Class 10: Circles – Exercise 18(d) – ICSE / ISC / CBSE Mathematics Portal for K12 Students

18d1 Question 1: In the given circle with diameter $AB$ , find the value of $x$ . [2003]

Answer:

$\angle ABD = \angle ACD = 30^o$ (angles in the same segment)

In $\triangle ADB$

$\angle BAD + \angle ADB + \angle ABD = 180^o$

Since $AB$ is the diameter,

$\angle ADB = 90^o$ (angle in the semi circle)

$x+ 90+30 = 180 \Rightarrow x = 60^o$

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18d2 Question 2: In the given figure, $O$ is the center of the circle with radius $5 \text{ cm }$ . $OP$ and $OQ$ are $\perp$ to $AB$ and $CD$ respectively. $AB = 8\text{ cm }$ and $CD = 6\text{ cm }$ . Determine the length of $PQ$ .

Answer:

Given: $Radius = 5 \text{ cm } , \ OP \perp AB \ and \ OQ \perp CD$

$AB = 8 \text{ cm } \ and \ CD = 6 \text{ cm }$

$P \ and \ Q$ are mid point of $AB \ and \ CD$ respectively ( $\perp$ from the center to a chord will bisect the chord)

In $\triangle OAP$

$OA^2= OP^2+AP^2$

$25 = OP^2 + 16 \Rightarrow OP = 3 \text{ cm }$

Similarly, in $\triangle OCQ$

$OC^2= OQ^2+CQ^2$

$25 = OQ^2+9 \Rightarrow OQ = 4 \text{ cm }$

Hence $PQ = OP + OQ = 3 + 4 = 7 \text{ cm }$

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18d3 Question 3: The given figure shows two circles with centers $A$ and $B$ ; and radius $5\text{ cm }$ and $3\text{ cm }$ respectively, touching each other internally. If the perpendicular bisector of $AB$ meets the bigger circle in $P$ and $Q$ , find the length of $PQ$ .

Answer:

$AB = AC - BC = 5 - 3 = 2 \text{ cm }$

$PQ$ bisects $AB$ (let the point be $R$ )

$AR = 1 \text{ cm }$

In $\triangle APR$

$AP^2= RP^2+AR^2$

$\Rightarrow 25 = RP^2+1 \Rightarrow RP = 24 = 2\sqrt{6} \text{ cm }$

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18d4 Question 4: In the given figure, $\triangle ABC$ in which $\angle BAC = 30^o$ . Show that $BC$ is equal to the radius of the circumcircle of the $\triangle ABC$ , whose center is $O$ .

Answer:

Given $\angle BAC = 30^o$

$\angle BOC = 2 \angle BAC = 2 . 30 = 60^o$ (angle at the center of the circle is twice that of the angle subtend at the circumference by the same chord)

In $\triangle OBC$

$OB = OC$ (radius of the same circle)

$\angle OBC + \angle OCB + \angle BOC = 180^o$

$\angle OBC + \angle OCB = 180-60 = 120^o$

$2 \angle OBC = 120^o \Rightarrow \angle OBC = 60^o$

Therefore $\angle OBC = \angle OCB = \angle BOC = 60^o$

Therefore $\triangle BOC$ is equilateral

$\Rightarrow BC = OB = OC$

Hence $BC$ is equal to the radius of the circle.

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Question 5: Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.

Answer:

Given: Isosceles $\triangle ABC$ , where $AB = AC$ .

$AB$ is the diameter of the circle.

$\angle ADB = 90^o$ (angle in the semi circle)

$\angle ADC = 180-90=90^o$

Consider $\triangle ADB$ and $\triangle ADC$

$AD$ is common

$\angle ADC = \angle ADB$

$AB = AC$

Therefore $\triangle ADB \cong \triangle ADC (SAS \ axiom)$

Therefore $BD = DC$

Hence $D$ is the midpoint of $BC$ .

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18d5 Question 6: In the given figure, chord $ED$ is parallel to diameter $AC$ of the circle. Given $\angle CBE = 65^o$ , calculate $\angle DEC$ .

