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Maths 4 Marks Questions

Circles · Gujarat Board (GSEB) STD 10 (GSEB - English Medium). 15 questions.

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Question 14 Marks
Prove that the perpendicular at the point of contact of the tangent to a circle passes through the centre.
Answer

Let $A B$ be the tangent to the circle at point $P$ with centre $O$.
To prove: $PQ$ passes through the point $O$.
Construction: Join $OP.$
Through $O$, draw a straight line $C D$ parallel to the tangent $A B$.
Proof: Suppose that $P Q$ doesn't passes through point $O$. $P Q$ intersect $C D$ at $R$ and also intersect $A B$ at $P$.
$A S, C D \| A B, P Q$ is the line of intersection,
$\angle ORP =\angle RPA$ (Alternate interior angles)
But also,
$\angle\text{RPA}=90^\circ(\text{OP}\perp\text{AB})$
$\Rightarrow\angle\text{ORP}=90^\circ$
$\angle\text{ROP}+\angle\text{OPA}=180^\circ$ (Co interior angles)
$\Rightarrow\angle\text{ROP}+90^\circ=180^\circ$
$\Rightarrow\angle\text{ROP}=90^\circ$
Thus, the $ \triangle\text{ORP}$ has 2 right angles i.e. $ \angle\text{ORP} $ and $ \angle\text{ROP} $ which is not possible.
Hence, our supposition is wrong.
$\therefore P Q$ passes through the point $O$.
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Question 24 Marks
A quadrilateral is drawn to circumscribe a circle. Prove that the sume of opposite sides are equal.
Answer

Given: $ABCD i$s a quadrilateral in which a circle is inscribed.
To prove: $AB + CD = AD + BC$
proof:
We know that the langths of tangent drawn fro an external point to circle are equal.
$\therefore AP = AS ....(i)$
$BP = BQ ....(ii)$
$CR = CQ ....(iii)$
$DR = DQ ....(iv)$
$\Rightarrow AB + CD = (AP + BP) + (CR + DR)$
$= (AS + BQ) + (CQ + DS) ....(from (i), (ii), (iii), (iv))$
$= (AS + DS) + (CQ + BQ)$
$= AD + BC$
Hence proved.
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Question 34 Marks
In the given figure, $O$ is the centre of a circle $PT$ and $PQ$ are tangents to the circle from an external point P. If $\text{TPQ} = 70^\circ$ then$ \angle\text{TRQ} .$
Answer
Construction: Join $OQ$ and $OT$

We know that the radius and tangent are perperpendular at their point of contact
$\therefore\angle\text{OTP}=\angle\text{OQP}=90^\circ$
Now, In quadrilateral OQPT
$\angle\text{QOT}+\angle\text{OTP}+\angle\text{OQP}+\angle\text{TPQ}=360^\circ$[Angle sum property of a quadrilateral]
$\Rightarrow \angle\text{QOT}+90^\circ+90^\circ+70^\circ=360^\circ$
$\Rightarrow 250^\circ+\angle\text{QOT}=360^\circ$
$\Rightarrow \angle\text{QOT}=110^\circ$
We know that the angle subtended by an arc at the centre is double the angle subtended by the arc at any point on the remaining part of the circle.
$\therefore\text{TRQ}=\frac{1}{2}(\angle\text{QOT})=55^\circ$
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Question 44 Marks
From an external point $P$, tangents $P A$ and $P B$ are drawn to a circle with centre $O$. If $C D$ is the tangent to the circle at a point $E$ and $P A=14 cm$, find the perimeter of $\triangle P C D$.
Answer
Given, $PA$ and $PB$ are the tangents to a circle with centre $O$ and $CD$ is a tangent at $E$ and $PA = 14cm.$
Tangent drawn from an external point are equal.
$\therefore\text{PA}=\text{PB},\text{CA}=\text{CE}$ and $\text{DB}=\text{DE}$
Perimeter of $\triangle\text{PCD}=\text{PC}+\text{CD}+\text{PD}$
$=(\text{PA}-\text{CA})+(\text{CE}+\text{DE})+(\text{PB}-\text{DB})$
$=(\text{PA}-\text{CE})+(\text{CE}+\text{DE})+(\text{PB}-\text{DE})$
$=(\text{PA}+\text{PB})$
$=2\text{PA } (\therefore\text{PA}=\text{PB})$
$=(2\times14)\text{cm}$
$=28\text{cm}$
$\therefore\text{Perimeter of }\triangle\text{ PCD}=28\text{cm}.$
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Question 54 Marks
$PQ$ is a chord of length $16\ cm$ of a circle of radius $10\ cm$. The tangent at $P$ and $Q$ intersect at point $T$ as shown in the figure.
Finf the length of $TP.$
Answer

