[5 marks sum]

Question 15 Marks

AB is a line segment and M is its mid – point three semi circles are drawn with AM, MB and AB as diameter on the same side of the line AB. A circle with radius r unit is drawn so that it touches all the three semi circles show that : AB = 6 × r

Answer

Let $O, P$ and $Q$ be the centers of the circle and semicircles. Join $O P$ and $O Q$.
$
O R=O S=r
$
and $AP = PM = MQ = QB =\frac{A B}{4}$
Now, $OP = OR + RP = r +\frac{A B}{4}$ (since $PM = RP =$ radii of same circle)Similarly, $OQ = OS + SQ = r +\frac{A B}{4}$
$
OM = LM -; OL =\frac{A B}{2}- r
$Now in Rt. $\triangle O P M$,
$
\begin{aligned}
& O P^2=P M^2+O M^2 \\
& \Rightarrow\left(r=\frac{A B}{4}\right)^2=\left(\frac{A B}{4}\right)^2+\left(\frac{A B}{2}-r\right)^2
\end{aligned}
$$
\begin{aligned}
& \Rightarrow r^2+\frac{ AB ^2}{16}+\frac{ rAB }{2}=\frac{ AB ^2}{16}+\frac{ AB ^2}{4}+r^2-r A B \\
& \Rightarrow \frac{ rAB }{2}=\frac{ AB ^2}{4}-r A B \\
& \Rightarrow \frac{ AB ^2}{4}=\frac{ rAB }{2}+r A B \\
& \Rightarrow \frac{ AB ^2}{4}=\frac{3 rAB }{2} \\
& \Rightarrow \frac{ AB }{4}=\frac{3}{2} r \\
& \Rightarrow A B=\frac{3}{2} r \times 4=6 r
\end{aligned}
$Hence $A B=6 \times r$

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Question 25 Marks

The given figure shows a semi-circle with centre O and diameter PQ. If PA = AB and ∠BCQ =140°; Find measures of angles PAB and AQB. Also, show that AO is parallel to BQ.

Answer

Join PB.
i) In cyclic quadrilateral $PBCQ$,
$
\begin{aligned}
& \angle BPQ +\angle BCQ =180^{\circ} \\
& \Rightarrow \angle BPQ +140^{\circ}=180^{\circ} \\
& \Rightarrow \angle BPQ =40^{\circ} \ldots \ldots . .(1)
\end{aligned}
$Now in $\triangle PBQ$,
$
\begin{aligned}
& \angle PBQ + BPQ +\angle BQP =180^{\circ} \\
& \Rightarrow 90^{\circ}+40^{\circ}+\angle BQP =180^{\circ} \\
& \Rightarrow \angle BQP =50^{\circ}
\end{aligned}
$In cyclic quadrilateral $PQBA$,
$
\begin{aligned}
& \angle PQB +\angle PAB =180^{\circ} \\
& \Rightarrow 50^{\circ}+\angle PAB =180^{\circ} \\
& \Rightarrow \angle PAB =130^{\circ}
\end{aligned}
$
ii) Now in $\triangle PAB$,
$
\begin{aligned}
& \angle PAB +\angle APB + ABP =180^{\circ} \\
& \Rightarrow 130^{\circ}+\angle APB +\angle ABP =180^{\circ} \\
& \Rightarrow \angle APB +\angle ABP =50^{\circ} \\
& \text { But } \\
& \angle APB =\angle ABP (\because PA = PB ) \\
& \therefore \angle APB =\angle ABP =25^{\circ} \\
& \angle BAQ =\angle BPQ =40^{\circ} \\
& \angle APB =25^{\circ}=\angle AQB \text { (angles in the same segment) }
\end{aligned}
$$
\therefore \angle AQB =25^{\circ}
$
iii) Arc AQ subtends $\angle A O Q a t$ the centre and $\angle A P Q$ at the remaining part of the circle.
We have,
$
\angle APQ =\angle APB +\angle BPQ
$From (1), (2) and (3), we have
$
\begin{aligned}
& \angle APQ =25^{\circ}+40^{\circ}=65^{\circ} \\
& \therefore \angle AOQ =2 \angle APQ =2 \times 65^{\circ}=130^{\circ}
\end{aligned}
$Now in $\triangle A O Q$,
$
\angle O A Q=\angle O Q A=(\because O A=O Q)
$But
$
\begin{aligned}
& \angle OAQ +\angle OQA +\angle AOQ =180^{\circ} \\
& \Rightarrow \angle OAQ +\angle OAQ +130^{\circ}=180^{\circ} \\
& \Rightarrow 2 \angle OAQ =50^{\circ} \\
& \Rightarrow \angle OAQ =25^{\circ} \\
& \therefore \angle OAQ =\angle AQB
\end{aligned}
$
But these are alternate angles.
Hence, $AO$ is parallel to $BQ$.

