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Mathematics [4 marks sum]

Tangents and Intersecting Chords · ICSE STD 10 (ICSE - English Medium). 57 questions.

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[4 marks sum]

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Question 14 Marks
Show that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.
Answer

Join AD.
AB is the diameter.
∴ ∠ADB = 90º (Angle in a semi-circle)
But, ∠ADB + ∠ADC = 180º (linear pair)
⇒ ∠ADC = 90º
In ΔABD and ΔACD,
∠ADB = ∠ADC (each 90º)
AB = AC (Given)
AD = AD (Common)
ΔABD ≅ ΔACD (RHS congruence criterion)
⇒ BD = DC (C.P.C.T)
Hence, the circle bisects base BC at D.
 
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Question 24 Marks
In cyclic quadrilateral ABCD, ∠A = 3 ∠C and ∠D = 5 ∠B. Find the measure of each angle of the quadrilateral.
Answer

$A B C D$ is a cyclic quadrilateral.
$
\begin{aligned}
& \therefore \angle A +\angle C =180^{\circ} \\
& \Rightarrow 3 \angle C +\angle C =180^{\circ}
\end{aligned}
$
$
\begin{aligned}
& \Rightarrow 4 \angle C=180^{\circ} \\
& \Rightarrow \angle C=45^{\circ} \\
& \because \angle A=3 \angle C \\
& \Rightarrow \angle A=3 \times 45^{\circ} \\
& \Rightarrow \angle A=135^{\circ}
\end{aligned}$
Similarly,$\begin{aligned}& \therefore \angle B+\angle D=180^{\circ} \\& \Rightarrow \angle B+5 \angle B=180^{\circ} \\& \Rightarrow 6 \angle B=180^{\circ} \\& \Rightarrow \angle B=30^{\circ} \\& \because \angle D=5 \angle B \\& \Rightarrow \angle D=5 \times 30^{\circ} \\& \Rightarrow \angle D=150^{\circ}\end{aligned}$Hence, $\angle A=135^{\circ}, \angle B=30^{\circ}, \angle C=45^{\circ}, \angle D=150^{\circ}$
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Question 34 Marks
ABCD is a cyclic quadrilateral in which BC is parallel to AD, angle ADC = 110° and angle BAC = 50°. Find angle DAC and angle DCA.
Answer

$A B C D$ is a cyclic quadrilateral in which $A D \| B C$
$
\begin{aligned}
& \angle ADC =110^{\circ}, \angle BAC =50^{\circ} \\
& \angle B +\angle D =180^{\circ}
\end{aligned}
$
(Sum of opposite angles of a quadrilateral)
$
\begin{aligned}
& \Rightarrow \angle B +110^{\circ}=180^{\circ} \\
& \Rightarrow \angle B =70^{\circ}
\end{aligned}
$
Now in $\triangle ABC$,
$
\begin{aligned}
& \angle BAC +\angle ABC +\angle ACB =180^{\circ} \\
& \Rightarrow 50^{\circ}+70^{\circ}+\angle ACB =180^{\circ} \\
& \Rightarrow \angle ACB =180^{\circ}-120^{\circ}=60^{\circ} \\
& \because AD I BC \\
& \therefore \angle DAC =\angle ACB =60^{\circ} \text { (alternate angles) }
\end{aligned}
$
Now in $\triangle A D C$,
$
\begin{aligned}
& \angle DAC +\angle ADC +\angle DCA =180^{\circ} \\
& \Rightarrow 60^{\circ}+110^{\circ}+\angle DCA =180^{\circ} \\
& \Rightarrow \angle DCA =180^{\circ}-170^{\circ}=10^{\circ}
\end{aligned}
$
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Question 44 Marks
Two chords AB and AC of a circle are equal. Prove that the centre of the circle lies on the bisector of angle BAC.
Answer

Given: AB and AC are two equal chords of C (O, r).
To prove: Centre, O lies on the bisector of ∠BAC.
Construction: Join BC. Let the bisector of ∠BAC intersects BC in P.
Proof:
In ΔAPB and ΔAPC,
AB = AC (Given)
∠ BAP = ∠CAP (Given)
AP = AP (Common)
∴ Δ APB ≅ ΔAPC (SAS congruence criterion)
⇒ BP = CP and ∠APB = ∠APC (CPCT)
∠APB + ∠APC = 180° (Linear pair)
⇒ 2 ∠APB = 180° (∠APB = ∠APC)
⇒ ∠APB = 90°
Now, BP = CP and ∠APB = 90°
∴ AP is the perpendicular bisector of chord BC.
⇒ AP passes through the centre, O of the circle.
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Question 54 Marks
In the figure given below, O is the center of the circle of the circle and SP is a tangent. if ∠SRT=65°, find the value of x, y and Z.
Answer
$
\begin{aligned}
& T S \perp S P, \\
& \Rightarrow \angle T S R=90^{\circ} \\
& \text { In } \triangle T S R \\
& \angle T S R+\angle T R S+\angle R T S=180^{\circ} \\
& \Rightarrow 90^{\circ}+65^{\circ}+x=180^{\circ} \\
& \Rightarrow x=180^{\circ}-90^{\circ}-65^{\circ} \\
& \Rightarrow x=25^{\circ}
\end{aligned}
$
Now, $y=2 x$ (Angle subtended at the center is double that of the angle subtended by the arc at the same centre)
$
\begin{aligned}
& \Rightarrow y=2 \times 25^{\circ} \\
& \Rightarrow y=50^{\circ} \\
& \operatorname{In} \triangle O S P \\
& \angle O S P+\angle S P O+\angle P O S \equiv 180^{\circ} \\
& \Rightarrow 90^{\circ}+z+50^{\circ}=180^{\circ} \\
& \Rightarrow z=180^{\circ}-140^{\circ} \\
& \Rightarrow z=40^{\circ}
\end{aligned}
$
Hence, $x=25^{\circ}, y=50^{\circ}$ and $Z=40^{\circ}$
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Question 64 Marks
In the given circle with center o, ∠ABC=100°, ∠ACD=40° and CT is tangent to the circle at C. find ∠ADC and ∠DCT. 

