10th Maths 10.2

NCERT Class 10th solution of Exercise 10.1

NCERT Class 10th Maths Projects

Exercise 10.2

In Q. 1 to 3, choose the correct option and give justification.

1. From a point

$Q,$ the length of the tangent to a circle is

$24 cm$ and the distance of

$Q$ from the centre is

$25 cm.$ The radius of the circle is

$7 cm$ B)

$12 cm$ C)

$15 cm$ D)

$24.5 cm$

Sol. :

$triangle OTQ$

by Pythagoras theorem

$OQ^2 = OT^2 + QT^2$

$(25)^2 = (24)^2 + QT^2$

$625 = 576 + QT^2$

$625 - 576 = QT^2$

$49 = QT^2$

$QT^2 = (7)^2$

$QT = 7$

Answer:

$A)$

$7 cm$

2. In figure, If

$TP$ and

$TQ$ are the two tangents to a circle with centre

$O$ so that

$angle POQ = 110^circ$ , then

$angle PTQ$ is equal to

$60^circ$ B)

$70^circ$ C)

$80^circ$ D)

$90^circ$

Sol. :

By angle sum property of a quadrilateral

$angle POQ + angle OPT + angle OQT + angle PTQ = 360^circ$

$110^circ + 90^circ + 90^circ + angle PTQ = 360^circ$

$290^circ + angle PTQ = 360^circ$

$angle PTQ = 360^circ - 290^circ$

$angle PTQ = 70^circ$

Answer:

$B)$

$70^circ$ .

3. If tangents

$PA$ and

$PB$ from a point

$P$ to a circle with centre

$O$ are each other at angle of

$80^circ,$ then

$anglePOA$ is equal to

$50^circ$ B)

$60^circ$ C)

$70^circ$ D)

$80^circ$

Sol. :

The line joining the centre of the circle to the external points bisect the angle between the two tangent drawn from the external point.

$angle APO = 1/2 angle APB = 1/2 times 80^circ = 40^circ$

$angle POA = 180^circ - angle APO - angle OPA$

$angle POA = 180^circ - 90^circ - 40^circ$

$angle POA = 180^circ - 130^circ$

$angle POA = 50^circ$

Answer:

$A)$

$50^circ$

Q4. Prove that the tangents drawn at the ends of a diameter of a circle are parallel.

Sol. :

Given:

Circle

$C(O, r),$

$PQ$ is the diameter,

$AB, CD$ are the tangent

To Prove:

$AB ∥ CD$

Proof:

Radius

$OP ⊥$ tangent

$APB$ i.e.

$OP ⊥ AB$

$angle APO = 90^circ$

Radius

$OQ ⊥$ tangent

$DQO$ i.e.

$OQ ⊥ CD$

$angle DQO = 90^circ$

$angle APO = angle DQO$

These are alternate angles

$AB ∥ CD$

The tangents are drawn at the endpoints of a diameter of a circle are parallel.

Proved.
Q5. Prove that the perpendicular at the point of contact to the tangent to a circle passes through the centre.

Sol. :

Given:

Circle

$C(O, r)$

tangent

$AB$

$QP ⊥ AB$

To Prove:

$PQ$ passes through the point

$O$ .

Proof:

Suppose

$PQ$ is not passing through

$O$ .

Then join

$OP$ .

$OP ⊥ AB$ [Radius

$OP ⊥$ tangent

$APB$ ]

$QP ⊥ AB$ [Given]

This is contradiction i.e. two perpendiculars is not possible through a point on a line on the same side hence

$OP$ and

$QP$ coincides.

Proved.

Q6. The length of a tangent from a point

$A$ at distance

$5 cm$ from the centre of the circle is

$4 cm.$ Find the radius of the circle.

Sol. :

$triangle APO,$

$angle P = 90^circ$

By Pythagoras theorem

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

$(5)^2 = OP^2 + (4)^2$

$OP^2 = 25 - 16$

$OP^2 = 9$

$OP^2 = (3)^2$

$OP = 3 cm$

$r = 3 cm$

Answer:

The Radius of the circle is

$3 cm$ .

Q7. Two concentric circles are of radii

$5 cm$ and

$3 cm.$ Find the length of the chord of the larger circle which touches the smaller circle.

Sol. :

$triangle QMO$

$angle M = 90^circ$

By Pythagoras theorem

$OM^2 = OM^2 + QM^2$

$(5)^2 = (3)^2 + QM^2$

$25 = 9 + QM^2$

$QM^2 = 25 - 9$

$QM^2 = 16$

$QM^2 = (4)^2$

$QM =4 cm$

The perpendicular is drawn from the centre of the circle to chord bisect the chord.

