Class 10: Circles – ICSE Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: In the figure, given below, $AB$ and $CD$ are two parallel chords and $O$ is the center. If the radius of the circle is $15\ cm$ , find the distance $MN$ between the two chords of lengths $24\ cm$ and $18\ cm$ respectively. [2010]

Answer:

We know that perpendicular drawn from the center of the circle will bisect the chord.

Therefore

$MO^2 = 15^2-12^2 = 225 - 144 = 81 \Rightarrow MO = 81 \ cm$

$ON^2 = 15^2-9^2 = 225 - 81 = 12 \Rightarrow ON = 12 \ cm$

Hence $MN = 9 + 12 = 21 \ cm$

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Question 2: 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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Question 3: In the following figure shows a circle with $PR$ as its diameter. If $PQ =7\ cm$ and $QR=3RS=6cm$ , find the perimeter of the cyclic quadrilateral $PQRS$ . [1992]

Answer:

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

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

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

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

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Question 4: 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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Question 5: 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 6: In the given diagram, $BD$ is the side of a regular hexagon, $DC$ is the side of a regular pentagon and $AD$ is a diameter. Calculate:

(i) $\angle ADC$

(ii) $\angle BDA$

(iii) $\angle ABC$

(iv) $\angle AEC$ [1984]

Answer:

$BD$ is a side of a regular hexagon

$\displaystyle \text{(i) } \angle BOD = \frac{360}{6} = 60^o$

$DC$ is a side of regular pentagon

$\displaystyle \angle DOC = \frac{360}{5} = 72^o$

(ii) In $\triangle BOD, \angle BOD = 60^o$

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

$\therefore \angle OBD = \angle ODB = 60^o$

(iii) In $\triangle OCD, \angle COD = 72^o \ and \ OC = OD$

$\displaystyle \angle ODC = \frac{1}{2} (180-72) = 54^o$

(iv) In cyclic quadrilateral $AECD$

$\angle AEC + \angle ADC = 180^o$ (opposite angles of a cyclic quadrilateral are supplementary)

$\angle AEC = 180-54 = 126^o$

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Question 7: In the diagram, $O$ is the center of the circle and the length of $\widehat{AB} = 2 \times \widehat{BC}$ . If $\angle AOB=108^o$ find:

(i) $\angle CAB$

(ii) $\angle ADB$ [1996]

Answer:

Given $\widehat{AB} = 2 \times \widehat{BC}, \angle AOB = 108^o$

(i) $\angle AOB = 2 \angle BOC$

$\displaystyle \Rightarrow \angle BOC = \frac{108}{2} = 54$

(ii) $\displaystyle \angle BAC = \frac{1}{2} \angle BOC$ (angle subtended at the center is twice that subtended at the circumference by a chord)

$\displaystyle \angle BAC = \frac{1}{2} \times 54 = 27^o$

(iii) Similarly, $\displaystyle \angle ACB = \frac{1}{2} \angle AOB = \frac{108}{2} = 54$

In cyclic quadrilateral $ADBC$

$\angle ADB + \angle ACB = 180^o$

$\Rightarrow \angle ADB = 180-54=126^o$

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Question 8: In the given diagram, $AB=BC=CD$ and $\angle ABC = 132^o$ . Find:

(i) $\angle AEB$

(ii) $\angle AED$

(iii) $\angle COD$ [1993]

Answer:

Given $AB = BC = CD, ABC = 32^o$

$\Rightarrow \angle AOB = \angle BOC = \angle COD$ (equal arcs subtend equal angles at the center of a circle)

(i) In cyclic quadrilateral $ABCE$

$\angle ABC + \angle AEC = 180^o$ (opposite angles in a cyclic quadrilateral are supplementary)

$132+ \angle AEC = 180 \Rightarrow \angle AEC = 48^o$

Since $AB = BC \Rightarrow \angle AEB = \angle BEC$ (equal chords subtend equal angles on the circumference)

$\displaystyle \therefore \angle AEB = \frac{1}{2} \angle AEC = 24$

(ii) Since $AB = BC = CD$

$\angle AEB = \angle BEC = \angle CED$

$\angle AED = 24+24+24=72$

(iii) $\widehat{CD}$ subtends $\angle COD$ at the center and $\angle CED$ on the circumference

$\therefore \angle COD = 2 \angle CED = 2 \times 24 = 48^o$

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Question 9: In a regular pentagon $ABCDE$ inscribed in a circle, find the ratio of the $\angle EDA : \angle ADC$ [1990]

Answer:

Consider $\widehat{AE}$

It subtends $\angle AOE$ at the center and $\angle ADE$ at the circumference

Therefore $\displaystyle \angle ADE = \frac{1}{2} \angle AOE = \frac{1}{2} \times \frac{360}{5} = 36^o$

Similarly, for $\widehat{BC}$ we have $\angle BDC = 36^o$

$\angle ADC = \angle ADB + \angle BDC = 36+36=72^o$

$\angle ADE : \angle ADC = 36 :72 = 1:2$

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Question 10: In the given figure, $\angle BAD = 65^o, \angle ABD = 70^o$ and $\angle BDC = 45^o$ .