 
Answer
$\angle A B C+\angle A D C=180^{\circ}$
(Opposite angle of a cydic quadrilateral are supplementary)
$
\begin{aligned}
& \Rightarrow 100^{\circ}+\angle A D C=180^{\circ} \\
& \Rightarrow \angle A D C=80^{\circ}
\end{aligned}
$
Now, in $\triangle A C D$,
$
\begin{aligned}
& \angle A D C+\angle C A D+\angle A D C=180^{\circ} \\
& \Rightarrow 40^{\circ}+\angle C A D+80^{\circ}=180^{\circ} \\
& \Rightarrow \angle C A D=180^{\circ}-120^{\circ} \\
& \Rightarrow \angle C A D=60^{\circ}
\end{aligned}
$
Now $\angle D C T=\angle C A D$ (angles in the alternate segment are equal)
$
\therefore \angle D C T=60^{\circ}
$
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Question 74 Marks
In the figure, given below, O is the centre of the circumcircle of triangle XYZ.

Tangents at X and Y intersect at point T. Given ∠XTY = 80°, and ∠XOZ = 140°, calculate the  value of ∠ZXY.
Answer

In the figure, a circle with centre $O$, is the circum circle of triangle $X Y Z$.
$
\angle XOZ =140^{\circ}
$Tangents at $X$ and $Y$ intersect at point $T$, such that $\angle X T Y=80^{\circ}$
$
\therefore \angle X O Y=180^{\circ}-80^{\circ}=100^{\circ}
$But, $\angle XOY +\angle YOZ +\angle ZOX =360^{\circ}$ [Angles at a point]
$
\begin{aligned}
& \Rightarrow 100^{\circ}+\angle YOZ +140^{\circ}=360^{\circ} \\
& \Rightarrow 240^{\circ}+\angle YOZ =360^{\circ} \\
& \Rightarrow \angle YOZ =360^{\circ}-240^{\circ} \\
& \Rightarrow \angle YOZ =120
\end{aligned}
$
Now arc YZ subtends $\angle YOZ$ at the centre and $\angle YXZ$ at the remaining part of the circle.
$
\begin{aligned}
& \therefore \angle YOZ =2 \angle YXZ \\
& \Rightarrow \angle Y X Z=\frac{1}{2} \angle Y O Z \\
& \Rightarrow \angle Y X Z=\frac{1}{2} \times 120^{\circ}=60^{\circ}
\end{aligned}
$
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Question 84 Marks
Two circle with centres A and B, and radii 5 cm and 3 cm, touch each other internally. If the perpendicular bisector of the segment AB meets the bigger circle in P and Q; find the length of PQ.
Answer

If two circles touch internally, then distance between their centres is equal to the difference of their radii.
So, $A B=(5-3) cm =2 cm$.
Also, the common chord $PQ$ is the perpendicular bisector of $AB$.
Therefore, $AC = CB =\frac{1}{2} AB$
$
=1 cm
$
In right $\triangle ACP$, we have $AP ^2= AC ^2+ CP ^2$
$
\begin{aligned}
& \Rightarrow 5^2=1^2+ CP ^2 \\
& \Rightarrow CP ^2=25-1=24 \\
& \Rightarrow CP =\sqrt{24}=2 \sqrt{6} cm
\end{aligned}
$
Now, $P Q=2 C P$
$
\begin{aligned}
& =2 \times 2 \sqrt{6} cm \\
& =4 \sqrt{6} cm
\end{aligned}
$
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Question 94 Marks
In the figure, given below, AC is a transverse common tangent to two circles with centres P and Q and of radii 6 cm and 3 cm respectively.

Given that AB = 8 cm, calculate PQ.
Answer

Since $A C$ is tangent to the circle with center $P$ at point $A$.
$
\therefore \angle PAB =90^{\circ}
$
Similarly, $\angle Q C B=90^{\circ}$
In $\triangle PAB$ and $\triangle QCB$
$
\begin{aligned}
& \angle PAB =\angle OCB =90^{\circ} \\
& \angle PBA =\angle QBC \text { (vertically opposite angles) } \\
& \therefore \triangle PAB \sim \triangle QCB \\
& \Rightarrow \text { ""PA "/"QC" = "PB"/"QB" ........ (i) }
\end{aligned}
$
Also in Rt. $\triangle PAB$,
$
\begin{aligned}
& P B=\sqrt{P A^2+P B^2} \\
& \Rightarrow P B=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10 cm....(ii)
\end{aligned}
$From (i) and (ii)
$
\begin{aligned}
& \frac{6}{3}=\frac{10}{Q B} \\
& \Rightarrow Q B=\frac{3 \times 10}{6}=5 cm
\end{aligned}
$
Now,
$
{ }^{\prime} P Q=P B+Q B=(10+5) cm =15 cm
$
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Question 104 Marks
In the following figure, $PQ = QR , \angle RQP =68^{\circ}, PC$ and $CQ$ are tangents to the circle with centre $O$

(i) $\angle QOP$
(ii) $\angle QCP$
Answer
Consider two concentric circles with centres at $O$. Let $A B$ and $C D$ be two chords of the outer circle which touch the inner circle at the points $M$ and $N$ respectively.

