2 Marks Questions

Question 12 Marks

Two tangents $BC$ and $BD$ are drawn to a circle with centre $O$, such that $\angle\text{CBD} = 120^\circ.$ Prove that OB= 2BC.

Answer

In $\triangle\text{BCO}$ and $\triangle\text{BDO},$
$\angle\text{BCO}=\angle\text{BDO}=90^\circ$ $....($Since $BC$ and $BD$ are tangent to the circle$)$
$\text{OB}=\text{OB}$ ....(Common side)
$\text{OC}=\text{OD}$ ....(radii of the same circle)
$\Rightarrow\triangle\text{BCO}\cong\triangle\text{BDO}$$ ....($RHS congruence criterion$)$
$\angle\text{OBC}=\angle\text{OBD}=60^\circ$ ....(cpct)
So, $\angle\text{COB}=30^\circ$
So, $\triangle\text{BCO}$ is a $30-60-90$ triangle.
Side opposite $30^\circ=\frac{1}{2}\text{ hypotenus}$
$\Rightarrow\text{BC}=\frac{1}{2}\text{OB}$
$\Rightarrow\text{OB}=2\text{BC}$
Hence proved.

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Question 22 Marks

In the given figure, a circle inscribed in a triangle $ABC$, touches the sides $AB, BC$ and $AC$ at points $D, E$ and $F$ respectively. If $AB = 12cm, BC = 8\ cm$ and $AC = 10\ cm$, find the lengths of $AD, BE$ and $CF$.

Answer

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
$AD = AF, BD = BE$ and $CE = CF$
Now, $AD + BD = 12cm ....(1)$
$AF + FC = 10cm$
$\Rightarrow AD + FC = 10cm ....(2)$
$BE + EC = 8cm$
$\Rightarrow BD + FC = 8cm ....(3)$
Adding all these we get
$AD + BD + AD + FC + BD + FC = 30$
$\Rightarrow 2(AD + BD + FC) = 30$
$\Rightarrow AD + BD + FC = 15cm ....(4)$
Solving $(1)$ and $(4)$, we get
$FC = 3cm$
Solving $(2)$ and $(4)$, we get
$BD = 5cm$
Solving $(3)$ and $(4)$, we get
and $AD = 7cm$
$\therefore$ $AD = AF = 7\ cm, BD = BE = 5\ cm$ and $CE = CF = 3\ cm$

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Question 32 Marks

In the given figure, $PA$ and $PB$ are the tangents to a circle with centre $O$. Show that the points $A, O, B, P$ are concyclic.

Answer

$OA = OB ...($radii of the same circle$)$
Since $PA$ and $PB$ are tangent to the circle,
$\angle\text{OAP}=\angle\text{OBP}=90^\circ$
Consider,
$\angle\text{OAP}+\angle\text{OBP}$
$=90^\circ+90^\circ$
$=180^\circ$
In quad. AOBP,
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ$
$\Rightarrow180^\circ+\angle\text{AOB}+\angle\text{APB}=360^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{APB}=180^\circ$
Since the sum of the opposite angles of quad. $AOBP$ are supplementary, $AOBP$ are concyclic.
That is, a circle passes through $A, O, B$ and $P$.
Hence proved.

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Question 42 Marks

In the given figure, the chord $AB$ of the larger of the two concentric circles, with centre $O$, touches the smaller circle at $C$. Prove that $AC = CB$.

Answer

Construction: Join $OA, OC$ and $OB$

We know that the radius and tangent are perpendicular at their point of contact
$\therefore\angle\text{OCA}=\angle\text{OCB}=90^\circ$
Now, In $\triangle\text{OCA}$ and $\triangle\text{OCB}$
$\angle\text{OCA}=\angle\text{OCB}=90^\circ$
$\text{OA}=\text{OB}$ (Radii of larger circle)
$\text{OC}=\text{OC}$ (Common)
By $RHS$ congruency
$\triangle\text{OCA}\cong\triangle\text{OCB}$
$\therefore \text{CA}=\text{CB}$

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Question 52 Marks

In the given figure, an isosceles triangle $ABC$, with $AB = AC$, circumscribes a circle. Prove that the point of contact $P$ bisects the base $BC$.