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Question 35 Marks

In a square ABCD, its diagonals AC and BD intersect each other at point O. the bisector of angle DAO meets BD at point M and the bisector of angle ABD meets AC at N and AM at L. Show that:
∠ ONL + ∠ OML = 180

Answer

$A B C D$ is a square whose diagonals $A C$ and $B D$ intersect each other at right angles at $O$.

$
\therefore \angle AOB =\angle AOD =90^{\circ}
$In $\triangle ANB$,
$
\angle ANB =180^{\circ}-(\angle NAB +\angle NBA )
$
$\Rightarrow \angle ANB =180^{\circ}-\left(45^{\circ}+\frac{45^{\circ}}{2}\right)( NB$ is bisector of $\angle ABD )$
$
\Rightarrow \angle ANB =180^{\circ}-45^{\circ}-\frac{45^{\circ}}{2}=135^{\circ}-\frac{45^{\circ}}{2}
$But, $\angle LNO =\angle ANB$ (vertically opposite angles)
$
\therefore \angle LNO =135^{\circ}-\frac{45^{\circ}}{2}
$
Now in $\triangle AMO$,
$
\begin{aligned}
& \angle AMO =180^{\circ}-(\angle AOM +\angle OAM ) \\
& \Rightarrow \angle AMO =180^{\circ}-\left(90+\frac{45^{\circ}}{2}\right)( MA \text { is bisector of } \angle DAO ) \\
& \Rightarrow \angle AMO =180^{\circ}-90^{\circ}-\frac{45^{\circ}}{2}=90^{\circ}-\frac{45^{\circ}}{2}
\end{aligned}
$
Adding (i) and (ii)
$
\begin{aligned}
& \angle LNO +\angle AMO =135-\frac{45^{\circ}}{2}+90^{\circ}-\frac{45^{\circ}}{2} \\
& \Rightarrow \angle LNO +\angle AMO =225^{\circ}-45^{\circ}=180^{\circ} \\
& \Rightarrow \angle ONL +\angle OML =180^{\circ}
\end{aligned}
$

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Question 45 Marks

ABC is a triangle with AB = 10 cm, BC = 8 cm and AC = 6 cm (not drawn to scale). Three circle are drawn touching each other with the vertices as their centres. Find the radii of the three circles

Answer

E
Given: $A B C$ is a triangle with $A B=10 cm , B C=8 cm , A C=6 cm$.
Three circles are drawn with centre $A, B$ and $C$ touch each other at $P, Q$ and $R$ respectively.
We need to find the radii of the three circles.
Let
$
\begin{aligned}
& P A=A Q=x \\
& Q C=C R=y \\
& R B=B P=z \\
& \therefore x+z=10 \ldots(1) \\
& z+y=8 \ldots \ldots .(2) \\
& y+x=6 \ldots \ldots (3)
\end{aligned}
$
Adding all the three equations, we have
$
\begin{aligned}
& 2(x+y+z)=24 \\
& \Rightarrow x+y+=\frac{24}{2}=12.......(4)
\end{aligned}
$
Subtracting (1) (2) and (3) from (4)
$
\begin{aligned}
& y=12-10=2 \\
& x=12-8=4 \\
& z=12-6=6
\end{aligned}
$
Therefore, radii are $2 cm , 4 cm$ and $6 cm$

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Question 55 Marks

Prove that, of any two chords of a circle, the greater chord is nearer to the centre.

