Three non-collinear points determine one and only one circle.
$\therefore$ There should be only one circle. This contradicts the given, which shows that our assumption is wrong.
Hence, two distinct circles cannot intersect each other in more than two points.

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Question 134 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 144 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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