RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.2 is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Exercise 12.2.

## Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Ex 12.2

Question 1.

In the following, write (RBSESolutions.com) true/false and if possible give reason for your answer.

(i) AB = 3 cm and CD = 4 cm are two chords of a circle. Angles subtended by these chords at center are respectively 70° and 50°.

(ii) Two chords of length 10 cm and 8 cm are at a distance from center 8 cm and 5 cm respectively.

(iii) If two chords AB and CD of a circle are at same distance 4 cm from center then AB = CD.

(iv) Two congruent circles having centers O and O’ intersect each other at two points A and B, then ∠AOB = ∠AO’B.

(v) A circle can be drawn from three collinear points.

(vi) A circle of 4 cm radius, passing through two points A and B can be drawn of AB = 8 cm.

Solution :

(i) False, because longer chord subtends larger angle at the center, where as shorter chord subtends smaller angle at the center.

(ii) False, because larger chords are closer from the center.

(iii) True, because equal chords are equidistant from the center.

(iv) True, because chords of congruent circles, subtend equal angle at the center.

(v) False, because a circle cannot be drawn though three collinear points.

(vi) True, because AB is a diameter.

Question 2.

If radius of circle is 13 cm, (RBSESolutions.com) and length of its one chord is 10 cm, then find the length of the chord from the center.

Solution :

Let O is the center of given circle and radius is 13 cm, chord AB = 10 cm

OM ⊥ AB.

We know that perpendicular drawn from center bisect the chord.

Thus AM = MB

∵ AB = 10 cm

then AM = MB = \(\frac { 1 }{ 2 }\) × AB

= \(\frac { 1 }{ 2 }\) × 10 = 5 cm

In right angled ΔOMA in (RBSESolutions.com) which ∠M is right angle, By Pythagoras theorem

OA^{2} = OM^{2} + AM^{2}

⇒ (13)^{2} = (OM)^{2} + (5)^{2}

⇒ 169 = OM^{2} + 25

⇒ OM^{2} = 169 – 25

⇒ OM^{2} = 144

⇒ OM = \(\sqrt { 144 }\)

⇒ OM= 12 cm

Thus, distance of chord AB from center is 12 cm

Question 3.

If AB = 6 cm, CD = 12 cm are two chords of a (RBSESolutions.com) circle and are parallel and lie at same side of center of circle of distance between AB and CD is 3 cm, then find radius of circle.

Solution :

Let O is center and r is radius of circle, whose chords AB = 6 cm, CD = 12 cm. Draw OM ⊥ AB which cuts CD at N

AB || CD

and ON ⊥ CD

OM ⊥ AB

AM = MB

(∵ Perpendicular drawn from (RBSESolutions.com) center of circle bisects the chord.)

AM = MB = \(\frac { 1 }{ 2 }\)AB ⇒ \(\frac { 1 }{ 2 }\) × 6 ⇒ 3 cm.

and ON ⊥ CD

CN = ND = \(\frac { 1 }{ 2 }\) × CD ⇒ \(\frac { 1 }{ 2 }\) × 12 ⇒ 6 cm.

Distance between chords AB and CD MN = 3 cm.

Let ON = x cm.

In right angled ΔOCN, by Pythagoras theorem

OC^{2} = ON^{2} + CN^{2}

(OC)^{2} = (x)^{2} + (6)^{2}

(OC)^{2} = x^{2} + 36 …..(i)

Now in right angled ∆OAM By Pythagoras

OA^{2} = AM^{2} + OM^{2}

⇒ OA^{2} = AM^{2} + (ON + MN)^{2}

⇒ OA^{2} = AM^{2} + (x + 3)^{2}

⇒ OA^{2} = (3)^{2} + (x + 3)^{2}

⇒ OA^{2} = 9 + x^{2} + 9 + 6x …..(ii)

∵ OA = OC

∴ OA^{2} = OC^{2}

x^{2} + 36 = 9 + x^{2} + 9 + 6x [from equation (i) and (ii)]

⇒ 36 = 6x + 18

⇒ 6x = 36 – 18

⇒ 6x = 18

⇒ x = \(\frac { 18 }{ 6 }\) = 3 cm.

Putting value (RBSESolutions.com) of x in equation (i)

(OC)^{2} = (3)^{2} + 36

= 9 + 36

= 45

OC = \(\sqrt { 45 }\)

OC = 3\(\sqrt { 5 }\) cm.