Answer

Given: $A$ circle with centre $O$ and radius $r . OM \perp AB$ and $ON \perp$ $C D$ Also $A B>C D$
To prove: $OM < ON$
Proof: Join OA and OC.
In Rt. $\triangle AOM$,
$
\begin{aligned}
& A O^2=A M^2+O M^2 \\
& r^2=\left(\frac{1}{2} A B\right)^2+O M^2 \\
& r^2=\frac{1}{4} A B^2+O M^2
.....(i)\end{aligned}
$
Again in Rt. $\triangle ONC$,
$
\begin{aligned}
& O C^2=N C^2+O N^2 \\
& \Rightarrow r^2=\left(\frac{1}{2} C D\right)^2+O N^2 \\
& \Rightarrow r^2=\frac{1}{4} C D^2+O N^2
......(ii)\end{aligned}
$
From (i) and (ii)
$
\begin{aligned}
& \frac{1}{4} A B^2+O M^2=\frac{1}{4} C D^2+O N^2 \\
& \text { But, } AB > CD \text { (given) } \\
& \therefore ON > OM \\
& \Rightarrow OM < ON
\end{aligned}
$
Hence, $A B$ is nearer to the centre than $C D$.

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Question 65 Marks

Prove that the perimeter of a right triangle is equal to the sum of the diameter of its incircle and twice the diameter of its circumcircle.

Answer

Join OL, OM and ON.
Let $D$ and $d$ be the diameter of the circumcircle and incircle. and let $R$ and $r$ be the radius of the circumcircle and incircle. In circumcircle of $\triangle ABC$,
$
\angle B =90
$
Therefore, $A C$ is the diameter of the circumcircle i.e. $A C=D$
Let radius of the incircle $=r$
$
\therefore OL = OM = ON = r
$
Now, from $B, B L, B M$ are the tangents to the incircle.
$
\therefore BL = BM = r
$
Similarly,
$AM = AN$ and $CL = CN = R$
(Tangents from the point outside the circle)
Now,
$
\begin{aligned}
& AB + BC + CA = AM + BM + BL + CL + CA \\
& = AN + r + r + CN + CA \\
& = AN + CN +2 r + CA \\
& = AC + AC +2 r \\
& =2 AC +2 r \\
& =2 D + d
\end{aligned}
$

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Question 75 Marks

In the adjoining figure, O is the centre of the circle and AB is a tangent to it at point B. ∠BDC = 65° Find ∠BAO.

Answer

$A B$ is a straight line.
$
\begin{aligned}
& \therefore \angle ADE +\angle BDE =180^{\circ} \\
& \Rightarrow \angle ADE +65^{\circ}=180^{\circ} \\
& \Rightarrow \angle ADE =115^{\circ}
\end{aligned}
$
$A B$ i.e. $D B$ is tangent to the circle at point $B$ and $B C$ is the diameter.
$
\begin{aligned}
& \therefore \angle DB =90^{\circ} \\
& \operatorname{In} \triangle BDC \\
& \angle DBC +\angle BDC +\angle DCB =180^{\circ} \\
& \Rightarrow 90^{\circ}+65^{\circ}+\angle DCB =180^{\circ} \\
& \Rightarrow \angle DCB =25^{\circ}
\end{aligned}
$
Now, $O E=O C$ (radii of the same circle)
$
\therefore \angle DCB \text { or } \angle OCE =\angle OEC =25^{\circ}
$
Also,
$
\angle OEC =\angle DEC =25^{\circ}
$
(vertically opposite angles)
$
\begin{aligned}
& \text { In } \triangle ADE \\
& \angle ADE +\angle DEA +\angle DAE =180^{\circ} \\
& \text { From (i) and (ii) } \\
& 115^{\circ}+25^{\circ}+\angle DAE =180^{\circ} \\
& \Rightarrow \angle DAE \text { or } \angle BAO =180^{\circ}-140^{\circ}=40^{\circ} \\
& \therefore \angle BAO =40^{\circ}
\end{aligned}
$

