18c10Question 1: In the given diagram, chord AB = chord BC and AB = AC.

(i) What is the relation between \widehat{AB} and \widehat{BC} ?

(ii) What is the relation between \angle AOB and \angle BOC ?

(iii) If \widehat{AD}  is greater than \widehat{ABC} , then what is the relation between the chord AD and AC ?

(iv) If \angle AOB = 50^o , find the measure of  \angle BAC .

c1Answer:

Given: AB = BC

(i) \widehat{AB}=\widehat{BC}  (Since equal chords subtend equal arcs)

(ii) \angle AOB = \angle BOC  (equal arcs will subtend equal angles at the center)

(iii) If \widehat{AD} > \widehat{ABC} 

Therefore AD > AC 

(iv) Given \angle AOB = 50^o  

\angle ABC = \frac{1}{2} Reflex \ \angle AOC = \frac{1}{2} . 260 = 130^o 

Therefore \angle BAC = \frac{1}{2} (180-130) = 25^o 

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Question 2: In \triangle ABC , the \perp from vertices A and B on their opposite sides meet (when produced) the circumference of the triangle at points D and E respectively. Prove that: \widehat{CD} = \widehat{CE} 

c2.jpgAnswer:

Consider \triangle AOM  & \triangle BPN 

\angle AMO = \angle BNO = 90^o  (given)

\angle NBO = \angle AOM  (vertically opposite angles)

\therefore \triangle AOM \sim \triangle BPN  (AAA postulate)

Therefore \angle NBO = \angle MAO 

Therefore Arcs that subtend equal angle at circumference are equal

Therefore \widehat{CD} = \widehat{CE}  

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Question 3: In a cyclic trapezium, prove that non parallel sides are equal and the diagonals are also equals.

c3Answer:

ABCD   is a cyclic trapezium. AB \parallel DC  

Chord AD   subtends \angle ABD   at circumference

Chord BC    subtends \angle BDC   at circumference

But \angle ABD = \angle BDC   (vertically opposite angles)

Therefore Chord \ AD = Chord \ BC  

\Rightarrow AD = BC  

Now in \triangle ADC   & \triangle BDC  

DC   is common

\angle CAD = \angle DBC   (angles in the same segment)

AD = BC   (proved above)

\therefore \triangle ADC \cong \triangle BDC   (SAS axiom)

\therefore AC = DB  

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18c9Question 4: In the given diagram, AD is the diameter of the circle with center O . Chords AB,=BC = CD .  If \angle DEF = 110^o , find (i) \angle AEF (ii) \angle FAB

Answer:

Given AB = BC = CD

(i) AOD is the diameter

\angle AED = 90^o (angle in semicircle)

\angle AEF + \angle AED = \angle FED c4.jpg

\angle AEF = 110-90=20^o

(ii) Since AB = BC = CD (given)

\angle AOB = \angle BOC = \angle COD (equal arcs subtend equal angles at center)

We know \angle AOD = 180^o

\Rightarrow \angle AOB = \angle BOC = \angle COD = 60^o

In \triangle AOB

OA = OB (radius of the same circle)

Therefore \angle OAB = \angle OBA

\angle OAB + \angle OBA = 180-\angle AOB = 180-60 = 120^o

\Rightarrow \angle OAB = \angle OBA = 60^o

ADEF is a cyclic quadrilateral

\angle DEF + \angle DAF = 180^o (opposite angles are supplementary)

\angle DAF = 180-110=70^o

Now \angle FAB = \angle DAF + \angle OAB = 70+60=130^o

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18c8Question 5: In the given diagram, if \widehat{AB} = \widehat{CD}  of circle with center O , prove that quadrilateral ABCD is an isosceles trapezium.

Answer:

Given \widehat{AB}=\widehat{CD} 

\angle ADB = \angle DBC 

(Equal arcs subtend equal angles at the circumference)c5

\Rightarrow AD \parallel BC  (alternate angles are equal)

Therefore ABCD  is a trapezium

Also Chords AB = Chard CD \ (since \ \widehat{AB}=\widehat{CD} )

Therefore ABCE  is an Isosceles trapezium.

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Question 6:  In a given figure \triangle ABC is an isosceles triangle and O is the center of the circumcircle. Prove AP bisects \angle BPC .

Answer:

Given AB = AC  (Since \triangle ABC is an isosceles triangle)c6

\angle APB = \angle APC  (equal chords subtend equal angles on the circumference)

Therefore AP  bisects \angle BPC .