To prove the given question, it is sufficient to prove $A B=C D$.
For this join OM, ON, OB and OD.
Let the radius of outer and inner circles be $R$ and $r$ respectively.
$A B$ touches the inner circle at $M$.
$A B$ is a tangent to the inner circle
$
\begin{aligned}
& \therefore OM \perp AB \\
& \Rightarrow BM =\frac{1}{2} AB \\
& \Rightarrow AB =2 BM
\end{aligned}
$Similarly $ON \perp CD$, and $CD =2 DN$
Using Pythagoras theorem in $\triangle OMB$ and $\triangle OND$
$
\begin{aligned}
& O B^2=O M^2+B M^2, O D^2=O N^2+D M^2 \\
& \Rightarrow B M=\sqrt{R^2-r^2}, D N=\sqrt{R^2-r^2}
\end{aligned}
$Now,
$
\begin{aligned}
& A B=2 B M=2 \sqrt{R^2-r^2}, C D=2 D N=2 \sqrt{R^2-r^2} \\
& \therefore AB = CD
\end{aligned}
$Hence proved.
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Question 114 Marks
In two concentric circles, prove that all chords of the outer circle, which touch the inner circle, are of equal length.
Answer
$
\begin{aligned}
& \text { i) } PQ = RQ \\
& \therefore \angle PRQ =\angle QPR \text { (opposite angles of equal sides of a triangle) } \\
& \Rightarrow \angle PRQ +\angle QPR +68^{\circ}=180^{\circ} \\
& \Rightarrow \angle 2 PRQ =180^{\circ}-68^{\circ} \\
& \Rightarrow \angle PRQ =\frac{112^{\circ}}{2}=56^{\circ}
\end{aligned}
$Now, $\angle QOP =2 \angle PRQ$ (angle at the centre is double)
$
\Rightarrow QOP =2 \times 56^{\circ}=112^{\circ}
$
ii) $\angle PQC =\angle PRQ$ (angles in alternate segments are equal) $\angle QPC =\angle PRQ$ (angles in alternate segments)
$
\begin{aligned}
& \therefore \angle PQC =\angle QPC =56^{\circ}\left(\because \angle PRQ =56^{\circ} \text { from }( i )\right) \\
& \angle PQC +\angle QPC +\angle QCP =180^{\circ} \\
& \Rightarrow 56^{\circ}+56^{\circ}+\angle QCP =180^{\circ} \\
& \Rightarrow \angle QCP =68^{\circ}
\end{aligned}
$
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Question 124 Marks
In the given figure, AB is the diameter. The tangent at C meets AB produced at Q.

If ∠ CAB = 34° , find : ∠ CQB
Answer
QC is tangent to the circle
∴ ∠ CAB  = ∠ QCB
Angle between tangent and chord = angle in alternate segment
∴ ∠ QCB =  34°
ABQ is a straight line
⇒ ∠ ABC + ∠ CBQ = 180°
⇒ 56° + ∠ CBQ = 180°
⇒ CBQ = 124°
Now,
∠ CQB  = 180° - ∠ QCB  = CBQ
⇒ ∠ CQB  = 180° - 34°- 124°
⇒ ∠ CQB   =  22°
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Question 134 Marks
In the given figure, QAP is the tangent at point A and PBD is a straight line.

If ∠ACB = 36° and ∠APB = 42°, find:
(i) ∠BAP (ii) ∠ABD (iii) ∠QAD (iv) ∠BCD
Answer
$PAQ$ is a tangent and $AB$ is a chord of the circle.
i) $\therefore \angle B A P=\angle A C B \quad 36$ (angles in alternate segment)
ii) In $\triangle APB$Ext $\angle ABD =\angle APB +\angle BAP$
$\Rightarrow$ Ext $\angle A B D=42^{\circ}+36^{\circ}=78^{\circ}$
iii) $\angle ADB =\angle ACB =36^{\circ}$ (angles in the same segment)Now in $\triangle PAD$
Ext. $\angle QAD =\angle APB +\angle ADB$
$\Rightarrow$ Ext $\angle QAD =42^{\circ}+36^{\circ}=78^{\circ}$
iv) PAQ is the tangent and $A D$ is chord
$\therefore QAD =\angle ACD =78^{\circ}$ (angles in alternate segment)
And $\angle B C D=\angle A C B+\angle A C D$
$\therefore \angle BCD =36^{\circ}+78^{\circ}=114^{\circ}$
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Question 144 Marks
In the given figure, XY is the diameter of the circle and PQ is a tangent to the circle at Y.

If ∠AXB = 50° and ∠ABX = 70° and ∠BAY and ∠APY
Answer
In $\triangle AXB$,
$
\begin{aligned}
& \left.\angle XAB +\angle AXB +\angle ABX =180^{\circ} \text { [Triangle property }\right] \\
& \Rightarrow \angle XAB +50^{\circ}+70^{\circ}=180^{\circ} \\
& \Rightarrow \angle XAB =180^{\circ}-120^{\circ}=60^{\circ} \\
& \Rightarrow \angle XAY =90^{\circ}[\text { Angle of semi-circle }] \\
& \therefore \angle BAY =\angle XAY -\angle XAB =90^{\circ}-60^{\circ}=30^{\circ}
\end{aligned}
$
and $\angle B X Y=\angle B A Y=30^{\circ}$ [Angle of same segment]
$\angle A C X=\angle B X Y+\angle A B X[$ External angle $=$ Sum of two interior angles]
$
\begin{aligned}
& =30^{\circ}+70^{\circ} \\
& =100^{\circ}
\end{aligned}
$
also,
$
\begin{aligned}
& \angle XYP =90^{\circ}[\text { Diameter } \perp \text { tangent }] \\
& \angle APY =\angle ACX -\angle CYP \\
& \angle APY =100^{\circ}-90^{\circ} \\
& \angle APY =10^{\circ}
\end{aligned}
$
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Question 154 Marks
In the following figure, a circle is inscribed in the quadrilateral ABCD.

If BC = 38 cm, QB = 27 cm, DC = 25 cm and that AD is perpendicular to DC, find the radius of the circle.
 