Answer:

Given $AD \parallel AC$

$\angle CBE = 65^o$

$\widehat{EC}$ subtends $\angle EOC$ at center and $\angle EBC$ at circumference.

$\angle ECO = 2 \angle EBC = 2 \times 65 = 130^o$

In $\triangle EOC$ ,

$EO = OC$ (radius of the same circle)

$\angle OEC = \angle OCE$

$\angle OEC + \angle OCE + \angle EOC = 180^o$

$2 \angle OEC + 130 = 180 \Rightarrow \angle OEC = 25^o$

Given $ED \parallel AC$

$\Rightarrow \angle DEC = \angle OEC \Rightarrow \angle DEC = 25^o$

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18d6 Question 7: Chords $AB$ and $CD$ of a circle intersect each other at point $P$ such that $AP = CP$ . Show that: $AB = CD$ .

Answer:

Given $AB$ and $CD$ intersect at $P$

$AP = CP$ (given)

Therefore $AP \times PB = CP \times PD$

$\displaystyle \Rightarrow \frac{AP}{CP} = \frac{PD}{PB}$

$\Rightarrow PD = PB$

Therefore $AP + PB = CP + PD \Rightarrow AB = CD$

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Question 8: The quadrilateral formed by the angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Answer:

Given: $ABCD$ is a cyclic quadrilateral and $PQRS$ is a quadrilateral

In $\triangle APD$ :

$PAD + ADP + APD = 180^o$ … … … … … (i)

In $\triangle BQC$

$QBC + BCQ + BQC = 180^o$ … … … … … (ii)

Adding (i) and (ii)

$PAD + ADP + APD + QBC + BCQ + BQC = 360^o$

$\displaystyle \frac{1}{2} (BAD + ADC + DCB + CBA) + APD + BQC = 360^o$

$180 + APD + BQC = 360^o \Rightarrow APD + BQC = 180^o$

Therefore $PQRS$ is a cyclic quadrilateral as the opposite angles are supplementary.

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18d7 Question 9: In the given diagram, $\angle DBC = 58^o$ , $BD$ is a diameter of the circle. Calculate: (i) $\angle BDC$ (ii) $BEC$ (iii) $\angle BAC$ [2014]

Answer:

Given $BD$ is diameter

$\angle DBC = 58^o$

(i) $\angle DCB = 90^o$ (angle in semi circle)

In $\triangle BDC$

$58+\angle BDC+90=180^o$

$\angle BDC = 32^o$

(ii) $ABEC$ is a cyclic quadrilateral

$\angle BDC = \angle BAC = 32^o$ (angle in the same segment)

Therefore $\angle BEC + 32 = 180^o$

$\angle BEC = 148^o$

(iii) $\angle BAC = 32^o$ (angle in the same segment)

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Question 10: $D$ and $E$ are points on equal sides $AB$ and $AC$ of an isosceles $\triangle ABC$ such that $AD = AE$ . Prove that the points $B, C, E$ and $D$ are concyclic.

Answer:

In $\triangle ABC AB = AC$ . Also given $AD = AE$

Therefore $\angle ABC = \angle ACB = x \ (say)$

In $\triangle ADE, AD = AE$

Now in $\triangle ABC$

$\displaystyle \frac{AD}{AB} =\frac{AE}{AC}$

$\Rightarrow DE \parallel BC$

$\Rightarrow \angle BCE = \angle DEA = \angle ADE = x$

Therefore $\angle BCE + \angle BDE = x + (180-x) = 180$

$\Rightarrow BDCE$ is a cyclic quadrilateral.

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18d8 Question 11: In the given figure, $ABCD$ is a cyclic quadrilateral. $AF \parallel CB$ and $DA$ is produced to point $E$ . If $\angle ADC = 92^o, \angle FAE= 20^o$ ; Determine $\angle BCD$ . Give reason in support of your answer.

