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 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°.
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 64 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 74 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 84 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 94 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 104 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 114 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 124 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 134 Marks

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

Answer

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

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Question 144 Marks

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

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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