(i) Prove that $AC$ is a diameter of the circle.

(ii) find $\angle ACB$ [2013]

Answer:

Given $\angle BAD = 65^o, \angle ABD = 70^o$ and $\angle BDC = 45^o$ .

(i) In $\triangle ABD$

$65^o+70^o+ \angle ADB = 180^o$

$\Rightarrow \angle ADB = 180^o-135^o = 45^o$

Therefore $\angle ADC = 45^o+45^o = 90^o$

Therefore $AC$ is the diameter (Theorem 11)

(ii) $\angle ACB = \angle ADB$ (angles in the same segment)

Therefore $\angle ACB = 45^o$ since $\angle ADB = 45^o$

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Question 11: In each of the following figures, $O$ is the center of the circle. Find the values of $a, b, c \ and \ d$ . [2007]

(i)	(ii)
(iii)	(iv)

Answer:

(i) $BD$ is the diameter

$\therefore \angle DAB = 90^o$

$\therefore \angle ADB = 180^o-90^o-3^o5=55^o$

Since $\angle ADB = \angle ACB = 55^o$ (angle in same segment) Therefore $a = 55^o$

(ii) $\angle ADB = \angle ACB$ (angle in same segment)

In $\triangle ECB$ ,

$\angle BEC = 180^o-120^o = 60^o$

$\therefore \angle ACB = 180^o - 60^o - 25^o = 95^o$

$\therefore \angle ADB = 95^o= b$

(iii) In $\triangle AOB$

$AO = OB =$ radius

$2 \angle ACB = \angle AOB$

$\therefore \angle AOB = 100^o$

$\therefore 2c+ 100^o= 180^o \Rightarrow c = 40^o$

(iv) Since $AB$ is the diameter

$\therefore \angle BAP = 180^o-90^o-45^o = 45^o$

$\angle PAB = \angle PCB = 45^o \Rightarrow d = 45^o$ (angles in the same segment)

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Question 12: In the given figure $AB = AC = CD$ and $\angle ADC = 38^o$ , calculate (i) $\angle ABC$ (ii) $\angle BEC$ [1995]

Answer:

(i) $AC = CD$

$\angle CAD = \angle CDA = 38^o$

$\angle ACD = 180^o-38^o - 38^o = 104^o$

$\angle ACB = 180^o-104^o = 76^o$

$AB = AC$

$\angle ABC = \angle ACB = 76^o$

(ii) $\angle BAC = 180^o-76^o-76^o = 38^o$

$\angle BAC = \angle BEC = 38^o$ (angles in the same segment)

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Question 13: In the given figure $AC$ is the diameter of the circle with center $O$ . Chord $BD \perp AC$ . Write down the angles $p, q$ and $r$ in terms of $x$ . [1996]

Answer:

$\angle AOB = 2 \angle ACB = 2 \angle ADB$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\displaystyle x = 2q \Rightarrow q=\frac{x}{2}$

$\displaystyle \angle ADB = \frac{x}{2}$

$\angle ADC = 90^o$ (angles in a semicircle)

$\displaystyle r +\frac{x}{2} = 90^o \Rightarrow r = 90-\frac{x}{2}$

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

$\displaystyle p = 90^o-q \Rightarrow p = 90^o-\frac{x}{2}$

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Question 14: In the given figure $AC$ is the diameter of the circle with center $O$ . $CD \parallel BE$ . $\angle AOB = 80^o$ and $\angle ACE = 110^o$ . Calculate (i) $\angle BEC$ (ii) $\angle BCD$ (iii) $\angle CED$ [1998]

Answer:

(i) $\angle BOC = 180^o-80^o=100^o$ (Since $AC$ is a straight line)

$\angle BOC = 2 \angle BEC$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\displaystyle \Rightarrow \angle BEC = \frac{100}{2} = 50^o$

(ii) $DC \parallel EB$

$\angle DCE = \angle BEC = 50^o$

$\angle AOB = 80^o$ (given)

$\displaystyle \Rightarrow \angle ACB = \frac{1}{2} \angle AOB = 40^o$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\angle BCD = \angle ACB + \angle ACE + \angle DCE = 40^o+10^o+50^o = 100^o$

(iii) $\angle BED = 180^o-\angle BCD = 180^o-100^o = 80^o$ (cyclic quadrilateral)