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Question 85 Marks

In the figure; $PA$ is a tangent to the circle, $PBC$ is secant and $AD$ bisects angle BAC. Show that triangle PAD is an isosceles triangle. Also, show that:
$
\angle C A D=\frac{1}{2}(\angle P B A-\angle P A B)
$

Answer

i) $PA$ is the tangent and $AB$ is a chord
$
\therefore \angle PAB =\angle C.....(i)
$
(angles in the alternate segment)
$A D$ is the bisector of $\angle B A C$
$
\begin{aligned}
& \therefore \angle 1=\angle 2 \ldots \ldots . . . \text { (ii) } \\
& \text { In } \triangle ADC \\
& \text { Ext. } \angle ADP =\angle C +\angle 1 \\
& \Rightarrow \text { Ext } \angle ADP =\angle PAB +\angle 2=\angle PAD
\end{aligned}
$
Therefore, $\triangle PAD$ is an isosceles triangle.
ii) In $\triangle A B C$,
$
\begin{aligned}
& \text { Ext. } \angle PBA =\angle C +\angle BAC \\
& \angle BAC =\angle PBA -\angle C \\
& \Rightarrow \angle 1+\angle 2=\angle P B A-\angle PAB
\end{aligned}
$
(fom (i) part)
$\begin{aligned} & 2 \angle 1=\angle P B A-\angle P A B \\ & \angle 1=\frac{1}{2}(\angle P B A-\angle P A B) \\ & \Rightarrow \angle C A D=\frac{1}{2}(\angle P B A-\angle P A B)\end{aligned}$

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Question 95 Marks

Three circles touch each other externally. A triangle is formed when the centres of these circles are joined together. Find the radii of the circle, if the sides of the triangle formed are 6 cm, 8 cm and 9 cm

Answer

$
A B=6 cm , A C=8 cm \text { and } B C=9 cm
$Let radii of the circles having centers $A, B$ and $C$ be $r_1, r_2$ and $r_3$ respectively
$
\begin{aligned}
& r_1+r_3=8 \\
& r_3+r_2=9 \\
& r_2+r_1=6
\end{aligned}
$
adding
$
\begin{aligned}
& r_1+r_3+r_3+r_2+r_2+r_1=8+9+6 \\
& 2\left(r_1+r_2+r_3\right)=23 \\
& r_1+r_2+r_3=11.5 cm \\
& r_1+9=11.5\left\{\left(\sin \text { cer }_2+r_3=9\right)\right. \\
& r_1=2.5 cm
\end{aligned}
$$
\begin{aligned}
& r_2+6=11.5\left(\sin c e r_1+r_3=6\right) \\
& r_2=5.5 cm \\
& r_3+8=11.5\left(\sin c e r_2+r_1=8\right) \\
& r_3=3.5 cm
\end{aligned}
$
Hence, $r_1=2.5 cm , r_2=5.5 cm$ and $r_3=3.5 cm$.

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Question 105 Marks

ABC is a right angles triangle with AB = 12 cm and AC = 13 cm. A circle, with centre O, has been inscribed inside the triangle.
Calculate the value of x, the radius of the inscribed circle.

Answer

$
\text { In } \triangle A B C, \angle B=90^{\circ}
$
$OL \perp AB , OM \perp BC$ and $ON \perp AC$LBNO is a square
$
\begin{aligned}
& \angle B=B N=O L=O M=O N=x \\
& \therefore A L=12-x \\
& \therefore A L=A N=12-x
\end{aligned}
$
since $A B C$ is a right triangle
$
\begin{aligned}
& A C^2=A B^2+ BC ^2 \\
& \Rightarrow 13^2=12^2+ BC ^2 \\
& \Rightarrow 169=144+ BC ^2 \\
& \Rightarrow BC ^2=25 \\
& \Rightarrow BC =5
\end{aligned}
$
$\begin{aligned} & \therefore M C=5-X \\ & \text { but } C M=C N \\ & \therefore C N=5-X \\ & \text { Now }, A C=A N+N C \\ & 13=(12-x)+(5-x) \\ & 13=17-2 x \\ & 2 x=4 \\ & x=2 cm \end{aligned}$

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Question 115 Marks

Two parallel tangents of a circle meet a third tangent at points P and Q. prove that PQ subtends a right angle at the centre.

