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Maths 3 Marks Question

Circles · Maharashtra Board STD 10 (MSBSHSE - Semi English Medium). 12 questions.

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Question 13 Marks
In the adjoining figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6cm, BC = 9cm and CD = 8cm. Find the length of AD.
Answer
We know that when a quadrilateral circumscribes a circle then sum of opposites sides is equal to the sum of other opposite sides
$\therefore$ AB + CD = AD + BC
⇒ 6 + 8 = AD + 9
⇒ AD = 5cm
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Question 23 Marks
A point P is 25cm away from the centre of a circle and the length of tangent drawn from P to the circle is 24cm. Find the radius of the circle.
Answer

Draw a circle and let P be a point such that OP = 25cm
Let TP be the tangent, so that TP = 24cm
Join OT, where OT is radius.
Now, tangent drawn from an external point is perpendicular to the radius at the point of contact.
$\therefore\text{OT}\perp\text{PT}$
In the right $\triangle\text{OTP},$ we have:
$\text{OP}^2=\text{OT}^2+\text{TP}^2$ [By Pythagoras' theorem]
$\text{OT}^2=\sqrt{\text{OP}^2-\text{TP}^2}$
$=\sqrt{25^2-24^2}$
$=\sqrt{625-576}$
$=\sqrt{49}$
$=7\text{cm}$
$\therefore$ The length of the radius is 7cm.
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Question 33 Marks
In the given figure, a circle touches all the four sides of a quadrilateral ABCD whose three sides are AB = 6cm, BC = 7cm and CD = 4cm. Find AD.
Answer