$PQ = 2 times 4$

$PQ = 8 cm$

Answer:

The length of the chord is

$8 cm.$

Q8. A quadrilateral

$ABCD$ is drawn to circumscribe a circle (see figure). Prove that

$AB + CD = AD + BC$

Sol. :

Given:

Quadrilateral

$ABCD$ and Circle touches it at

$P, Q, R, S$

To Prove:

$AB + CD = AD + BC$

Proof.

In quadrilateral

$ABCD$

The lengths of tangents drawn from an external point to a circle are equal.

$AB$ and

$AC$ are two tangents, and the external point is

$A$

So that

$AP = AS$ _______(1)

$BP$ and

$BQ$ are two tangents, and the external point is

$B$

So that

$BP = BQ$ ________(2)

$CR$ and

$CQ$ are two tangents, and the external point is

$C$

So that

$CR = CQ$ ________(3)

$DR$ and

$DS$ are two tangents, and the external point is

$D$

So that

$DR = DS$ ________(4)

From equations (1) + (2) + (3) + (4)

$AP + BP +CR + DR = AS + BQ + CS + DS$

$AB + CD + CR = AD + BC$

Proved.

Q9. In figure,

$XY$ and

$X'Y'$ are two parallel tangents to a circle with centre

$O$ and another tangent

$AB$ with point of contact

$C$ intersecting

$XY$ at

$A$ and

$X'Y'$ at

$B$ . Prove that

$angle AOB = 90^circ.$

Sol. :

Given:

$XY∥X'Y'$ are tangents intersected by tangent

$AB$

To Prove:

$angle AOB = 90^circ$

Proof:

$angle XAB + angleX'BA = 180^circ$ ____(1) [interior angles on the same side]

$AX$ and

$AB$ are two tangents drawn from the external point

$A$ to the circle.

$angle OAB = 1/2angle X'AB$ _____(2)

$BX'$ and

$BA$ are two tangents drawn from the external point

$B$ to the circle.

$angle OBA = 1/2 angle X'BA$ _____(3)

from equation (2) + (3)

$angle OAB + angle OBA = 1/2angle X' AB + 1/2 angle X'BA$ __(4)

$angle OAB + angle OBA = 1/2(angleX'AB + angleX'BA)$ __(5)

$angle OAB + angle OBA = 1/2 times 180^circ$

$angle OAB + angle OBA = 90^circ$ ___(6)[From eq (5) and (1)]

By angle sum property of

$triangle$

$angle OAB + angle OBA + angle AOB = 180^circ$

$90^circ + angle AOB = 180^circ$

$angle AOB = 180^circ - 90^circ$

$angle AOB = 90^circ$

Proved.

Q10. Prove that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line segment joining the points of contact at the centre.

Sol. :

Given:

$PQ$ and

$PR$ are tangents and Circle

$C(O, r)$

To Prove:

$angle QOR + angle QPR = 180^circ$

Construction:

Join

$OP$

Proof:

The tangent at any point of a circle is perpendicular to the radius through the point of contact.

$triangle PQO,$

$angle PQO = 90^circ$

Acute angles of right

$triangle$

$angle QOP + angle QPO = 90^circ$ ____(1)

$triangle ORP,$

$angleORP = 90^circ$

Acute angles of right

$triangle$

$angle POR + angle OPR = 90^circ$ _____(2)

From equation (1) + (2)

$angle QOP + angle QPO + angle POR + angle OPR = 90^circ + 90^circ$

$angle QOR + angle QPR = 180^circ$

Proved.

Q11. Prove that the parallelogram circumscribing a circle is a rhombus.

Sol. :

Given:

Quadrilateral

$ABCD$ and Circle touches it at

$P, Q, R, S$

Construction:

Join

$OA, OB, OC,$ and

$OD.$

To Prove:

The parallelogram is a rhombus.

Proof:

In quadrilateral

$ABCD$

$AB$ and

$AD$ are two tangents and external point

$A$ to a circle.

$angle BAO = angle DAO = 1/2 angle BAD$ _____(1)

$BA$ and

$BC$ are two tangents and external point

$B$ to a circle

$angle ABO = angle CBO = 1/2 angle ABC$ ______(2)

From equation (1) + (2)

$angle ABO + angle ABO = 1/2 angle BAD + 1/2 angle ABC$

$angle ABO + angle ABO = 1/2 (angle BAD + angle ABC)$ ___(3)

and

Adjacent angles of parallelogram

$angle BAD + angle ABC = 180^circ$ ____(4)

from equation (3) and (4)

$angle BAO + angle ABO = 1/2 times 180^circ$ ___(5)

By angle sum property of

$triangle$

$angle BAO + angle ABO + angle AOB = 180^circ$ __(6)

Equation (6) - (5)

$angle AOB = 90^circ$

Similarly

$angle BOC = angle COD = angle DOA = 90^circ$

$angle AOB + angle AOD = 180^circ$

$angle AOB + angle BOC = 180^circ$

Diagonal

$AC$ and

$BD$ are bisect each other at right angles.