$\angle CED + 50^o = 80^o \Rightarrow \angle CED = 30^o$

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Question 15: In the given figure, $AE$ is the diameter of the circle. Write down the numerical value of $\angle ABC + \angle CDE$ . Give reasons for your answer. [1998]

Answer:

$\displaystyle \angle AOC = \frac{180}{2} = 90^o$

$\angle AOC = 2 \angle AEC$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

$\displaystyle \Rightarrow \angle AEC = \frac{90}{2} = 45^o$

$ABCE$ is a cyclic quadrilateral

$\therefore \angle ABC + \angle AEC = 180 ^o$

$\Rightarrow \angle ABC = 180-45 = 135^o$

Similarly, $\angle CDE = 135^o$

$\therefore \angle ABC + \angle CDE = 135^o + 135^o = 270^o$

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Question 16: In the given figure $AOC$ is the diameter and $AC \parallel ED$ . If $\angle CBE = 64^o$ , calculate $\angle DEC$ . [1991]

Answer:

$\angle ABC = 90^o$ (angle in a semi circle)

$\angle ABE = 90^o-64^o = 26^o$

$\angle ABE = \angle ACE = 26^o$ (angles in the same segment)

$AC \parallel ED$

$\therefore \angle DEC = \angle ACE = 26^o$ (alternate angles)

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Question 17: Use the given figure below to find (i) $\angle BAD$ (ii) $\angle DQB$ [1987]

Answer:

(i) In $\triangle ADP$

$\angle BAD = 180^o - 85^o - 40^o = 55^o$

(ii) $\angle ABC = 180^o - \angle ADC = 180^o - 85^o = 95^o$

$\angle AQB = 180^o - 95^o - 55^o = 30^o$

$\Rightarrow \angle DQB = 30^o$

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Question 18: In the given figure $AOB$ is the diameter and $DC \parallel AB$ . If $\angle CAB = x^o$ , find in terms of $x$ , the values of: (i) $\angle COB$ (ii) $\angle DOC$ (iii) $\angle DAC$ (iv) $\angle ADC$ [1991]

Answer:

(i) $\angle OCB = 2 \angle CAB = 2x$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

(ii) $\angle OCD = \angle COB = 2 x$ (alternate angles)

In $\triangle OCD$

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

$\angle ODC = \angle OCD = 2x$

$\angle DOC = 180^o-2x-2x = 180^o-4x$

(iii) $\displaystyle \angle DAC = \frac{1}{2} \angle DOC = \frac{1}{2} (180^o-4x) = 90^o-2x$ (angles subtended by an chord on the center is double that subtended by the same chord on the circumference)

(iv) $DC \parallel AO$ (given)

$\therefore \angle ACD = \angle OAC = x$ (alternate angles)

$\therefore \angle ADC = 180^o - \angle DAC-\angle ACD = 180^o-(90^o-2x) - x = 90^o+x$

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Question 19: In the figure $AB$ is the diameter of the circle with center $O$ . $\angle BCD = 130^o$ . Find (i) $\angle DAB$ (ii) $\angle DBA$ [2012]

Answer:

(i) $\angle DAB = 180^o-\angle DCB = 180^o-130^o = 50^o \ (ABCD$ is a cyclic quadrilateral)

(ii) In $\triangle ADB$

$\angle DAB + \angle ADB + \angle DBA = 180^o$

$\Rightarrow \angle DBA = 180^o-50^o-90^o = 140^o$

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Question 20: In the given figure $PQ$ is the diameter of the circle whose center is $O$ . Given $\angle ROS=42^o$ , calculate $\angle RTS$ . [1992]

Answer:

Join P and S as shown in the diagram.

$\angle PSQ = 90^o$ (angle in a semi circle)

$\displaystyle \angle SPR = \frac{1}{2} \angle ROS$

$\displaystyle \Rightarrow \angle SPT = \frac{1}{2} \times 42^o = 21^o$

In $\triangle PST$

$\angle PST = 90^o-\angle SPT$

$\Rightarrow \angle RTS = 90^o-21^o = 69^o$

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Question 21: In the given figure $PQ$ is the diameter. Chord $SR \parallel PQ$ . Given the $\angle PQR = 58^o$ , calculate (i) $\angle RPQ$ (ii) $\angle STP$ [1989]

Answer:

Join P and R as shown in the diagram.

(i) $\angle PRQ = 90^o$ (angles in the semi circle)

$\therefore \angle RPQ = 90^o- \angle PQR = 90^o-58^o = 32^o$

(ii) $SP \parallel PQ$ (given)

$\angle PSR = \angle RPQ = 32^o$ (alternate angles)

$\angle SPT = 180^o-\angle PSR = 180^o-32^o = 148^o$ ( $PTSR$ is a cyclic quadrilateral)

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Question 22: $AB$ is the diameter of the circle with center $O$ . $OD \parallel BC$ and $\angle AOD = 60^o$ . Calculate the numerical values of (i) $\angle ABD$ (ii) $\angle DBC$ (iii) $\angle ADC$ [1987]

Answer:

Join B and D as shown in the diagram.