Answer
From the figure we see that $BQ = BR =27 cm$ (since length of the tangent segments from an external point are equal)
$
\begin{aligned}
& \text { As } B C=38 cm \\
& \Rightarrow C R=C B-B R=38-27 \\
& =11 cm
\end{aligned}
$
Again,
$C R=C S=11 cm$ (length of tangent segments from an external point are equal)
Now, as DC $=25 cm$
$
\begin{aligned}
& \therefore D S=D C-S C \\
& =25-11 \\
& =14 cm
\end{aligned}
$
Now, in quadrilateral DSOP,
$\angle PDS =90^{\circ}$ (given)
$\angle OSD =90^{\circ}, \angle OPD =90^{\circ}$ (since tangent is perpendicular to the radius through the point of contact)
$\Rightarrow$ DSOP is a parallelogram
$\Rightarrow OP \| SD$ and $\Rightarrow PD \| OS$
Now, as OP $=$ OS (radii of the same circle)
$\Rightarrow$ OPDS is a square. $\therefore D S=O P=14 cm$
$\therefore$ radius of the circle $=14 cm$
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Question 164 Marks
Prove that any four vertices of a regular pentagon are concylic (lie on the same circle)
Answer

$ABCDE$ is a regular pentagon.
$
\begin{aligned}
& \therefore \angle BAF =\angle ABC =\angle BCD =\angle CDE =\angle DEA = \\
& \left(\frac{5-2}{5}\right) \times 180^{\circ}=180^{\circ}
\end{aligned}
$
In $\triangle AED$,
$A E=E D$ (Sides of regular pentagon $A B C D E$ )
$
\therefore \angle EAD =\angle EDA
$
In $\triangle AED$,
$
\begin{aligned}
& \angle AED +\angle EAD +\angle EDA =180^{\circ} \\
& \Rightarrow 108^{\circ}+\angle EAD +\angle EAD =180^{\circ} \\
& \Rightarrow 2 C \angle EAD =180^{\circ}-108^{\circ}=72^{\circ} \\
& \Rightarrow \angle EAD =36^{\circ} \\
& \therefore \angle EDA =36^{\circ} \\
& \angle BAD =\angle BAE -\angle EAD =108^{\circ}-36^{\circ}=72^{\circ}
\end{aligned}
$
In quadrilateral $A B C D$,
$
\angle BAD +\angle BCD =108^{\circ}+72^{\circ}=180^{\circ}
$
$\therefore A B C D$ is a cyclic quadrilateral
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Question 174 Marks
Two circles intersect in points P and Q. A secant passing through P intersects the circles in Aand B respectively. Tangents to the circles at A and B intersect at T. Prove that A, Q, B and T lie on a circle.
Answer

Join PQ.
AT is tangent and AP is a chord.
$\therefore \angle TAP =\angle AQP$ (angles in alternate segments).....(i)
Similarly, $\angle TBP =\angle BQP$
Adding (i) and (ii)
$
\angle TAP +\angle TBP =\angle AQP +\angle BQP....(ii)$
$
\Rightarrow \angle TAP +\angle TBP =\angle AQB$
Now in $\triangle T A B$,
$
\begin{aligned}
& \angle ATB +\angle TAP +\angle TBP =180^{\circ} \\
& \Rightarrow \angle ATB +\angle AQB =180^{\circ}
\end{aligned}$
Therefore, $A Q B T$ is a cyclic quadrilateral.
Hence, A, Q, B and T lie on a circle.
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Question 184 Marks
In the given figure, PAT is tangent to the circle with centre O at point A on its circumference and is parallel to chord BC. If CDQ is a line segment , show that:
(i) ∠BAP = ∠ADQ
(ii) ∠AOB = 2∠ADQ
(iii) ∠ADQ = ∠ADB
Answer
i) Since PAT || BC
$\therefore \angle PAB =\angle ABC$ (alternate angles)
In cyclic quadrilateral $A B C D$,
Ext $\angle A D Q=\angle A B C$
From (i) and (ii)
$
\angle PAB =\angle ADQ
$
ii) Arc $A B$ subtends $\square A O B$ at the centre and $\triangle A D B$ at the remaining part of the circle.
$
\begin{aligned}
& \therefore \angle AOB =2 \angle ADB \\
& \Rightarrow \angle AOB =2 \angle PAB \text { (angles in alternate segments) } \\
& \Rightarrow \angle AOB =2 \angle ADQ \text { (proved in (i) part) }
\end{aligned}
$
iii)
$
\begin{aligned}
& \therefore \angle BAP =\angle ADB \text { (angles in alternate segments) } \\
& \text { But } \\
& \angle BAP =\angle ADQ \text { (proved in (i) part) } \\
& \therefore \angle ADQ =\angle ADB
\end{aligned}
$
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Question 194 Marks
The given figure shows a circle with centre O such that chord RS is parallel to chord QT, angle PRT = 20° and angle POQ = 100°. Calculate:
(i) angle QTR
Answer

Join $P Q, R Q$ and $S T$.
i)
$
\begin{aligned}
& \angle POQ +\angle QOR =180^{\circ} \\
& \Rightarrow 100^{\circ}+\angle QOR =180^{\circ} \\
& \Rightarrow \angle QOR =80^{\circ}
\end{aligned}
$
Arc RQ subtends $\angle QOR$ at the centre and $\angle QTR$ at the remaining part of the circle.
$
\begin{aligned}
& \therefore \angle Q T R=\frac{1}{2} \angle Q O R \\
& \Rightarrow \angle Q T R=\frac{1}{2} \times 80^{\circ}=40^{\circ}
\end{aligned}
$
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Question 204 Marks
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
(ii) If the area of the rhombus is 32 3 cm2 find the radius of the circle.
Answer

Area of rhombus $=32 \sqrt{3} cm ^2$
But area of rhombus $O A B C=2 \times$ area of $\triangle O A B$.
Area of rhombus $OABC =2 \times \frac{\sqrt{3}}{4} r^2$
Where $r$ is the side of the equilateral triangle $O A B$.
$
\begin{aligned}
& 2 \times \frac{\sqrt{3}}{4} r^2=32 \sqrt{3} \\
& \Rightarrow \frac{\sqrt{3}}{2} r^2=32 \sqrt{3} \\
& \Rightarrow r^2=64 \\
& \Rightarrow r=8
\end{aligned}
$
Therefore, radius of the circle $=8 cm$
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Question 214 Marks
In a square ABCD, its diagonal AC and BD intersect each other at point O. The bisector of angle DAO meets BD at point M and bisector of angle ABD meets AC at N and AM at L. Show that -  ∠ BAM = ∠ BMA
Answer
$A B C D$ is a square whose diagonals $A C$ and $B D$ intersect each other at right angles at $O$.