Answer

Join $O P, O Q, O A, O B$ and $O C$.
In $\triangle OAP$ and $\triangle OCP$
$O A=O C$ (Radii of the same circle)
$O P=O P$ (Common)
PA $= PC$ (Tangents from P)
$\therefore$ By side - side - side criterion of congruence,
$\triangle OAP \cong \triangle OCP$ (SSS postulate)
The corresponding parts of the congruent triangles are congruent.
$
\Rightarrow \angle APO =\angle CPO \quad( cpct )
$
Similarly, we can prove that
$
\begin{aligned}
& \therefore \triangle OCQ \cong \triangle OBQ \\
& \Rightarrow \angle CQO =\angle BQO .
\end{aligned}
$
$\therefore \angle APC =2 \angle CPO$ and $\angle CQB =2 \angle CQO$
But,
$
\angle APC =\angle CQB =180
$
(Sum of interior angles of a transversal)
$
\therefore 2 \angle CPO +2 \angle CQO =180^{\circ}
$
$
\Rightarrow \angle CPO +\angle CQO =90^{\circ}
$Now in $\triangle POQ$,
$
\begin{aligned}
& \angle CPO +\angle CQO +\angle POQ =180^{\circ} \\
& \Rightarrow 90^{\circ}+\angle POQ =180^{\circ} \\
& \therefore \angle POQ =90^{\circ}
\end{aligned}
$

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Question 125 Marks

Tangents AP and AQ are drawn to a circle, with centre O, from an exterior point A. prove that: PAQ = 2 ∠OPQ

Answer

In quadrilateral $OPAQ$
$
\begin{aligned}
& \angle OPA =\angle OQA =90^{\circ} \\
& (\because OP \perp PA \text { and } OQ \perp QA )
\end{aligned}
$
$
\begin{aligned}
& \therefore \angle POQ +\angle PAQ +90^{\circ}+90^{\circ}=360^{\circ} \\
& \Rightarrow \angle POQ +\angle PAQ =360^{\circ}-180^{\circ}=180^{\circ}
\end{aligned}
$
In triangle $OPQ$,
$
\begin{aligned}
& OP = OQ \text { (Radii of the same circle) } \\
& \therefore OPQ =\angle OQP
\end{aligned}
$
$
\begin{aligned}
& \text { But } \\
& \angle POQ +\angle OPQ +\angle OQP =180^{\circ} \\
& \Rightarrow \angle POQ +\angle OPQ +\angle OPQ =180^{\circ} \\
& \Rightarrow \angle POQ +2 \angle OPQ =180^{\circ}
\end{aligned}
$
From (i) and (ii)
$
\begin{aligned}
& \angle POQ +\angle PAQ =\angle POQ +2 OPQ \\
& \Rightarrow \angle PAQ =2 \angle OPQ
\end{aligned}
$

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Question 135 Marks

From a point P outside the circle, with centre O, tangents PA and PB are drawn. Prove that:
OP is the perpendicular bisector of chord AB.

Answer

In Δ OAM and Δ OBM
OA = OB (Radii of the same circle)
∠ AOM = ∠ BOM (Proved ∠AOP = ∠BOP)
OM = OM (Common)
∴ By side angle side criterion ofcongruence,
Δ OAM ≅ Δ OBM
The corresponding parts of the congruent triangles are congruent.
⇒ AM = MB
and ∠OMA = ∠OMB = 180°
∴ ∠OMA = ∠OMB = 90°
Hence , OM or OP is the perpendicular bisector of chord AB.

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Mathematics [5 marks sum]

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