Answer:

Given: $ABCD$ is a cyclic quadrilateral

$AF \parallel CB. \angle ADC = 92^o$ and $\angle FAE = 20^o$

$\angle CDA + \angle CBA = 180^o$

$\angle CBA = 180-92 = 88^o$

Since $AF \parallel CB$

$\angle CBA = \angle BAF$ (alternate angles)

$\angle BAE = 88 + 20 = 108^o$

Therefore $\angle DBA = 180-108 = 72^o$

Therefore $\angle BCD = 180-72 = 108^o$

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18d9 Question 12: If $I$ is the incentre $\triangle ABC$ and $AI$ when produced meets the circumcircle of $\triangle ABC$ at point $D$ . If $\angle BAC = 66^o$ and $\angle ABC = 80^o$ . Calculate (i) $\angle DBC$ , (ii) $\angle IBC$ , (iii) $\angle BIC$

Answer:

Given $\angle BAC = 66^o$ and $\angle ABC = 80^o$

(i) $\angle DBC = \angle DAC$ (angles in the same segment)

Since $I$ is the incenter

$\displaystyle \angle DAC = \frac{1}{2} \angle BAC = \frac{1}{2} \times 66 = 33^o$

Therefore $\angle DBC = 33^o$

(ii) Similarly, $\displaystyle \angle IBC = \frac{1}{2} \angle ABC = \frac{1}{2} \times 80 = 40^o$

(iii) In $\triangle ABC, \angle ACB = 180-\angle ABC - \angle BAC = 180-80-66 = 34^o$

Also $\displaystyle \angle ICB = \frac{1}{2} \angle BCA = \frac{1}{2} \times 34 = 17$

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18d10 Question 13: ln the given figure, $AB = AD = DC = PB$ and $\angle DBC = x^o$ . Determine, in terms of $x$ : (i) $\angle ABD$ , (ii) $\angle APB$ . Hence or otherwise, prove that $AP \parallel DB$ .

Answer:

Given $AB = AD =DC = PB$

$\angle DBC = x^o$

$\angle DAC = \angle DBC$ (angles in the same segment)

Since $AC = DC$

$\angle DAC = \angle DCA = x$

Similarly, $\angle ABD = \angle DAC$ ( $Since ABD = ACD \Rightarrow ACD = DAC$ )

In $\triangle ABP, \ \angle ABC = \angle BAP + \angle APB$

Since $AB = BP$

$\angle BAP = \angle APB$

$2x = \angle APB + \angle APB = 2 \angle APB$

$\Rightarrow \angle APB = x^o$

Therefore $\angle APB = \angle DBC = x^o$ and hence $AP \parallel DB$ (alternate angles)

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18d11 Question 14: In the given figure; $ABC, AEQ$ and $CEP$ are straight lines. Show that $\angle APE$ and $\angle CQE$ are supplementary.

Answer:

Given $ABC, AEQ \ and \ CEP$ are straight lines.

In cyclic quadrilateral $ABFP$

$\angle APE + \angle ABE = 180^o$ … … … … … (i)

Similarly in cyclic quadrilateral $BCQE$

$\angle CQE+\angle CBE = 180^o$ … … … … … (ii)

Adding (i) and (ii)

$\angle APE + \angle ABE + \angle CQE+\angle CBE = 360$

$\angle APE + \angle CQE + 180 = 360$

$\Rightarrow \angle APE + \angle CQE = 180$

Hence they are supplementary.

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18d12 Question 15: In the given figure, $AB$ is the diameter of the circle with center $O$ . If $\angle ADC = 32^o$ , find $\angle BOC$ .

Answer:

$\widehat{AC}$ subtends $AOC$ at the center and $ADC$ at the circumference of the circle.

$\angle AOC = 2 \angle ADC$

$\Rightarrow \angle AOC = 2 \times 32 = 64^o$

$\angle AOC + \angle BOC = 180 \Rightarrow \angle BOC = 180-64=116^o$

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d21 Question 16: In a cyclic-quadrilateral $PQRS, \angle PQR = 135^o$ . Sides $SP$ and $RQ$ produced meet at point $A$ whereas sides $PQ$ and $SR$ produced meet at point $B$ . lf $\angle A: \angle B = 2 : 1$ ; find $\angle A$ and $\angle B$ .

Answer:

Given $PQRS$ is a cyclic quadrilateral.