Construction: Join $OQ.$
In $\triangle\text{TPO}$ and $\triangle\text{TQO},$
$TP = TQ ...$.(tangent from an external point to the circle are equal)
$OT = OT ....$(Common side)
$OP = OQ ...$.(radii of the same circle)
$\Rightarrow\triangle\text{TPO}\cong\triangle\text{TQO}$ ....(SSS congruence criterion)
$\Rightarrow\angle\text{PTO}=\angle\text{QTR}$ ....(cpct)$ .....(i)$
In $\triangle\text{TRP}$ and $\triangle\text{TRQ},$
$TP = TQ ...$.(tangent from an external point to the circle are equal)
$TR = TR ....$(Common side)
$\angle\text{PTR}=\angle\text{QTR}\dots\text{(from (i)})$
$\Rightarrow\triangle\text{TRP}\cong\triangle\text{TRQ} ....(SAS$ congruence criterion)
$\Rightarrow\angle\text{TRP}=\angle\text{TRQ}$
Since PRQ is a straight line segment,
$\angle\text{TRP}=\angle\text{TRQ}=180^\circ$
$\Rightarrow\angle\text{TRP}=\angle\text{TRQ}=90^\circ$
So, $\text{OR}\perp\text{PQ}$
We know that the perpendicular from the centre to the chord of a circle bisects the chord.
So, $PR = 8cm$
In $\triangle\text{ORP},$
$OR^2 = OP^2 - RP^2 ...$(By Pythagoras theorem)
$\Rightarrow OR^2 = 10^2 - 8^2$
$\Rightarrow OR^2 = 36$
$\Rightarrow OR = 6cm$
In right $\triangle\text{PRT},$
$PT^2 = TR^2 + PR^2$
$\Rightarrow PT^2 = TR^2 + 8^2 ....(i)$
In right $\triangle\text{POT},$
$\Rightarrow OT^2 = PT^2 + OP^2$
$\Rightarrow (TR+ 6)^2 = PT^2 + OP^2$
$\Rightarrow TR^2 + 12TR + 36 = PT^2 + 10^2 ....(ii)$
Solving $(i)$ and $(ii)$, we get
$TP = 10.7cm$
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Question 64 Marks
In the given figure, common tangents $AB$ and $CD$ to the two circles with centres $O_1$ and $O_2$ intersect at E. Prove that $AB = CD.$
Answer
We know that tangent segments to a circle from the same external point are congruent.
So, we have
$EA = EC$ for the circle having centre $O_1$
and
$ED = EB$ for the circle having centre $O_1$
Now, Adding ED on both sides in $EA = EC,$ we get
$EA + ED = EC + ED$
$\Rightarrow EA + EB = EC + ED$
$\Rightarrow AB = CD$
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Question 74 Marks
In the given figure, $O$ is the centre of the circle. $PA$ and $PB$ are tangents. Show that $AOBP$ is a cyclic quadrilateral.
Answer
We know that the radius and tangent are perpendicular at their point of contact
$\therefore\angle\text{OBP}=\angle\text{OAP}=90^\circ$
Now, In quadrilateral $AOBP$
$\angle\text{APB}+\angle\text{AOB}+\angle\text{OBP}+\angle\text{OAP}=360^\circ$[Angle sum property of a quadrilateral]
$\Rightarrow\angle\text{APB}+\angle\text{AOB}+90^\circ+90^\circ=360^\circ$
$\Rightarrow\angle\text{APB}+\angle\text{AOB}=180^\circ$
Also, $\angle\text{OBP}+\angle\text{OAP}=180^\circ$
Since, the sum of the opposite angles of the quadrilateral is 80º
Hence, $AOBP$ is a cyclic quadrilateral.
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Question 84 Marks
In the given figure, $PA$ and $PB$ are two tangents to the circle with centre O. If $\angle\text{APB}=60^\circ$ then find the measure of $\angle\text{OAB}.$
Answer