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Question 62 Marks

Two concentric circle are of radii $6.5cm$ and $2.5cm$. Find the length of the chord of the larger circle which touches the smaller circle.

Answer

We know that the radius and tangent are perpendicular at their point of contact
In right triangle $AOP$
$ A O^2=O P^2+P A^2 $
$ \Rightarrow(6.5)^2=(2.5)^2+P A^2$
$ \Rightarrow P A^2=36$
$\Rightarrow PA = 6\ cm$
Since, the perpendicular drawn from the centre bisect the chord.
$\therefore PA = PB = 6\ cm$
Now, $AB = AP + PB = 6 + 6 = 12\ cm$
Hence, the length of the chord of the larger circle is $12\ cm$

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Question 72 Marks

Prove that the line segment joining the points of contact of two parallel tangents of a circle, passes through its centre.

Answer

Suppose $CD$ and $AB$ are two parallel tangents of a circle with centre $O$
Construction: Draw a line parallel to $CD$ passing through $O$ i.e, $OP$
We know that the radius and tangent are perperpendular at their point of contact.
$\angle\text{OQC}=\angle\text{ORA}=90^\circ$
Now, $\angle\text{OQC}=\angle\text{POQ}=180^\circ$ (co-interior angles)
$\Rightarrow\angle\text{POQ}=180^\circ-90^\circ=90^\circ$
Similarly, Now, $\angle\text{ORA}+\angle\text{POR}=180^\circ$ (co-interior angles)
$\Rightarrow\angle\text{POR}=180^\circ-90^\circ=90^\circ$
Now, $\angle\text{POR}+\angle\text{POQ}=90^\circ+90^\circ=180^\circ$
Since, $\angle\text{POR}$ and $\angle\text{POQ}$ are linear pair angles whose sum is $180^\circ$
Hence, $QR$ is a straight line passing through centre $O$.

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Question 82 Marks

In the given figure, $PA$ and $PB$ are two tangents from an external point $P$ to a circle with centre $O$. If $\angle\text{PBA} = 65^\circ$ find $\angle\text{OAB}$ and $\angle\text{APB}.$

Answer

Since PB is a tangent to the circle, $\angle\text{OBP}=90^\circ.$
Now,
$\angle\text{OBA}=\angle\text{OBP}-\angle\text{ABP}$
$=90^\circ-65^\circ$
$=25^\circ$
Since $ \text{OB} = \text{OA},\angle\text{OAB}=\angle\text{OBA}=25^\circ.$
In $\triangle\text{AOB},$
$\angle\text{AOB}+\angle\text{ABO}+\angle\text{OAB}=180^\circ$
$\Rightarrow\angle\text{AOB}+25^\circ+25^\circ=180^\circ$
$\Rightarrow\angle\text{AOB}=130^\circ$
In quad. AOBP
$\angle\text{PAO}+\angle\text{PBO}+\angle\text{AOB}+\angle\text{APB}=360^\circ$
$\Rightarrow90^\circ+90^\circ+130^\circ+\angle\text{APB}=360^\circ$
$\Rightarrow\angle\text{APB}=50^\circ$
Thus, $\angle\text{OAB}=25^\circ$ and $\angle\text{APB}=50^\circ.$

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Question 92 Marks

In the given figure, a quadrilateral $ABCD$ is drawn to circumscribe a circle such that its side $AB, BC, CD$ and $AD$ touch the circle at $P, Q, R$ and $S$ respectively. If $AB = x\ cm, BC = 7\ cm, CR = 3\ cm$ and $AS = 5\ cm$ , find $x$.