Thus, radius of circle is 3\(\sqrt { 5 }\) cm

Question 4.

In fig., two equal chords AB and CD (RBSESolutions.com) intersect each other at E. Prove that arc DA = arc CB

Solution :

Given :

Two equal chords AB and CD intersecting of point E. Let O be the centre.

To Prove :

arc DA = arc CB

⇒ ∠AOD = ∠COB

Construction : Through O join OA, OB, OC and OD

Proof : We know that (RBSESolutions.com) corresponding arcs of equal chords are equal.

We know that same arcs subtends equal angles at center.

Question 5.

Q. 5. In fig., AB and CD are equal (RBSESolutions.com) chords of a circle, with center O OM ⊥ AB and ON ⊥ CD then Prove that ∠OMN = ∠ONM

Solution :

Given :

In circle C (O, r) chord AB = chord CD

and OM ⊥ AB and ON ⊥ CD

To prove : ∠OMN = ∠ONM

Proof : We know that equal chord of (RBSESolutions.com) a circle are equidistant from center. Thus OM = ON ΔOMN will be an isosceles triangle. In isosceles triangle

angles opposite to equal sides are equal

Thus ∠OMN = ∠ONM

Question 6.

In fig., O and O’ are centres of given circle AB || OO’ Prove that AB = 2OO’

Solution :

Given O and O’ are centre of two (RBSESolutions.com) circles and AB || OO’ and OD ⊥ AB and O’E ⊥ AB

To Prove : AB = 2OO’

Proof :

∵ AB || OO’ and ED || OO’ …..(i)

Thus, ED || OO’ and OD ⊥ AB, O’E ⊥ AB

OD = O’E

⇒ OO’ ED is a parallelogram whose one angle is right angle.

⇒ OO’ ED is a rectangle so ED = OO’ …(ii)

We know that perpendicular drawn from center bisects the chord. Then

Then CD = AD, CE = BE

AB = BE + CE + CD + AD

= CE + CE + CD + CD

= 2CE + 2CD

= 2(CE + CD) = 2ED

AB = 2OO’ ((i)and(ii))

Question 7.

It AB = 10 cm, CD = 24 cm are two (RBSESolutions.com) chords of a circle. AB || CD and distance between AB and CD is 17 cm. Find the radius of circle.

Solution :

Let O be center and r be radius of the circle. OP ⊥ AB and OQ ⊥ AB

∵ AB || CD so P, O, Q will lie on same line

PQ = 17 cm

AB = 10 cm

CD = 24 cm

∵ OP ⊥ AB and

OQ ⊥ CD

∴ AP = BP = \(\frac { 1 }{ 2 }\)AB

= \(\frac { 1 }{ 2 }\) × 10 = 5 cm

and CQ = QD = \(\frac { 1 }{ 2 }\) × CD

= \(\frac { 1 }{ 2 }\) × 24 = 12 cm

∵ PQ = 17 cm

Let OP = x cm then OQ = 17 – x cm

In right angled ∆OPA

by Pythagoras theorem

OA^{2} = OP^{2} + AP^{2}

r^{2} = x^{2} + (5)^{2} = x^{2} + 25 …(i)

Similarly on (RBSESolutions.com) right angled ∆OQC

OC^{2} = OQ^{2} + CQ^{2}

r^{2} = (17 – x)^{2} + (12)^{2}

= (17 – x)^{2} + 144 …(ii)

Equating value of r’ from equations (i) and (ii)

x^{2} + 25 = (17 – x)^{2} + 144

= 289 – 34x + x^{2} + 144

25 = 289 – 34x + 144

34x = 289 + 144 – 25

= 433 – 25

34x = 408

x = 12 cm

Putting value of x in equation (i)

r^{2} = x^{2} + 25 = (12)^{2} + 25

= 144 +25

r^{2} = 169

r = \(\sqrt { 169 }\) = 13 cm

Thus, radius of circle = 13 cm.

Question 8.

In a circle of radius 10 cm, length of (RBSESolutions.com) two parallel chords are 12 cm and 16 cm respectively. Find the distance between AB and CD of chords are (a) same side of center (b) on opposite sides of center.