We know that the radius and tangent are perperpendular at their point of contact.
$\therefore\angle\text{OBP}=\angle\text{OAP}=90^\circ$
Now, In quadrilateral AOBP
$\angle\text{AOB}+\angle\text{OBP}+\angle\text{APB}+\angle\text{OAP}=360^\circ$ [Angle sum property of a quadrilateral]
$\Rightarrow\angle\text{AOB}=90^\circ+60^\circ+90^\circ=360^\circ$
$\Rightarrow240^\circ+\angle\text{AOB}=360^\circ$
$\Rightarrow\angle\text{AOB}=120^\circ$
Now, In isoceles triangle AOB
$\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$[Angle sum property of a triangle]
$\Rightarrow120^\circ+2\angle\text{OAB}=180^\circ$ $\big[\therefore\angle\text{OAB}=\angle\text{OBA}\big]$
$\Rightarrow\angle\text{OAB}=30^\circ$
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Question 94 Marks
In two concentric circles, a chord of length $8\ cm$ of the larger circle touches the smaller circle. If the radius of the larger circle is $5\ cm$ then Find the radius of the smaller circle.
Answer

We know that the radius and tangent are perperpendular at their point of contact
Since, the perpendicular drawn from the centre bisect the chord.
$\therefore\text{AP}=\text{PB}=\frac{\text{AB}}{2}=4\text{cm}$
In right triangle AOP
$ A O^2=O P^2+P A^2 $
$ \Rightarrow 5^2=O P^2+4^2 $
$ \Rightarrow O P^2=9 $
$ \Rightarrow O P=3 \mathrm{~cm}$
Hence, the radius of the smaller circle is $3cm.$
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Question 104 Marks
In the given figure, $PA$ and $PB$ are two tangents to the circle with centre $O$. If $ \text{APB} = 50^\circ$ then what is the measure of $\angle\text{OAB}.$
Answer
Construction: Join $OB$

We know that the radius and tangent are perperpendular at their point of contact
$\therefore\angle\text{OBP}=\angle\text{OAP}=90^\circ$
Now, In quadrilateral $AOBP$
$\angle\text{AOB}+\angle\text{OBP}+\angle\text{APB}+\angle\text{OAP}=360^\circ$[Angle sum property of a quadrilateral]
$\Rightarrow \angle\text{AOB}+90^\circ+50^\circ+90^\circ=360^\circ$
$\Rightarrow 230^\circ+\angle\text{BOC}=360^\circ$
$\Rightarrow \angle\text{AOB}=130^\circ$
Now, In isoceles triangle $AOB$
$\angle\text{AOB}+\angle\text{OAB}+\angle\text{OBA}=180^\circ$[Angle sum property of a triangle]
$\Rightarrow130^\circ+2\angle\text{OAB}=180^\circ$$[\therefore\angle\text{OAB}=\angle\text{OBA}]$
$\Rightarrow\angle\text{OAB}=25^\circ$
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Question 114 Marks
QUESIION In the given figure, a triangle $A B C$ is drawn to circumscribe a circle of radius $2 \ cm$ such that the segments $B D$ and $D C$ into which $B C$ is divided by the point of contact $D$, are of lengths $4 \ cm$ and $3 \ cm$ respectively. If the area of $\triangle A B C=21 cm^2$ then find the lengths of sides of $A B$ and $A C$.
Answer
Construction: Join $\text{OA},\text{OB},\text{OC},\text{OE}\perp\text{AB}$ at $\text{OF}\perp\text{AC}$ at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
$AE = AF, BD = BE = 4cm$ and $CD = CF = 3cm$
Now, $\text{Area}(\triangle\text{ABC})=\text{Area}(\triangle\text{BOC})+\text{Area}(\triangle\text{AOB})+\text{Area}(\triangle\text{ABC})+\text{Area}(\triangle\text{AOC})$
$\Rightarrow21=\frac{1}{2}\times\text{BC}\times\text{OD}+\frac{1}{2}\times\text{AB}\times\text{OE}\frac{1}{2}\times\text{AC}\times\text{OF}$
$\Rightarrow42=7\times2+(4+\text{x})\times2+(3+\text{x})\times2$
$\Rightarrow21=7+4+\text{x}+3+\text{x}$
$\Rightarrow21=14+2\text{x}$
$\Rightarrow2\text{x}=7$
$\Rightarrow\text{x}=3.5\text{cm}$
$\therefore$ $AB = 4 + 3.5 = 7.5 cm$ and $AC = 3 + 3.5 = 6.5 cm$
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Question 124 Marks
Two concentric circles are of radii $5\ cm$ and $3\ cm$, respectively. Find the length of the chord of the larger circle (in cm) which touches the smaller circle.
Answer