Answer

We know that tangent is from an external point to the circle are equal.
$AP = AS = 5cm$
$CQ = CR = 3cm$
$BQ = BP$
Now,
$BC = CQ + BQ$
$BQ = BC - CQ$
$= 7 - 3$
$= 4cm$
So, $BQ = PB = 4cm$
Thus, $AB = AP + PB = 5 + 4 = 9cm$

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Question 102 Marks

In the given figure, a circle with centre $O$, is inscribed in a quadrilateral $ABCD$ such that it touches the side $BC, AB, AD$ and $CD$ at points $P, Q, R$ and $S$ respectively. If $AB = 29\ cm, AD = 23\ cm$, $\angle\text{B} = 90^\circ$and $DS = 5\ cm$ then find the radius of the circle.

Answer

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
$DS = DR, AR = AQ$
Now, $AD = 23cm$
$\Rightarrow AR + RD = 23$
$\Rightarrow AR = 23 - RD$
$\Rightarrow AR = 23 - 5$ $[\therefore$ $DS = DR = 5$$]$
$\Rightarrow AR = 18cm$
Again, $AB = 29cm$
$\Rightarrow AQ + QB = 29$
$\Rightarrow QB = 29 - AQ$
$\Rightarrow QB = 29 - 18$ $[\therefore$ $AR = AQ = 18$$]$
$\Rightarrow QB = 11cm$
Since all the angles are in a quadrilateral $BQOP$ are right angles and $OP = BQ.$
Hence, $BQOP$ is a square.
We know that all the sides of square are equal.
Therefore, $BQ = PO = 11\ cm$
Hence, the radius of the circle is $11\ cm$.

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Question 112 Marks

In the given figure, $PA$ and $PB$ are the tangent segments to a circle with centre $O$. Show that the points $A, O, B$ and $P$ are concyclic.

Answer

Here, $OA = OB$
And $\text{OA}\perp\text{AP},\text{OA}\perp\text{BP}$ (Since tangent drawn from an external point are perpendicular to the radius at the point of contact)
$\therefore\angle\text{OAP}=90^\circ,\angle\text{OBP}=90^\circ$
$\therefore\angle\text{OAP}+\angle\text{OBP}=90^\circ+90^\circ=180^\circ$
$\therefore\angle\text{AOB}+\angle\text{APB}+\text{PB}=360^\circ)$
Sum of opposite angle of a quadrilateral is $180^\circ$.
Hence, $A, O B$ and $P$ are concyclic.

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Question 122 Marks

In the given figure, $O$ is the center of two concentric circle of radii $4\ cm$ and $6\ cm$ respectively. $PA$ and $PB$ are tangle to the outer and inner circle respectively. If $PA = 10\ cm$, find the length of $PB$ up to one place of decimal.

Answer

Given, $O$ is the centre of two concentric circles of radii $OA = 6\ cm$ and $OB = 4\ cm$.$PA$ and $PB$ are the two tangents to the outer and inner circles respectively and $PA = 10 \ cm$.
Now, tangent drawn from an external point is perpendicular to the radiusat the point of contact.
$\therefore\angle\text{OAP}=\angle\text{OBP}=90^\circ$
$\therefore$ From right- angled $\triangle\text{OAP},\text{OP}^2=\text{OA}^2+\text{PA}^2$
$=>\text{OP}=\sqrt{\text{OA}^2+\text{PA}^2}$
$=>\text{OP}=\sqrt{\text{6}^2+\text{10}^2}$
$=>\text{OP}=\sqrt{136}\text{cm}.$
$\therefore$ From right- angled $\triangle\text{OAP},\text{OP}^2=\text{OB}^2+\text{PB}^2$
$=>\text{PB}=\sqrt{\text{OP}^2-\text{OB}^2}$
$=>\text{PB}=\sqrt{\text{366}-\text{16}}$
$=>\text{PB}=\sqrt{120}\text{cm}.$
$=>\text{OP}=10.9\text{cm}.$
$\therefore$ The length of $PB$ is $10.9\ cm$.

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