$\begin{aligned}
& \angle BAM =\angle BAO +\angle OAM \\
& \Rightarrow \angle BAM =45^{\circ}+\frac{45^{\circ}}{2}=67 \frac{1^{\circ}}{2}
\end{aligned}
$
And
$
\begin{aligned}
& \Rightarrow \angle BMA =180^{\circ}-(\angle AOM +\angle OAM ) \\
& \Rightarrow \angle BMA =180^{\circ}-90^{\circ}-\frac{45^{\circ}}{2}=90^{\circ}-\frac{45^{\circ}}{2}=67 \frac{1^{\circ}}{2} \\
& \therefore \angle BAM =\angle BMA
\end{aligned}
$
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Question 224 Marks
OABC is a rhombus whose three vertices A, B and C lie on a circle with centre O.
(i) If the radius of the circle is 10 cm, find the area of the rhombus
Answer

$\begin{aligned} & \text { i) Radius }=10 cm \\ & \text { In rhombus } O A B C \text {, } \\ & O C=10 cm \\ & \therefore O E=\frac{1}{2} \times O B=\frac{1}{2} \times 10=5 cm \\ & \text { In Rt. } \triangle O C E \text {, } \\ & O C^2=O E^2+E C^2 \\ & \Rightarrow 10^2=5^2+E C^2 \\ & \Rightarrow E C^2=100-25=75 \\ & \Rightarrow E C=5 \sqrt{3} \\ & \therefore A C=2 \times E C=2 \times 5 \sqrt{3}=10 \sqrt{3} \\ & \text { Area of rhombus }=\frac{1}{2} \times O B \times A C \\ & =\frac{1}{2} \times 10 \times 10 \sqrt{3} \\ & =50 \sqrt{3} cm ^2 \approx 86.6 cm ^2(\sqrt{3}=1.73)\end{aligned}$
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Question 234 Marks
ABCDE is a cyclic pentagon with centre of its circumcircle at point O such that AB = BC = CD and angle ABC = 120°
Calculate :  ∠BEC 
Answer

$
\begin{aligned}
& \text { Join } O C \text { and } O B \text {. } \\
& A B=B C=C D \text { and } \angle A B C=120^{\circ} \\
& \therefore \angle B C D=\angle A B C=120^{\circ}
\end{aligned}
$
$O B$ and $O C$ are the bisectors of $\angle A B C$ and
$\angle BCD$ respectively.
$
\therefore \angle OBC =\angle BCO =60^{\circ}
$
In $\triangle BOC$,
$
\angle BOC =180^{\circ}-(\angle OBC +\angle BOC )
$
$
\begin{aligned}
& \Rightarrow \angle BOC =180^{\circ}-\left(60^{\circ}+60^{\circ}\right) \\
& \Rightarrow \angle BOC =180^{\circ}-120^{\circ}=60^{\circ}
\end{aligned}
$
Arc $BC$ subtends $\angle BOC$ at the centre and $\angle BEC$ at the remaining part of the circle.
$
\therefore \angle BEC =\frac{1}{2} \angle BOC =\frac{1}{2} \times 60^{\circ}=30^{\circ}
$
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Question 244 Marks
In the given figure, AC = AE Show that:
(i) CP = EP
(ii) BP = DP

 
Answer
In $\triangle A D C$ and $\triangle A B E$,
$\angle A C D=\angle A E B$ (angles in the same segment)
$AC = AE$ (Given)
$\angle A=\angle A($ common)
$\therefore \triangle ADC \cong \triangle ABE$ (ASA postulate)
$\Rightarrow A B=A D$
But $A C=A E$
$\therefore A C-A B=A E-A D$
$\Rightarrow BC = DE$
In $\triangle BPC$ and $\triangle DPE$
$\angle C =\angle E$ (angles in the same segment)
$BC = DE$
$\angle CBP =\angle CDE$ (angles in the same segment)
$\therefore \triangle BPC \cong \triangle DPE$ (ASA Postulate)
$\Rightarrow BP = DP$ and $CP = PE$ (cpct)
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Question 254 Marks
$ABC$ is a right triangle with angle $B = 90^\circ . A$ circle with $BC$ as diameter meets by hypotenuse $AC$ at point $D.$
Prove that $- BD^2 = AD \times DC.$
Answer

$\text { In } \triangle ADB ,$
$ \angle D =90^{\circ}$
$ \therefore \angle A +\angle ABD =90^{\circ}$
But in $\triangle ABC , \angle B =90$
$\therefore \angle A +\angle C =90$
From $(i)$ and $(ii)$
$\angle C=\angle ABD$
Now in $\triangle A B D$ and $\triangle C B D$
$\angle BDA =\angle BDA =90^{\circ}$
$ \angle ABD =\angle BCD$
$ \therefore \triangle ABD \sim \triangle CBD \text { (AA postulate) }$
$ \therefore \frac{ BD }{ DC }=\frac{ AD }{ DP }$
$ \Rightarrow BD ^2=A D \times D C$
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Question 264 Marks
The given figure shows a circle with centre O and BCD is tangent to it at C. Show that: ∠ACD + ∠BAC = 90°
Answer

Join OC.
$B C D$ is the tangent and $O C$ is the radius.
$
\begin{aligned}
& \therefore OC \perp BD \\
& \Rightarrow \angle OCD =90^{\circ} \\
& \Rightarrow \angle OCA +\angle ACD =90^{\circ}
\end{aligned}
$
But in $\triangle OCA$
$O A=O C$ (radii of same circle)
$
\therefore \angle OAC +\angle OAC
$
Substituting (i)
$
\begin{aligned}
& \angle OAC +\angle ACD =90^{\circ} \\
& \Rightarrow \angle BAC +\angle ACD =90^{\circ}
\end{aligned}
$
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Question 274 Marks
In the figure, AB is the chord of a circle with centre O and DOC is a line segment such that BC = DO. If ∠C = 20°, find angle AOD.
Answer