Given: Two circles have the same centre $O$ and $AB$ is a chord of the larger circle touching the
smaller circle at $C$; also, $OA = 5\ cm$ and $OC = 3\ cm.$
In
$\triangle\text{OAC},\text{OA}^2=\text{OC}^2-\text{AC}^2$
$\therefore\text{AC}^2=\text{OA}^2-\text{OC}^2$
$\Rightarrow\text{AC}^2=5^2-3^2$
$\Rightarrow\text{AC}^2=25-9$
$\Rightarrow\text{AC}^2=16$
$\Rightarrow\text{AC}=4\text{cm}$
$\therefore\text{AB}=2\text{AC}$(since perpendicular drawn from the centre of the circlebisects the chord)
$\therefore\text{AB}=2\times4=8\text{cm}$
The length of the chord of the larger circle is $8\ cm.$
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Question 134 Marks
 In the given figure, a triangle $A B C$ is drawn to circumscribe a circle of radius $3 \ cm$ such that the segments $B D$ and $D C$ into which $B C$ is divided by the point of contact $D$ are, of lengths $6 \ cm$ and $9 \ cm$ respectively. If the area of $\triangle A B C=54 cm^2$ then find the lengths of sides of $A B$ and $A C$.
Answer
Construction: Join $\text{OA},\text{OB},\text{OC},\text{OE}\perp\text{AB}$ at E and $\text{OF}\perp\text{AC}$ at F

We know that tangent segments to a circle from the same external point are congruent.
Now, we have
$AE = AF, BD = BE = 6cm$ and $CD = CF = 9cm$
Now,
$\text{Area}(\triangle\text{ABC})$
$=\text{Area}(\triangle\text{BOC})+\text{Area}(\triangle\text{AOB})+\text{Area}(\triangle\text{AOC})$
$\Rightarrow54=\frac{1}{2}\times\text{BC}\times\text{OD}+\frac{1}{2}\times\text{AB}\times\text{OE}+\frac{1}{2}\times\text{AC}\times\text{OF}$
$\Rightarrow108=15\times3+(6+\text{x})\times3+(9+\text{x})\times3$
$\Rightarrow36=15+6+\text{x}+9+\text{x}$
$\Rightarrow36=30+2\text{x}$
$\Rightarrow2\text{x}=6$
$\Rightarrow\text{x}=3\text{cm}$
$\therefore$ $AB = 6 + 3 = 9cm$ and $AC = 9 + 3 = 12cm$
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Question 144 Marks
Prove that the opposite sides of a quadrilateral circumscribing a circle subtend supplementary angles at the centre of the circle.
Answer

Given A quad. $ABCD$ circumscibes a circle with centre $O.$
To prove: $\angle\text{AOB}+\angle\text{COD}=180^\circ$
and $\angle\text{AOD}+\angle\text{BOC}=180^\circ$
Construction: Join $OP, OQ, OR$ and $OS.$
Proof:
We know that tangents drawn from an external point to a circle subtend equal angles at the centre.
$\therefore\angle1=\angle2,\angle3=\angle4,\angle5=\angle6$ and $\angle7=\angle8$
and $\angle1+\angle2+\angle3+\angle4+\angle5+\angle6+\angle7+\angle8=360^\circ$
$\Rightarrow2(\angle2+\angle3)+2(\angle6+\angle7)=360^\circ$ and$2(\angle1+\angle8)+2(\angle4+\angle5)=360^\circ$
$\Rightarrow\angle2+\angle3+\angle6+\angle7=180^\circ$ and $\angle1+\angle8+\angle4+\angle5=180^\circ$
$\Rightarrow\angle\text{AOB}+\angle\text{COD}=180^\circ$ and $\angle\text{AOD}+\angle\text{BOC}=180^\circ$
Hence proved.
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Question 154 Marks
Prove that the angle between the two tangent drawn from an external point to a circle is supplenentary to the angle subtended by the line segments joining the points of contact at the centre.
Answer

Given $PA$ and $PB$ are the tangent drawn from a point P to a circle with centre $O$. Also, the line segments $OA$ and $OB$ are shown.
To prove: $\angle\text{APB}+\angle\text{AOB}=180^\circ$
Proof:
We know that the tangent is perpendicular to the radius through the point of contact.
$\therefore\text{PA}\perp\text{OA}\Rightarrow\angle\text{OAP}=90^\circ$
$\therefore\text{PB}\perp\text{OB}\Rightarrow\angle\text{OBP}=90^\circ$
$\therefore\angle\text{OBP}+\angle\text{OBP}=90^\circ+90^\circ=180^\circ\dots\text{(i)}$
But, we know that tha sum of all the angles of a quadrilateral is 360º.
$\therefore\angle\text{OAP}+\angle\text{APB}+\angle\text{AOB}+\angle\text{OBP}=360^\circ\dots\text{(ii)}$
From (i) and (ii), we get
$\angle\text{APB}+\angle\text{AOB}=180^\circ$
Hence proved.
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