$\angle A = \angle B = 2 :1$ or if $\angle B = x$ , then $\angle A = 2x$

$\angle PQR +\angle PSR = 180^o$ (sum of the opposite angles in a cyclic quadrilateral is $180^o$ )

$\angle PSR = 180-135 = 45^o$

In $\triangle PBS; \angle SPB + 45+x = 180^o$

$\Rightarrow \angle SPB = 135-x$ … … … (i)

In $\triangle PQA; \angle APQ = 180-2x-45 = 135-2x$

Therefore $\angle SPB = 180 - (135-2x) = 45+2x$ … … … (ii)

Hence $35-x = 45+2x \Rightarrow x = 30^o$

Therefore $\angle B = 30^o \ and \ \angle A = 60^o$

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18d13 Question 17: In the following figure, $AB$ is the diameter of a circle with center $O$ and $CD$ is the chord with length equal to radius $OA$ . If $AC$ produced and $BD$ produced meet at point $P$ ; show that $\angle APB = 60^o$ .

Answer:

Given: $AB$ is diameter, $CD = OA = Radius$

In $\triangle CDO, CD = OD = OC$

$\Rightarrow \triangle CDO$ is equilateral

Therefore $\angle OCD = \angle OCD = \angle COD = 60^o$

In $\triangle AOC$

$OA = OC$ (radius of the same circle)

Therefore $\angle OCA = \angle CAO$

Similarly in $\triangle BOD$

$OD = OB$ (radius of the same circle)

Therefore $\angle ODB = \angle BDO$

Since $ABCD$ is cyclic quadrilateral

Therefore $\angle ACD + \angle OBD = 180^o$ (opposite angles of a cyclic quadrilateral are supplementary)

$60+ \angle ACO + \angle OBD = 180^o$

$\Rightarrow \angle ACO + \angle OBD = 120^o$

In $\triangle APB$

$\angle APB + \angle PBA + \angle BAP = 180^o$

$\angle APB + 120 = 180^o$

$\Rightarrow \angle APB = 60^o$ . Hence proved.

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Question 18: In the following figure, $ABCD$ is a cyclic quadrilateral in which $AD$ is parallel to $BC$ . If the bisector of $\angle A$ meets $BC$ at point $E$ and the given circle at point $F$ , prove that: (i) $EF = FC$ (ii) $BF = DF$

Answer:

Given: $ABCD$ is a cyclic quadrilateral

$AD \parallel BC$

$\angle BAE = \angle EAD \ (AE$ is the angle bisector) … … … (i)

(i) $\angle AEB = \angle EAD$ (alternate angles) … … … (ii)

In $\triangle ABE$

$\angle ABE = 180^o - \angle BAE - \angle AEB$

Using (i) and (ii) we get $\angle ABE = 180^o - 2 \angle AEB$

$\angle CEF = \angle AEB$ (vertically opposite angles)

$\angle ADC = 180^o - \angle ABC = 180^o -(180^o-2 \angle AEB)$

$\Rightarrow \angle ADC = 2 \angle AEB$

$\angle AFC = 180^o - \angle ADC = 180^o - 2 \angle AEB (since AFCD$ is a cyclic quadrilateral)

$\angle ECF = 180^o - (\angle AFC + \angle CEF) \\ = 180^o - (180^o-2 \angle AEB + \angle AEB) = \angle AEB$

Also $\angle AEB = \angle CEF \Rightarrow EF = EC$

(ii) $\widehat{BF}$ subtends $\angle BAF$ on circumference

$\widehat{OF}$ subtends $\angle FAD$ on circumference

Given $\angle BAF = \angle FAD$

Therefore $\widehat{BF} = \widehat{OF}$ (equal arcs subtends equal angles on circumference)

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Question 19: $ABCD$ is a cyclic quadrilateral. Sides $AB$ and $DC$ produced meet at point $E$ ; whereas sides $BC$ and $AD$ produced meet at point $F$ . If $\angle DCF: \angle F: \angle E = 3: 5 : 4$ , find the angles of the cyclic quadrilateral $ABCD$ .