$
\begin{aligned}
& \text { Join } O B, \\
& \text { In } \triangle O B C, \\
& B C=O D=O B \text { (Radii of the same circle) } \\
& \therefore \angle B O C=\angle B C O=20^{\circ}
\end{aligned}
$
And Ext. $\angle ABO =\angle BCO +\angle BOC$
$\Rightarrow$ Ext. $\angle ABO =20^{\circ}+20^{\circ}=40^{\circ} \ldots .$. (i)
In $\triangle OAB$,
$
\begin{aligned}
& OA = OB \text { (radii of the same circle) } \\
& \therefore \angle OAB =\angle OBA =40^{\circ} \text { (from (i) } \\
& \angle AOB =180^{\circ}-\angle OAB -\angle OBA \\
& \Rightarrow \angle AOB =180^{\circ}-40^{\circ}-40^{\circ}=100^{\circ}
\end{aligned}
$
Since DOC is a straight line
$
\begin{aligned}
& \therefore \angle AOD +\angle AOB +\angle BOC =180^{\circ} \\
& \Rightarrow \angle AOD +100^{\circ}+20^{\circ}=180^{\circ} \\
& \Rightarrow \angle AOD =180^{\circ}-120^{\circ} \\
& \Rightarrow \angle AOD =60^{\circ}
\end{aligned}
$
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Question 284 Marks
Bisectors of vertex angles $A, B$, and $C$ of a triangle $A B C$ intersect its circumcircle at the points $D, E$ and $F$ respectively. Prove that angle EDF $=90^{\circ}-\frac{1}{2} \angle A$
Answer

We have,Given that, $BE$ is the bisector of $B$.
$
\therefore \angle A B E=\frac{\angle B}{2}
$
$\angle A D E=\angle A B E$
(Angles in the same segment for chord AF )Similarly, $\angle A C F=\angle A D F=\frac{\angle C}{2} \quad$
(Angle in the same segment for chord AF )
$\begin{aligned}
& \angle D=\angle A D E+\angle A D F \\
& \angle D=\frac{\angle B}{2}+\frac{\angle C}{2} \\
& =\frac{1}{2}(\angle B+\angle C) \\
& =\frac{1}{2}\left(180^{\circ}-\angle A\right) \quad\left(\therefore \angle A+\angle B+\angle C=180^{\circ}\right) \\
& \angle D=90^{\circ}-\frac{\angle A}{2}
\end{aligned}
$
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Question 294 Marks
In a cyclic quadrilateral ABCD, the diagonal AC bisects the angle BCD. Prove that the diagonal BD is parallel to the tangent to the circle at point A.
Answer

$\angle ADB =\angle ACB$.....(i) (Angles in same segement)
Similarly ,
$
\angle ABD =\angle ACD....(ii)$
But, $\angle ACB =\angle ACD ( AC$ is bisector of $\angle BCD )$
$\therefore \angle A D B=\angle A B D$ (from (i) and (ii))
TAS is a tangent and $A B$ is a chord
$\therefore \angle BAS =\angle ADB$ (angles in alternate segment)
But, $\angle ADB =\angle ABD$
$
\therefore \angle BAS =\angle ABD$
But these are alternate angles
Therefore, TS || BD
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Question 304 Marks
Two circles touch each other internally at a point P. A chord $A B$ of the bigger circle intersects the other circle in $C$ and $D$. Prove that $\angle CPA =\angle DPB$.
Answer

Draw a tangent $TS$ at $P$ to the circles given.
Since TPS is the tangent, PD is the chord.
$\therefore \angle PAB =\angle BPS....(i)$ (Angles in alternate segment)
Similarly,
$
\begin{aligned}
& \angle PCD =\angle DPS \ldots . . \text { (ii) } \\
& \text { Subtracting (i) from (ii) } \\
& \angle PCD -\angle PAB =\angle DPS -\angle BPS \\
& \text { But in } \angle PAC \\
& \text { Ext. } \angle PCD =\angle PAB +\angle CPA \\
& \therefore \angle PAB +\angle CPA -\angle PAB =\angle DPS -\angle BPS \\
& \Rightarrow \angle CPA =\angle DPB
\end{aligned}$
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Question 314 Marks
Tangent at P to the circumcircle of triangle PQR is drawn. If the tangent is parallel to side, QR show that ΔPQR is isosceles.
Answer

$D E$ is the tangent to the circle at $P$.
DE \| QR (Given)
$\angle EPR =\angle PRQ$ (Alternate angles are equal) $\angle DPQ =\angle PQR$ (Alternate angles are equal) ..... (i)
Let $\angle DPQ = x$ and $\angle EPR = y$
Since the angle between a tangent and a chord through the point of contact is equal to the angle in the alternate segment
$\therefore \angle DPQ =\angle PRQ....(ii)$(ii) (DE is tangent and $P Q$ is chord)
from (i) and (ii)
$\angle PQR =\angle PRQ$
$\Rightarrow PQ = PR$
Hence, triangle PQR is an isosceles triangle.
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Question 324 Marks
AB is the diameter and AC is a chord of a circle with centre O such that angle BAC = 30°. The tangent to the circle at C intersects AB produced in D. show that BC = BD.
Answer

Join OC,
$
\angle BCD =\angle BAC =30^{\circ} \text { (angles in alternate segment) }
$
$\operatorname{Arc} BC$ subtends $\angle DOC$ at the centre of the circle and $\angle BAC$ at the remaining part of the circle.
$
\therefore \angle BOC =2 \angle BAC =2 \times 30^{\circ}=60^{\circ}
$
Now in $\triangle OCD$,
$
\begin{aligned}
& \angle BOC \text { or } \angle DOC =60^{\circ} \\
& \angle OCD =90^{\circ}( OC \perp CD ) \\
& \therefore \angle DCO +\angle ODC =90^{\circ} \\
& \Rightarrow 60^{\circ}+\angle ODC =90^{\circ} \\
& \Rightarrow \angle ODC =90^{\circ}-60^{\circ}=30^{\circ}
\end{aligned}
$
Now in $\triangle B C D$,
$
\begin{aligned}
& \because \angle ODC \text { or } \angle BDC =\angle BCD =30^{\circ} \\
& \therefore BC = BD
\end{aligned}
$
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Question 334 Marks
If PQ is a tangent to the circle at R; calculate:
(i) ∠PRS
(ii) ∠ROT 