Let the circle touch the sides of the quadrilateral AB, BC, CD and DA at P, Q, R and S respectively.
Given, AB = 6cm, BC = 7cm and CD = 4cm.
Tangents drawn from an external point are equal.
$\therefore\text{AP}=\text{AS},\text{BP}=\text{BQ},\text{CR}=\text{CQ}$ and $\text{DR}=\text{DS}$
Now, $\text{AB}+\text{CD}=(\text{AP}+\text{BP})+(\text{CR}+\text{DR})$
$=>\text{AB}+\text{CD}=(\text{AS}+\text{BQ})+(\text{CQ}+\text{DS})$
$=>\text{AB}+\text{CD}=(\text{AS}+\text{DS})+(\text{CQ}+\text{CQ})$
$=>\text{AB}+\text{CD}=\text{AD}+\text{BC}$
$=>\text{AD}=(\text{AB}+\text{CD})-\text{BC}$
$=>\text{AD}=(6+4)-7$
$=>\text{AD}=3\text{cm}$
$\therefore$ The length of AD is 3cm.
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Question 43 Marks
In the given figure, O is the centre of the circle and TP is the tangent to the circle from an external point T. If $\angle\text{PBT} = 30^\circ,$prove that
BA : AT = 2 : 1.
Answer
AB is the chord passing through the centre
So, AB is the diameter
Since, angle in a semi circle is a right angle
$\therefore\angle\text{APB} = 90^\circ$
By using alternate segment theorem
We have $\angle\text{APB} = \angle\text{PAT} = 30^\circ $
Now, in $\triangle\text{APB}$
$\angle\text{BAP} + \angle\text{APB} + \angle\text{BAP} = 180^\circ $ (Angle sum property of triangle)
$\Rightarrow\angle\text{BAP} = 180^\circ − 90^\circ − 30^\circ = 60^\circ$
Now, $\angle\text{BAP}=\angle\text{APT}+\angle\text{PTA}$ (Exterior angle property)
$\Rightarrow 60^\circ = 30^\circ + \angle\text{PTA}$
$\Rightarrow \angle\text{PTA}=60^\circ - 30^\circ =30^\circ $
We know that sides opposite to equal angles are equal.
$\therefore\text{ AP} = \text{AT}$
In right triangle ABP
$\sin\angle\text{ABP}=\frac{\text{AP}}{\text{BA}}$
$\Rightarrow\sin30^\circ=\frac{\text{AT}}{\text{BA}}$
$\Rightarrow\frac{1}{2}=\frac{\text{AT}}{\text{BA}}$
$\therefore\text{BA}:\text{AT}=2:1$
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Question 53 Marks
In the given figure, PQ is a chord of a circle with centre O and PT is a tangent. If $\angle\text{QPT} = 60^\circ ,$ find $\angle\text{PRQ}.$
Answer
We know that the radius and tangent are perperpendular at their point of contact.
$\therefore\angle\text{OPT}=90^\circ$
Now, $\angle\text{OPT}=\angle\text{OPT}-\angle\text{OPT}=90^\circ-60^\circ=30^\circ$
Since, OP = OQ as both are radius
$\therefore\angle\text{OPQ}=\angle\text{OQP}=30^\circ$(Angles opposite to equal sides are equal)
Now, In isosceles $\triangle\text{POQ}$
$\angle\text{POQ}+\angle\text{OPQ}+\angle\text{OQP}=180^\circ$(Angle sum property of a triangle)
$\Rightarrow\angle\text{POQ}=180^\circ-30^\circ-30^\circ=120^\circ$
Now, $ \angle\text{POQ}+\text{reflex }\angle\text{POQ}=360^\circ$(Complete angle)
$\Rightarrow\text{reflex }\angle\text{POQ}=360^\circ-120^\circ=240^\circ$
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.
$\therefore\angle\text{PRQ}=\frac{1}{2}(\text{reflex }\angle\text{POQ})=120^\circ.$
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Question 63 Marks
A point $P$ is at a distance of $29\ cm$ from the center of a circle of radius $20\ cm$. Find the length of the tangent drawn from $P$ to the circle.
Answer
Since tangent is always perpendicular to the radius, so a right triangle will be formed...
So, $(29)^2 = (20)^2 + x^2$
$841 = 400 + x^2$
$x^2 = 441$
$x = 21cm$
Hence, Length of the tangent is $21\ cm$.
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Question 73 Marks
A circle is inscribed in a $\triangle\text{ABC}$ touching AB, BC and AC at P, Q and R respectively. If AB = 10cm, AR = 7cm and CR = 5cm, find the length of BC.
Answer
Given, a circle inscribed in triangle ABC, such that the circle touches the sides of the triangle at P, Q, R.Tangent drawn to a circle from an external point are equal.
$\therefore\text{AP}=\text{AR}=7\text{cm},\text{CQ}=\text{CR}=5\text{cm}.$
Now, $\text{BP}=(\text{BP}-\text{BP})=(10-7)=3\text{cm}$
$\therefore\text{BP}=\text{BQ}=3\text{cm}$
$\therefore\text{BC}=(\text{BQ}+\text{QC})$
$=>\text{BC}=3+5$
$=>\text{BC}=8$
$\therefore$ The length of BC is 8cm.
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Question 83 Marks
In the given figure, two tangents RQ and RP are drawn from an external point R to the circle with centre O. If $\angle\text{PRQ} = 120^\circ$ then prove that OR = PR + RQ.
Answer

Construction: Join PO and OQ
In $\triangle\text{POR}$ and $\triangle\text{QOR}$
OP = OQ (Radii)
RP = RQ (Tangents from the external point are congruent)
OR = OR (Common)
By SSS congruency, $\triangle\text{POR}\cong\triangle\text{QOR}$
$\angle\text{PRO}=\angle\text{QRO}$ (C.P.C.T)
Now, $\angle\text{PRO}+\angle\text{QRO}=\angle\text{PRQ}$
$\Rightarrow2\angle\text{PRO}=120^\circ$
$\Rightarrow\angle\text{PRO}=60^\circ$
Now In $\triangle\text{POR}$
$\cos60^\circ=\frac{\text{PR}}{\text{OR}}$
$\Rightarrow\text{OR}=2\text{PR}$
$\Rightarrow\text{OR}=\text{PR}+\text{PR}$
$\Rightarrow\text{OR}=\text{PR}+\text{RQ}$
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Question 93 Marks
In the given figure, a circle inscribed in a triangle ABC touches the sides AB, BC and CA at points D, E and F respectively. If AB = 14cm, BC = 8cm and CA = 12cm. Find the lengths 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 = 14cm ....(1)
AF + FC = 12cm
⇒ AD + FC = 12cm ....(2)
BE + EC = 8cm
⇒ BD + FC = 8cm ....(3)
Adding all these we get
AD + BD + AD + FC + BD + FC = 34
⇒ 2(AD + BD + FC) = 34
⇒ AD + BD + FC = 17cm .....(4)
Solving (1) and (4), we get
FC = 3cm
Solving (2) and (4), we get
BD = 5cm = BE
Solving (3) and (4), we get
and AD = 9cm.
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Question 103 Marks
PQ is a chord of length $4.8\ cm$ of a circle of radius $3\ cm$. The tangent at P and Q intersect at a point T as shown in the figure. Find the length of TP.
Answer