Answer: d23

Given $DCF : F : E = 3:4:5$

In $\triangle CDF: \angle ADC = 3x+5x = 8x$

In $\triangle BCE: \angle ABC = 3x+4x = 7x$

$\angle ABC + \angle ADC = 180^o$

or $7x+8x= 180 \Rightarrow x = 12^o$

Therefore

$\angle DAC = 180 -12x = 180 - 144 = 36^o$

$\angle CBA = 7x = 7 \times 12 = 84^o$

$\angle DCB = 180-3x = 180-36 = 144^o$

$\angle ADC = 8x = 8 \times 12 = 96^o$

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18d15 Question 20: In the following figure shows a circle with $PR$ as its diameter. If $PQ =7\text{ cm }$ and $QR=3RS=6cm$ , find the perimeter of the cyclic quadrilateral $PQRS$ . [1992]

Answer:

Given: $PQ = 7 \ cm, QR = 3RS = 6 \text{ cm }$

$PR = \sqrt{7^2+6^2} = \sqrt{85}$

$SR = \sqrt{PR^2-PS^2} = \sqrt{85-4} =9$

Therefore the perimeter of $PQRS = 2+9+6+7 = 24 \text{ cm }$

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18d16 Question 21: In the following figure, $AB$ is the diameter of a circle with center $O$ . If $chord AC = chord AD$ , prove that: (i) $arc BC = arc DB$ (ii) $AB$ is bisector of $\angle CAD$ . Further, if the length of $arc AC = 2 \times arc BC$ , find : (a) $\angle BAC$ (b) $\angle ABC$ .

Answer:

Given: $AC = AD$

Consider $\triangle ABC$ and $\triangle ABD$

$AB$ is common

$\angle ACB = \angle ADB = 90^o$ (angles in semi circle)

$AC = AD$

Therefore $\triangle ABC \cong \triangle ABD$

(i) $BC = BD$ (corresponding parts of congruent triangles)

(ii) $\angle CAB = \angle BAD$ (equal chords subtend equal angles on the circumference of the same circle)

Therefore $AB$ bisects $\angle CAD$

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Question 22: In cyclic quadrilateral $ABCD$ ; $AD = BC, \angle BAC = 30^o$ and $\angle CBD = 70^o$ ; find: (i) $\angle BCD$ (ii) $\angle BCA$ (iii) $\angle ABC$ (iv) $\angle ADC$

Answer:

Given $AD = BC, \angle BAC = 30^o$ and $\angle CBD = 70^o$

$\angle BDC = \angle BAC = 30^o$ (angles in the same segment)

$\angle DAC = \angle DBC = 70^o$ (angles in the same segment)

$\angle BCD + 30^o + 70^o = 180^o \Rightarrow \angle BCD = 80^o$ (opposite angles in a cyclic quadrilateral are supplementary)

$BC = AD \Rightarrow \angle BDC = \angle ADC = 30^o$

Therefore $\angle BCA = 50^o$

$70^o + 30^o + 30^o +\angle BDA = 180 \Rightarrow \angle BDA = 50^o$

$50^o+ 30^o + 70^o+\angle ABD = 180 \Rightarrow \angle ABD = 30^o$

$\angle ABC = 100^o, \angle BAD = 100^o, \angle BCD = 80^o \ and \ \angle CDA = 80^o$

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18d17 Question 23: In the given figure $\angle ACE = 43^o$ and $\angle CAF = 62^o$ ; find the values of $a, b$ and $c$ . [2007]

Answer:

Given $\angle ACE = 43^o$ and $\angle CAF = 62^o$

$\angle BAE + \angle BDE = 180^o$ (ABDE is a cyclic quadrilateral)

$\angle BDE = 118^o$

$\angle c + 118^o = 180^o$ (straight line) $\Rightarrow c = 62^o$

In $\triangle ACE$ :

$62^o+43^o+\angle CEA = 180^o \Rightarrow \angle CEA = 75^o$

Therefore $\angle DEF = 180^o - 75^o = 105^o$

Therefore $\angle a + 75^o = 180^o \Rightarrow \angle a = 105^o \ (ABDE$ is a cyclic quadrilateral)

In $\triangle DEF$ :