Given :  O is the centre of the circle and angle TRQ = 30°
Answer
PQ is a tangent and OR is the radius.
∴ OR ⊥  PQ
∴ ∠ ORT = 90°
⇒∠ TRQ = 90° - 30° = 60°
But in Δ OTR ,
OT = OR (Radii of the same circle)
∴ ∠ OTR = 60° Or ∠ STR = 60°
But,
∠ PRS = ∠ STR = 60 (Angle in the alternate segment)
In ΔORT,
∠ ORT  = 60°
∠ OTR =60°
∴ ∠ ROT = 180° - (60° +60° ) 
∴ ∠ ROT = 180°  - 120° = 60°
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Question 344 Marks
Circles with centres P and Q intersect at points A and B as shown in the figure. CBD is a segment and EBM is tangent to the circle with centre Q, at point B. If the circle are congruent; show that
CE = BD
Answer

Join $A B$ and $A D$
EBM is a tangent and $B D$ is a chord.
$\angle DBM =\angle BAD$ (angles in alternate segments)
But, $\angle DBM =\angle CBE$ (Vertically opposite angles)
$
\therefore \angle BAD =\angle CBE
$
Since in the same circle or congruent circles, if angles are equal, then chords opposite to them are also equal.
Therefore, $CE = BD$
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Question 364 Marks
Two circles intersect each other at points A and B. their common tangent touches the circles at points P and Q as shown in the figure. Show that the angles PAQ and PBQ are supplementary.
Answer

Join AB.
$P Q$ is the tangent and $A B$ is a chord
$
\therefore \angle QPA =\angle PBA.....(i)
$(angles in alternate segment)
Similarly,
$
\angle PQA =\angle QBA.....(ii)
$
Adding (i) and (ii)
$
\begin{aligned}
& \angle QPA +\angle PQA =\angle PBA +\angle QBA \\
& \text { But, in } \triangle PAQ , \\
& \angle QPA +\angle PQA =180^{\circ}-\angle PAQ \ldots \ldots(iii)
\end{aligned}
$
And $\angle PBA +\angle QBA =\angle PBQ.....(iv)$
From (iii) and (iv)
$
\begin{aligned}
& \angle PBQ =180^{\circ}-\angle PAQ \\
& \Rightarrow \angle PBQ +\angle PAQ =180^{\circ} \\
& \Rightarrow \angle PBQ+\angle PBQ =180^{\circ}
\end{aligned}
$
Hence $\angle PAQ$ and $\angle PBQ$ are supplementary
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Question 374 Marks
Two circles intersect each other at points A and B. A straight line PAQ cuts the circles at P and Q. If the tangents at P and Q intersect at point T; show that the points P, B, Q and T are concyclic
Answer

Join $A B, P B$ and $B Q$
TP is the tangent and PA is a chord
$\therefore \angle TPA =\angle ABP \ldots .$. (i) (angles in alternate segment)
Similarly,
$
\angle TQA =\angle ABQ \ldots \ldots . \text { (ii) }
$
Adding (i) and (ii)
$
\angle TPA +\angle TQA =\angle ABP +\angle ABQ
$
But, $\triangle PTQ$,
$
\begin{aligned}
& \angle TPA +\angle TQA +\angle PTQ =180^{\circ} \\
& \Rightarrow \angle PBQ =180^{\circ}-\angle PTQ \\
& \Rightarrow \angle PBQ +\angle PTQ =180^{\circ} \ldots
\end{aligned}
$
But they are the opposite angles of the quadrilateral Therefore, PBQT are cyclic.
Hence, P, B, Q and T are concyclic.
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Question 384 Marks
In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If ∠BCG = 108° and O is the centre of the circle, find : Angle BCT

Answer

Join OC, OD and AC
∠ BCG + ∠ BCD = 180° (Linear pair)
⇒ 180 °+  ∠ BCD  = 180°
⇒ ∠ BCD = 180°  - 108°  =72°
BC = CD
∴ ∠ DCP = ∠ BCT
But, ∠ BCT + ∠ BCD + ∠ DCP = 180°
∴ ∠ BCT  + ∠ BCT + 72° = 180°
2 ∠ BCT = 180°  - 72°
∠ BCT = 54°
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Question 394 Marks
From the given figure, prove that:
$
AP + BQ + CR = BP + CQ + AR
$

Also show that:
$
A P+B Q+C R=\frac{1}{2} \times \text { Perimeter of } \triangle A B C \text {. }
$
Answer
Since from $B, B Q$ and $B P$ are the tangents to the circle
Therefore, $BQ = BP$
Similarly, we can prove that
$AP = AR$
and $C R=C Q$
Adding,
$
AP + BQ + CR = BP + CQ + AR
$
Adding $A P+B Q+C R$ to both sides
$
\begin{aligned}
& 2(A P+B Q+C R)=A P+P Q+C Q+Q B+A R+C R \\
& 2(A P+B Q+C R)=A B+B C+C A
\end{aligned}
$Therefore, $A P+B Q+C R=\frac{1}{2} \times(A B+B C+C A)$
$A P+B Q+C R=\frac{1}{2} \times$ perimeter of triangle $A B C$
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Question 404 Marks
If the sides of a parallelogram touch a circle in following figure , Prove that the parallelogram is a rhombus.
Answer