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Question 7: If two sides of a cyclic quadrilateral are parallel, prove that:

(i) its other two sides are equal

(ii) its diagonals are equal

Answer:c7

(i) Given AB \parallel DC 

Therefore \angle BAC = \angle DCA  (alternate angles)

Chord AD  subtends \angle DCA  on circumference.

Chord BC  subtends \angle CAB  on circumference.

Therefore AD = BC  (since equal chords subtend equal angles on the circumference)

(ii) Consider \triangle ADB \ \ \& \ \  \triangle ABC 

AB  is common

AD = CB  (proved above)

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

Therefore \triangle ADC \cong \triangle ABC  (by SAS axiom)

Therefore AC = DB  (corresponding parts of congruent triangles are equal)

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18c6Question 8: In the given diagram, circle with center O . PQ=QR=RS and \angle PTS=75^o . Calculate:

(i) \angle POS

(ii) \angle QOR

(iii) \angle PQR

Answer:

Given PQ = QR = RS c8.jpg

\Rightarrow \angle POQ = \angle QOR = \angle ROS (Equal chords subtend equal angles at the center of a circle)

\angle POS = 2 \ \angle PTS (angle subtended at the center is twice that of the one subtended on the circumference)

\angle POQ = \angle QOR = \angle ROS = \frac{150}{3} = 50^o

In \triangle OPQ ,

OP = OQ (radius of the same circle)

\angle OPQ = \angle OQP

Therefore \angle OPQ + \angle OQP + \angle POQ = 180^o

\angle OPQ + \angle OQP = 180-50=130^o

2 \angle OPQ = 130

\angle OPQ = 65^o

Similarly, in \triangle OQR, \angle OQR = \angle ORQ = 65^o

and in \triangle ORS, \angle ORS = \angle OSR = 65^o

Therefore

(i) \angle POS = 150^o

(ii) \angle QOR = 50^o

(iii) \angle PQR = \angle PQO + \angle OQR = 65+65 = 130^o

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18c5Question 9: In the given diagram, AB is the side of regular six-side polygon and AC is a side of a regular eight-sides polygon inscribed in a circle with center O . Calculate:

(i) \angle AOB

(ii) \angle ACB

(iii) \angle ABC

Answer:

(i) Since AB is the side of a regular hexagon

\angle AOB = 60^o

(ii) \widehat{AB} subtends \angle ACB at the circumference and \angle AOB at the center

We know \angle ACB = \frac{1}{2} \angle AOB = \frac{60}{2} = 30^o

(iii) Since AC is the side of a regular octagon

\angle AOC = \frac{360}{8} = 45^o

\widehat{AC} subtends \angle ABC at the circumference and \angle AOC at the center

Therefore \angle ABC = \frac{1}{2} \angle AOC = \frac{45}{2} = 22.5^o

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Question 10: In a regular pentagon ABCDE inscribed in a circle, find the ratio of the \angle EDA : \angle ADC [1990]

Answer:c10.jpg

Consider \widehat{AE}

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

Therefore \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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18c4Question 11: 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 c11.jpg

\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)

\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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c12.jpgQuestion 12: 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 

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

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

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

(iii) Similarly, \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 13: In the diagram given below, O is the center of the circle. AB is the side of regular pentagon and AC is the side of a regular hexagon. Find the angles of the \triangle ABC .

Answer:

AB is the side of a regular pentagon

\Rightarrow \angle AOB = \frac{360}{5} = 72^o c14

AC is a side of regular hexagon

\Rightarrow \angle AOC = \frac{360}{6} = 60^o

Now \angle AOB + \angle AOC + reflex \angle BOC = 360^o

\angle BOC = 360 - 72-60 = 228^o

\widehat{BC} subtends \angle BOC at the center and \angle BAC on the circumference

\angle BAC = \frac{1}{2} \angle BOC = \frac{228}{2} = 114^o

Similarly \angle ABC = \frac{1}{2} \angle AOC = \frac{60}{2} = 30^o

\angle ACB = \frac{1}{2} \angle AOB = \frac{1}{2} \times 72 = 36^o

Therefore \angle ABC = 30^o, \angle ACB = 36^o and \angle BAC = 114^o

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18c1Question 14: 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:c15.jpg

BD is a side of a regular hexagon

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

DC is a side of regular pentagon

\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

\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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