(i) $\displaystyle \angle ABD = \frac{1}{2} \angle AOD = \frac{1}{2} \times 60^o = 30^o$ (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)

(ii) $\angle BDA = 90^o$ (angle in semi circle)

Since $\angle OAD = 60^o \Rightarrow \triangle OAD$ is equilateral

$\angle ODB = 90^o-ODA = 90^o-60^o = 30^o$

Since $OD \parallel BC$

$\angle DBC = \angle ODB = 30^o$ (alternate angles)

(ii) $\angle ABC = \angle ABD + \angle DBC = 30^o +30^o = 60^o$

$\angle ADC = 180^o -\angle ABC = 180^o- 60^o = 120^o$ ( $ABCD$ is a cyclic quadrilateral)

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Question 23: In the given figure, the center $O$ of the smaller circle lies on the circumference of the bigger circle. If $\angle APB = 75^o$ and $\angle BCD = 40^o$ , find: (i) $\angle AOB$ (ii) $\angle ACB$ (iii) $\angle ABD$ (iv) $\angle ADB$ . [1984]

Answer:

Join A and B as shown in the diagram.

(i) $\angle AOB = 2 \angle APB = 2 \times 75^o = 150^o$ (angle subtended by an chord on the center is double that subtended by the same chord on the circumference)

(ii) $\angle ACB = 180^o-\angle AOB^o = 180^o-150^o = 30^o$ ( $AOBC$ is a cyclic quadrilateral)

(iii) $\angle ABD = 180^o - \angle ACD = 180^o-(40^o+30^o) = 110^o$ ( $ABDC$ is a cyclic quadrilateral)

(iv) $\angle ADB = 180^o - \angle AOB = 180^o-150^o= 30^o$ ( $ADBO$ is a cyclic quadrilateral)

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Question 24: $O$ is the center of the circle of radius $10 \ cm$ . $P$ is any point in the circle such that $OP=6 \ cm$ . $A$ is the point travelling along the circumference, $x$ is the distance from $A to P$ . What are the least and the greatest values of $x \ in \ cm$ . What is the position of the points $O, P \ and \ A$ at these values . [1992]

Answer:

When $P$ is on $OA$ , the least distance of $x = 4 \ cm$

When $P$ is on extended $OA$ , the greatest distance of $x = 16 \ cm$

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Question 25: In the given figure, $O$ is the center of the circle. $AB$ and $CD$ are two chords of circle. $OM \perp AB$ and $ON \perp CD$ . $AB = 24 \ cm, OM = 5 \ cm$ , $ON = 12 \ cm$ . Find the

(i) radius of the circle

(ii) length of chord $CD$ [2014]

Answer:

$AB = 24 \ cm$ , $OM = 5 \ cm$ , $ON =12 \ cm$

$OM \perp AB$

$\Rightarrow AM = BM$

Therefore $AM = MB = 12 \ cm$

(i) Consider $\triangle AOM$

$AO^2 = AM^2 +OM^2 = 12^2+5^2 = 169$

$\therefore AO = 13 \ cm =$ Radius of the circle.

(ii) Now consider $\triangle CNO$

$CO^2 = CN^2+ON^2$

$\Rightarrow CN^2 = CO^2-ON^2 = 13^2 - 12^2 = 169 - 144 = 25$

$\therefore CN= 5 \ cm$

$\\$

2011-1 Question 26: In the given figure $O$ is the center of the circle Tangent of $A$ and $B$ meet at $C$ if $\angle AOC = 30^o$ , find (i) $\angle BCO$ (ii) $\angle AOB$ (iii) $\angle APB$ [2011]

Answer:

Consider $\triangle AOC$ and $\triangle BOC$

$\angle OAC = \angle OBC = 90^o$

$OC$ is common

$AC = BC$ (two tangents drawn from a point on a circle are of equal lengths)

Therefore $\triangle AOC \cong \triangle BOC$

(i) $\angle ACO = \angle BCO = 30^o$

(ii) $\angle AOC = 180^o - 90^o - 30^o = 60^o$

$\angle BOC = 180^o - 90^o - 30^o = 60^o$

$\therefore \angle AOB = \angle AOC + \angle BOC = 120^o$

(iii) $\angle AOB = 2 \angle APB$ (chord subtends twice the angle at the center than that it subtends on the circumference)

$\Rightarrow \angle AOB = 2 \angle APB$

$\Rightarrow \angle APB = 60^o$

$\\$

Question 27: In the following figure O is the center of the circle and AB is a tangent to it at point B. $\angle BDC = 65^o$ . Find $\angle BAO$ . [2010]