From A, AP and AS are tangents to the circle.
Therefore, AP = AS.......(i)
Similarly, we can prove that:
BP = BQ .........(ii)
CR = CQ .........(iii)
DR = DS .........(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
But AB = CD and BC = AD.......(v) Opposite sides of a ||gm
Therefore, AB + AB = BC + BC
2AB = 2 BC
AB = BC ........(vi)
From (v) and (vi)
AB = BC = CD = DA
Hence, ABCD is a rhombus
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Question 414 Marks
If the sides of a quadrilateral ABCD touch a circle, prove that:
AB + CD = BC + AD
Answer

Let the circle touch the sides AB, BC, CD and DA of quadrilateral ABCD at P, Q, R and S respectively.
Since AP and AS are tangents to the circle from external point A
AP = AS .......(i)
Similarly, we can prove that:
BP = BQ .......(ii)
CR = CQ .......(iii)
DR = DS ........(iv)
Adding,
AP + BP + CR + DR = AS + DS + BQ + CQ
AB + CD = AD + BC
Hence, AB + CD = AD + BC
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Question 434 Marks
Two circle touch each other internally. Show that the tangents drawn to the two circles from any point on the common tangent are equal in length.
Answer

From Q, QA and QP are two tangents to the circle with centre O
Therefore, QA = QP .......(i)
Similarly, from Q, QB and QP are two tangents to the circle with centre O'
Therefore, QB = QP .......(ii)
From (i) and (ii)
QA = QB
Therefore, tangents QA and QB are equal.
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Question 444 Marks
In the given figure, O is the center of the circle and AB is a tangent at B. If AB = 15 cm and AC = 7.5 cm, calculate the radius of the circle.
Answer
$
A B=15 cm , A C=7.5 cm
$
Let ' $r$ ' be the radius of the circle.
$
\begin{aligned}
& \therefore O C=O B=r \\
& A O=A C+O C=7.5+r \\
& \ln \triangle A O B, \\
& A O^2=A B^2=O B^2 \\
& (7.5+r)^2=15^2+r^2 \\
& \Rightarrow\left(\frac{15+2 r}{2}\right)^2=225+r^2 \\
& \Rightarrow 225+4 r^2+60 r=900+4 r^2 \\
& \Rightarrow 60 r =675 \\
& \Rightarrow r =11.25 cm
\end{aligned}
$
Therefore, $r=11.25 cm$
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Question 454 Marks
In the given figure, PQ is a tangent to the circle at A. AB and AD are bisectors of ∠CAQ and ∠PAC. If ∠BAQ = 30°, prove that : BD is diameter of the circle.
Answer
∠CAB = ∠BAQ = 30°……(AB is angle bisector of ∠CAQ) 
∠CAQ = 2∠BAQ = 60°……(AB is angle bisector of ∠CAQ)
∠CAQ + ∠PAC = 180°……(angles in linear pair)
∴∠PAC = 120°
∠PAC = 2∠CAD……(AD is angle bisector of ∠PAC) 
∠CAD = 60° 
Now,
∠CAD + ∠CAB = 60 + 30 = 90° 
∠DAB = 90° 
Thus, BD subtends 90° on the circle
So, BD is the diameter of circle
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Question 464 Marks
In the given figure, O is the centre of the circumcircle ABC. Tangents at A and C intersect at P. Given angle AOB = 140° and angle APC = 80°; find the angle BAC.
Answer

Join OC.
Therefore, PA and PA are the tangents
$
\therefore OA \perp PA \text { and } OC \perp PC
$
In quadrilateral APCO,
$
\begin{aligned}
& \angle APC +\angle AOC =180^{\circ} \\
& \Rightarrow 80^{\circ}+\angle AOC =180^{\circ} \\
& \Rightarrow \angle AOC =100^{\circ} \\
& \angle BOC =360^{\circ}-(\angle AOB +\angle AOC ) \\
& \angle BOC \cong 360^{\circ}-\left(140^{\circ}+100^{\circ}\right) \\
& \angle B =360^{\circ}-240^{\circ}=120^{\circ}
\end{aligned}
$Now, arc $B C$ subtends $\angle B O C$ at the centre and $\angle B A C$ at the remaining part of the circle
$
\begin{aligned}
& \therefore \angle BAC =\frac{1}{2} \angle BOC \\
& \angle BAC =\frac{1}{2} \times 120^{\circ}=60^{\circ}
\end{aligned}
$
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Question 474 Marks
PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50° ; calculate : ∠ APT
Answer

Join AT and BT.
∠ BTX = ∠ TCB = 30° (Angles in the same segment)
∴ ∠ PTB = 180° - 30° = 150°
Now in Δ PTB ,
∠APT + ∠ PTB + ∠ ABT = 180°
⇒ ∠APT +  150° + 20° = 180°
⇒ ∠APT = 180° - 170° = 10°
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Question 484 Marks
PT is a tangent to the circle at T. If ∠ ABC = 70° and ∠ ACB = 50° ; calculate : ∠ APT
Answer

Join AT and BT.
∠ CBA = 70°
∴ ∠ABT = ∠CBT - ∠CBA = 90° - 70° = 20°
Now , ∠ ACT = ∠ABT = 20°  (Angles the same segment of the circle)
∴ ∠TCB = ∠ACB - ∠ACT = 50° - 20° = 30°
But , ∠TCB = ∠TAB (Angles in the same segment of the circle)
But , ∠TCB or ∠BAT = 30° 
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Question 494 Marks
In quadrilateral ABCD; angles D = 90°, BC = 38 cm and DC = 25 cm. A circle is inscribed in this quadrilateral which touches AB at point Q such that QB = 27 cm, Find the radius of the circle.
Answer

$B Q$ and $B R$ are the tangents from $B$ to the circle.
Therefore, $BR = BQ =27 cm$.
Also $RC =(38-; 27)=11 cm$
Since $C R$ and $C S$ are the tangents from $C$ to the circle
Therefore, $C S=C R=11 cm$
So, DS $=(25-11)=14 cm$
Now DS and DP are the tangents to the circle
Therefore, $DS = DP$
Now, $\angle PDS =90^{\circ}$ (given)
and $OP \perp AD , OS \perp DC$
therefore, radius $= DS =14 cm$
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