$62^o+ 105^o + \angle b = 180^o \Rightarrow \angle b = 13^o$

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18d18 Question 24: In the given figure, $AB \parallel DC, \angle BCE = 80^o$ and $\angle BAC = 25^o$ . Find (i) $\angle CAD$ (ii) $\angle CBD$ (iii) $\angle ADC$

Answer:

Given $\angle = 80^o$ and $\angle BAC = 25^o$ . Also $AB \parallel DC$

Therefore $\angle CAB = \angle ACD$ (alternate angles)

Hence $\angle ACD = 25^o$

Therefore $\angle ACB = 180^o - 80^o - 25^o = 75^o$

and $\angle DCB = 25^o + 75^o = 100^o$

$\angle DAB = 180^o - 100^o = 80^o$

Therefore $\angle DAC = 80^o - 25^o = 55^o$

In $\triangle ABC$ :

$\angle ABC = 180^o - 25^o - 75^o = 80^o$

Therefore $\angle ADC = 180^o = 80^o = 100^o$

$\angle CBD = \angle DAC = 55^o$ (angles in the same segment)

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Question 25: $ABCD$ is a cyclic quadrilateral of a circle center $O$ such that $AB$ is a diameter of the circle and the length of the chord $CD$ is equal to the radius of the circle. If $AD$ and $BC$ produced meet at $P$ , show that $\angle APB = 60^o$ .

Answer:

Given $CD = OC = Radius$

In $\triangle OCD$ :

$OD = OC = CD \Rightarrow \angle ODC \\ = \angle OCD = \angle DOC = 60^o$

In $\triangle AOD, OA = OD$ (Radius of the same circle)

Let $\angle OAD = \angle ADO = x$

Therefore $\angle AOD = 180^o -2x$ … … … (i)

In $\triangle COB, OC = OB$ (Radius of the same circle)

Let $\angle OBC = \angle OCB = y$

Therefore $\angle COB = 180^o - 2y$ … … … (ii)

Now, $180^o - 2x + 60^o + 180^o - 2y = 180^o$ (straight line angle)

$\Rightarrow x + y = 120^o$

In $\triangle APB$ :

$x + y + \angle APB = 180^o \Rightarrow \angle APB = 180^o - 120^o = 60^o$

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18d19 Question 26: In the figure, given alongside, $CP$ bisects $\angle ACB$ . Show that $DP$ bisects $\angle ADB$

Answer:

Given $CP$ bisects $\angle ACB$

Therefore $\angle ACP = \angle PCB$

$\angle ACP = \angle ADP$ (angles in the same segment)

$\angle PCB = \angle PDB$ (angles in the same segment)

Therefore $\angle ADP = \angle PDB$

Hence $DP \ bisects \ \angle ADB$

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18d20 Question 27: In the figure shown, $AD = BC, \angle BAC = 30^o$ and $\angle CBD=70^o$ . Find (i) $\angle BCD$ (ii) $\angle BCA$ (iii) $\angle ABC$ (iv) $\angle ADB$

Answer:

From $chord \ CD: \angle DBC = \angle DAC$ ( angles in the same segment)

$\Rightarrow \angle DAC = 70^o$

From $chord \ CB: \angle CAB = \angle CDB$ (angles in the same segment)

$\Rightarrow \angle CDB = 30^o$

In $\triangle \angle DCB: 70 + 30 + \angle DCB = 180 \Rightarrow \angle DCB = 80^o$

Since $AD = BC: \angle DCA = \angle CAB \Rightarrow \angle DCA = 30^o$

In $\triangle ADC: 70 + \angle ADB + 30 + 30 = 180 \Rightarrow \angle ADB = 50^o$

$ABCD$ is a cyclic quadrilateral. $\therefore 50 + 30 + \angle DBA + 70 = 180 \Rightarrow DBA = 30^o$

Hence

(i) $\angle BCD = 80^o$

(ii) $\angle BCA = 50^o$

(iii) $\angle ABC = 100^o$

(iv) $\angle ADB = 50^o$

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18d21 Question 28: In the figure, given below, $AB$ and $CD$ are two parallel chords and $O$ is the center. If the radius of the circle is $15\text{ cm }$ , find the distance $MN$ between the two chords of lengths $24\text{ cm }$ and $18\text{ cm }$ respectively. [2010]