Question 21: Prove that: (i) the parallelogram inscribed in a circle is a rectangle (ii) the rhombus, inscribed in a circle is a square.

Answer:

(i) Let ABCD be a parallelogram inscribed in the circle.

\therefore \angle BAD = \angle BCD (opposite angles of a parallelogram are equal)

\angle BAD + \angle BCD = 180^o

\Rightarrow \angle BAD = \angle BCD = 90^o

Similarly, \angle ABC = \angle ADC = 90^o

Therefore ABCD is a rectangle

(ii) ABCD is a rhombus (given) i.e. all four sides are equal.

\angle BAD = \angle BCD = 90^o

Similarly, \angle ABC = \angle ADC = 90^o

Therefore ABCD is a square

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c45Question 22: In the following figure AB = AC . Prove that DEBC is an isosceles trapezium.

Answer:

AB = AC (given)

\angle ABC = \angle ACB

\angle CBA+\angle CED = 180^o (BCED is a cyclic quadrialteral)

\angle ACB + \angle CED = 180^o

DE \parallel BC … … … … (i)

\angle ADE = \angle ABC (corresponding angles)

\angle AED = \angle ACB

Hence \angle ADE = \angle AED

\Rightarrow AD = AE

AB-AD = AC-AE

\Rightarrow BD = EC … … … … (ii)

Hence because of (i) and (ii), ABCD is an isosceles trapezium.

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Question 23: Two circles intersect at P and Q . Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

c46Answer:

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

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

Therefore \angle AQB = 180^o

Therefore A, Q \ and \  B are collinear.

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Question 24: ABCD is a quadrilateral inscribed in a circle. having \angle A = 60^o . O is the center of the circle. Show that: \angle OBD+\angle ODB= \angle CBD+\angle CDB .

c47Answer:

\angle BAD + \angle BCD = 180^o (cyclic quadrilateral)

\angle BCD = 180^o-60^o=120^o

\angle BOD = 2 \angle BAD

\angle CBD + \angle CDB = 180^o-120^o = 60^o (sum of the angles in a triangle is 180)

Similarly \angle OBD + \angle ODB = 180^o-120^o=60^o

Therefore \angle OBD + \angle ODB = \angle CBD + \angle CDB

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c44Question 25: The figure given below shows the circle with center O . Given \angle AOC = a and \angle ABC=b . (i) Find the relationship between a \ and  \ b . (ii) Find \angle OAB if OABC is a parallelogram.

Answer:

(i) \angle AOC = a and \angle CBA = b (given)

\angle ABC = \frac{1}{2} Reflex (\angle COA)

b = \frac{1}{2} (360^o-a)

2b = 360^o-a \ or \ a+2b=360^o

(ii) If OABC is a parallelogram

Therefore a = b (opposite angles are equal)

\therefore 3b = 360^o \Rightarrow b = 120^o

\therefore a = 120^o

\angle ACB = \angle BAC

\therefore \angle BAC = 30^o

OC = OA (radius of the same circle)

\therefore \angle OCA = \angle OAC 

\therefore \angle OAC = 30^o

Hence \angle OAB = 30^o+30^o=60^o

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Question 26: Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the center O is equal to twice the \angle APC .

c48Answer:

\angle AOC = 2 \angle ADC

Similarly, \angle BOD = 2 \angle BAD

Adding the two

\angle AOC + \angle BOD = 2(\angle ADC+\angle BAD) … … … … (i)

In  \triangle PAD

180^o-\angle APC + \angle PAD + \angle ADC = 180^o

\therefore \angle PAD + \angle ADC = \angle APC … … … … (ii)

Using (1) and (ii)

\angle AOC + \angle BOD = 2(\angle BAD + \angle ADC) = 2 \angle APC

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c43Question 27: In the given figure RS is a diameter of the circle. NM \parallel RS  and \angle MRS= 29^o . Find (i) \angle RNM (ii) \angle NRM

Answer:

(i) \angle RMS = 90^o (angle in a semi circle)

\therefore \angle RMS = 180^o-90^o-29^o = 61^o

\therefore \angle RNM + 61^o = 180^o \Rightarrow \angle RNM = 119^o

c411.jpg(ii) Given NM \parallel RS

\angle NMR = \angle MRS = 29^o (alternate angles)

\angle NMS  = 90^o+29^o = 119^o

\angle NRS + \angle NMS = 180^o

\angle NRM+\angle MRS + \angle NMR + \angle RMS = 180^o

\angle NRM +29^o+29^o+90^o = 180^o

\Rightarrow \angle NRM = 180^o-148^o = 32^o

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c42Question 28: In the given figure, AB \parallel CD and O is the center of the circle. If \angle ADC = 25^o , find the  \angle AEB . Give reasons.

Answer:

\angle CAD = 90^o and \angle CBD = 90^o (angles in a semi circle)

Since AB \parallel CD

c49\angle CDA = \angle DAB = 25^o

\angle BAC = \angle BAD + \angle CAD

= 25^o+90^o=115^o

\therefore \angle ADB = 180^o-25^o -\angle ABD

In \triangle \angle ACD +90^o+25^o = 180^o

\Rightarrow \angle ACD = 180^o-115^o = 65^o

\angle ACD + \angle ABD = 180^o (cyclic quadrilateral)

\angle ABD = 180^o-65^o=115^o

\angle ADB = 180^o-25^o-115^o=40^o

\therefore \angle ADB = \angle AEB = 40^o (angles in the same segment)

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c41Question 29: Two circles intersect at P and Q . Through P a straight line APB is drawn to meet the circles in A and B . Through Q , a straight line is drawn to meet the circles at C and D . Prove that AC is parallel to BD .

Answer:

APQC and BPQD are cyclic quadrilateral

c410\angle CAP + \angle PQC = 180^o … … … … (i)

\angle PQD + \angle PBD = 180^o … … … … (ii)

\angle PQC + \angle PQD = 180^o   … … … … (iii)

From (i) and (ii)

\angle CAP + 180^o - \angle PQD = 180^o

\angle CAP = \angle PQD   … … … … (iv)

From (iv) and (iii)

\angle CAP = 180^o -\angle PBD

\Rightarrow \angle CAP + \angle PBD = 180^o     … … … … (v)

Therefore AB \parallel BD by (v)

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Question 30: ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD . Prove that AD is parallel to BC .

c45Answer:

ABCD is a cyclic quadrilateral, AD \parallel BC and PA = PD (given)

\angle PAD = \angle PDA

\angle CDA = 180^o - \angle PDA

\angle ABC + \angle CDA = 180^o

\angle ABC + 180^o - \angle PAD = 180

\angle ABC = \angle PAD

\therefore AD \parallel BC

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