Let $TR = y$ and $TP = x$
We know that the perpendicular drawn from the centre to the chord bisects
$\therefore$ $PR = RQ$
Now, $PR + RQ = 4.8$
$\Rightarrow PR + PR = 4.8$
$\Rightarrow PR = 2.4$
Now, in right triangle POR
By Using Pyhthagoras theorem, we have
$PO^2 = OR^2 + PR^2$
$\Rightarrow 3^2 = OR^2 + (2.4)^2$
$\Rightarrow OR^2 = 3.24$
$\Rightarrow OR = 1.8$
Now, in right triangle TPR
By Using Pyhthagoras theorem, we have
$TP^2 = TR^2 + PR^2$
$\Rightarrow x^2 = y^2 + (2.4)^2$
$\Rightarrow x^2 = y^2 + 5.76 .....(1)$
Again, in right triangle TPQ
By Using Pyhthagoras theorem, we have
$TO^2 = TP^2 + PO^2$
$\Rightarrow (y + 1.8)^2 = x^2 + 3^2$
$\Rightarrow y^2 + 3.6y + 3.24 = x^2 + 9$
$\Rightarrow y^2 + 3.6y = x^2 + 5.76 .....(2)$
Solving (1) and (2), we get
$x = 4\ cm$ and $y = 3.2\ cm$
$\therefore$ $TP = 4\ cm$
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Question 113 Marks
If the angle between two tangents drawn from an external point P to a circle of radius a and centre O, is 60° then find the length of OP.
Answer
We know that tangent is always perpendicular to the radius at the point of contact.
So, $\angle\text{OAP}=90^\circ$
We know that if 2 tangents are drawn from an external point, then they are equally inclined to the line segment joining the centre to that point.
So, $\angle\text{OPA}=12\angle\text{APB}=12\times60^\circ=30^\circ$
According to the angle sum property of triangle
In $\triangle\text{AOP}\angle\text{AOP}+\angle\text{OAP}+\angle\text{OPA}=180^\circ$
$\Rightarrow\angle\text{AOP}+90^\circ+30^\circ=180^\circ$
$\Rightarrow\angle\text{AOP}=60^\circ$
So, in triangle AOP
tan angle $\text{AOP}=\frac{\text{AP}}{\text{OA}}$
$\sqrt{3}=\frac{\text{AP}}{\text{a}}$
Therefore, $\text{AP}=\sqrt{3\text{a}}$
Hence, proved.
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Question 123 Marks
If PT is a tangent to a circle with centre O and PQ is a chord of the circle such that $\angle\text{QPT} = 70^\circ$ then find the measure of $\angle\text{POQ}.$
Answer
We know that the radius and tangent are perperpendular at their point of contact.
$\therefore\angle\text{OPT} = 90^\circ$
Now, $\angle\text{OPQ}=\angle\text{OPT}-\angle\text{TPQ}=90^\circ-70^\circ=20^\circ$
Since, OP = OQ as both are radius
$\therefore\angle\text{OPQ}=\angle\text{OQP}=20^\circ$(Angles opposite to equal sides are equal)
Now, In isosceles $ \triangle\text{POQ}$
$\angle\text{POQ}+\angle\text{OPQ}+\angle\text{OQP}=180^\circ$ (Angle sum property of a triangle)
$\Rightarrow\angle\text{POQ} = 180^\circ − 20^\circ − 20^\circ = 140^\circ$
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