Solution :

Let O is center and r is radius of circle r = 10 cm chord AB = 12 cm and chord CD = 16 cm. Draw OP ⊥ AB which cuts chord CD at Q

Since AB || CD

Thus, OQ || CD

AP = BP

= \(\frac { 1 }{ 2 }\)AB

= \(\frac { 1 }{ 2 }\) × 12 = 6 cm

and CQ = QD = \(\frac { 1 }{ 2 }\) × CD = \(\frac { 1 }{ 2 }\) × 16 = 8 cm.

In right angled (RBSESolutions.com) triangle OPA By Pythagoras theorem

OA^{2} = AP^{2} + OP^{2}

(10)^{2} = (6)^{2} + (OP)^{2}

100 = 36 + (OP)^{2}

OP^{2} = 100 – 36 = 64

OP = \(\sqrt { 64 }\) = 8 cm

Similarly on right angled triangle OCQ By Pythagoras theorem

OC ^{2} = CQ^{2} + OQ^{2}

(10)^{2} = (8)^{2} + OQ^{2}

100 = 64 + OQ^{2}

OQ^{2} = 100 – 64 = 36

OQ = \(\sqrt { 36 }\) = 6 cm

(a) Hence, distance between AB and CD

PQ = OP – OQ = 8 – 6 cm = 2 cm

(b) Hence, distance (RBSESolutions.com) between two chords AB and CD

PQ = OP + OQ = 8 + 6 cm PQ = 14 cm

Thus, Distance between two chords is 14 cm

Question 9.

If the vertices of a quadrilateral lie on circle s.t. AB = CD, then prove that AC = BD.

Solution :

Given AB = CD and (RBSESolutions.com) vertices A, B, C, D of quadrilateral ABCD lie on circle.

To Prove : AC = BD

Construction : Join AC and BD

Proof : AB = CD (given)

Question 10.

If two equal chords of a circle (RBSESolutions.com) intersect each other, then prove that ordered parts of one chord are respectively equal to the corresponding parts of other chord.

Solution :

Given :

Let on a circle C (O, r) two equal chords AB and CD intersect each other at point P.

To Prove :

AP = CP

and PB = PD

Construction : Draw QL ⊥ AB and OM ⊥ CD and join OP

Proof : OL ⊥ AB and OM ⊥ CD

⇒ OL = OM …..(i) [∵ AB = CD] (Equal chords are (RBSESolutions.com) equidistant from center)

In right angled ∆OLP and ∆OMP

OL = OM [from eqn (i)]

∠OLP = ∠OMP [each 90°]

OP = OP [common side]

∴ ΔOLP = ΔOMP

⇒ LP = PM …..(ii)

But AB = CD

\(\frac { 1 }{ 2 }\) AB = \(\frac { 1 }{ 2 }\)CD

AL = CM ……(iii)

Adding equations (ii) and (iii)

LP + AL = PM + CM

⇒ AP = PC ……(iv)

thus AB = CD

AP = CP

Now subtracting (RBSESolutions.com) equation (iv) from (v)

AB – AP = CD – CF

PB = PD

Question 11.

Prove that line segment joining mid points of two parallel chords passes through center of the circle.

Solution :

Given Let AB and CD are two parallel chords of circle C(O, r), whose mid points are P and Q respectively. Join OP and OQ.

To Prove : Points P, O and Q are collinear

i.e., POQ is a straight line.

Construction : Draw line AOY parallel to AB

Proof : Since OP (RBSESolutions.com) bisects AB, Thus

OP ⊥ AB

⇒ ∠BPO = 90°

But XY || AB then

∠BPO + ∠YOP = 180°

90° + ∠YOP = 180°

⇒ ∠YOP = 180°- 90°

⇒ ∠YOP = 90°

Since Q is mid point of chord CD. Thus

OQ ⊥ CD

⇒ ∠OQD = 90°

CD || XY

∠OQD + ∠YOQ = 180°

90° + ∠YOQ = 180°

∠YOQ = 180° – 90°

∠YOQ = 180° – 90°

∠YOQ = 90° …(ii)

From equations (i) and (ii)

∠YOP + ∠YOQ = 90° + 90° = 180°

∠YOP + ∠YOQ= 180°

Thus, POQ is a (RBSESolutions.com) straight line.

We hope the given RBSE Solutions for Class 10 Maths Chapter 12 Circle Ex 12.2 will help you. If you have any query regarding Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Exercise 12.2, drop a comment below and we will get back to you at the earliest.

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