So that

$ABCD$ is rhombus.

Proved.

Q12. A triangle

$ABC$ is drawn to circumscribe a circle of radius

$4 cm$ such that the segments

$BD$ and

$DC$ into which

$BC$ is divided by the point of contact

$D$ are of lengths

$8 cm$ and

$6 cm$ respectively (see figure). Find the sides

$AB$ and

$AC$ .

Sol. :

Given:

$triangle ABC, C(O, r), OD = 4 cm, CD = 6 cm$ and

$BD = 8 cm$

To Find:

Side

$AB$ and

$AC$

Solve:

Let

$AE = AF = x cm$

$ar(OAB) = 1/2 times AB times OE$

$ar(OAB) = 1/2 times (x + 8) times 4$

$ar(OAB) = (2x + 16) cm^2$

$ar(OBC) = 1/2 times BC times OD$

$ar(OBC) = 1/2 times 14 times 4$

$ar(OBC) = 28 cm^2$

$ar(OCA) = 1/2 times CA times OF$

$ar(OCA) = 1/2 times (x + 6) times 4$

$ar(OCA) = (2x+12) cm^2$

$ar(ABC) = ar(OAB) + ar(OBC) + ar(OCA)$

$ar(ABC) = (2x + 16) + 28 + (2x + 12)$

$ar(ABC) = 4x + 56$

$S = (a+b+c)/2$

$=(14 + x + 6 + x + 8)/2 = (x + 14) cm$

$ar(ABC)$

$= sqrt((x+14)(x+14-14)(x+14-x-6)(x+14-x-8))$

$4(x + 14)= sqrt((x+14)(x)(8)(6))$

$4(x + 14)= sqrt(16(3x^2 + 42x))$

$4(x + 14)= 4sqrt(3x^2 + 42x)$

$x+14 = sqrt(3x^2 +42x)$

$(x + 14)^2 = 3x^2 +42x$

$x^2 +28x +196 =3x^2 +42x$

$2x^2+14x-196 =0$

$x^2+7x-98=0$

$x^2+14x-7x-98=0$

$x(x+14)-7(x+14)=0$

$(x-7)(x+14)=0$

$x =7$

$x= -14$ [ It is not possible]

$AB =x+8 =7+8 =15 cm$

$AC =x+6 =7+6 =13 cm$

Answer:

$AB = 15 cm$ and

$AC = 13 cm$

Q13. Prove that opposite sides of a quadrilateral circumscribing a circle subtend supplementary angle at the centre of the circle.

Sol. :

Given:

Quadrilateral

$ABCD$ and Circle touch its point

$P, Q, R, S$

To Prove:

$angle DOC + angle AOB =180^circ$

Construction:

Join

$DO, AO, OB, OC$

Proof:

The lengths of tangents drawn from an external point to a circle are equal.

$angleCOD = angleADO = 1/2angleD$ _____(1)

$angleDAO = angleBAO = 1/2angleA$ _____(2)

$angleABO = angleCBO = 1/2angleB$ ______(3)

$angleBCO = angleDCO = 1/2angleC$ ______(4)

By sum of interior angles of quadrilateral.

$angleA+angleB+angleC+angleD=360^circ$

$1/2angleA+1/2angleB+ 1/2angleC+1/2angleD = 180^circ$ ___(5)

$triangle DOC$

$angleDOC+angleCDO+angleDCO=180^circ$ ____(6)

By eq. (1), (4) and (6)

$angleDOC+1/2angleD+1/2angleC=180^circ$ ____(7)

$triangleAOB$

$angleAOB+angleABO+angleBAO=180^circ$ _____(8)

By eq. (2), (3) and (8)

$angleAOB+1/2angleB+angle1/2angleA=180^circ$ __(9)

$angleDOC+angleAOB$

By eq. (7) and (9)

$=1/2angleA+1/2angleB+1/2angleC+1/2angleD=360^circ$ __(10)

By eq. (10) - (5)

$angleDOC+angleAOB=180^circ$

Proved.

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10th Maths 10.2

NCERT Class 10th solution of Exercise 10.1

NCERT Class 10th Maths Projects

Exercise 10.2

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