Solved Examples

SOLVED EXAMPLES 1. In figure 8-50, Calculate the measure of ∠PQB , where O is the centre of the circle Sol:

∠ABP = 40o

(Given)

and ∠APB = 90o

(Since AOB is a diameter and angle in a semi-circle is 90o )

Now, in Triangle APB

∠APB + ∠ABP + ∠PAB = 180o or 90o + 40o + ∠PAB = 180o or 130o + ∠PAB = 180o or ∠PAB = 180o − 130o = 50o Now arc PB subtends ∠PAB and ∠PQB , in the same segment So, ∠PAB = ∠PQB

But, ∠PAB = 50o Therefore ∠PQB = 50o 2. In figure 8-51, O is the centre of the circle. If ∠BAC = 35o , find ∠OBC Sol:

Arc BC subtends ∠BOC at the centre and ∠BAC at the circumference of the circle.

Therefore ∠BOC = 2∠BAC or ∠BOC = 2 × 35o or ∠BOC = 70o In Triangle OBC , OB=OC

(Radii of the same circle)

Thus, Triangle OBC , is an isosceles triangle with OB=OC Therefore ∠OBC = ∠OCB Now, in Triangle OBC

(Base angles of an isosceles triangle are equal)

∠BOC + ∠OBC + ∠OCB = 180o

(Angle Sum Property of triangles)

or 70o + ∠OBC + ∠OBC = 180o

( Since ∠OCB = ∠OBC , proved above)

or 70o + 2∠OBC = 180o or 2∠OBC = 180o − 70o = 110o or ∠OBC =

110o = 55o 2

3. In figure 8-52, AB and CD are two chords of a circle such that AB=6cm, CD=12 cm and AB parallel to CD . The distance between AB and CD is 3cm. Find the radius of the circle. Sol:

Let the radius of the circle = r cm Draw perpendicular bisectors OP on AB and OQ on CD. Since, AB parallel to CD Therefore O, Q, P are collinear Join OA and OC. Let OP = x cm

OQ=OP-PQ or OQ = x − 3 cm Also AP =

( Since PQ = 3cm and OP = x cm)

6 12 = 3 cm and CQ = = 6 cm 2 2

Now, in right OAP

OA2 = OP 2 + AP 2

(Pythagoras theorem)

or r 2 = x 2 + 32 or r 2 = x 2 + 9

… (1)

Also in right Triangle OCQ

OC 2 = OQ 2 + CQ 2 or r 2 = ( x − 3) 2 + 62 or r 2 = x 2 − 6 x + 9 + 36

… (2)

From (1) and (2), we get

x 2 + 9 = x 2 − 6 x + 9 + 36 or 6 x = 36 or x =

36 = 6 cm 6

Substituting the value of x in (1), we have

r 2 = 36 + 9 = 45 or r = 45 = 3 5

Therefore Radius of the circle = r = 3 5 cm

4. In figure 8-53, ABCDE is a regular pentagon. Prove that A,B, C and E are con-cyclic Sol: ABCDE is a regular pentagon

Therefore Each angle =

or ∠ABC =

2x − 4 × 90o , where x = number of sides x

2×5 − 4 6 × 90o = × 90o = 108o 5 5

In Triangle ABC , ∠ABC = 108o Also AB=BC

Therefore ∠BAC = ∠ACB =

180 − 108 72 = = 36o 2 2

Similarly, ∠AEB = 36o , ∠DEC = 36o

Therefore ∠AEC = 108o − 36o = 72o Now in quadrilateral ABCE

∠ABC + ∠AEC = 108o + 72o = 180o

Therefore Opposite angles of a quadrilateral ABCE are supplementary. Hence ABCE is a cyclic quadrilateral. 5. In figure 8-54 a quadrilateral ABCD is inscribed in a circle so that AB is the diameter of the circle. If ∠ADC = 115o . Find ∠BAC Sol:

∠B + ∠D = 180o

(Opposite angles of a cyclic quadrilateral)

∠B + 115o = 180o or ∠B = 180o − 115o = 65o

∠ACB = 90o

(Angle in a semi-circle)

∠BAC + ∠ABC + ∠ACB = 180o Therefore ∠BAC + 65o + 90o = 180o or ∠BAC + 155o = 180o or ∠BAC = 180o − 155o

or ∠BAC = 25o 6. PT is a tangent to the circle with centre O. Determine the length of the tangent PT. Sol: We know that tangent is perpendicular to the radius at the point of contact.

Therefore Triangle OTP is a right triangle Using Pythagoras Theorem, we have

OT 2 + PT 2 = OP 2 or 42 + PT 2 = 52

or PT 2 = 52 − 42 = 25 − 16 = 9 or PT = 9 = 3 cm 7. In figure 8-56, ∠ABC = 45o . Find ∠CAD and ∠AEC Sol:

Since AD is the tangent to the circle at A and AC is a chord through the point of contact,

Therefore ∠CAD = ∠ABC

(Angles in the alternate segment)

But ∠ABC = 45o

Therefore ∠CAD = 45o

Now, quadrilateral AECB is cyclic

Therefore ∠AEC + ∠ABC = 180o or ∠AEC + 45o = 180o or ∠AEC = 180o − 45o = 135o 8. In figure 8-57, Find x and y Sol:

∠AOB = 180o − (45o + 45o ) = 180o − 90o = 90o 1 Now y = × 90o = 45o 2

(Angle subtended on the circumference)

x = 45o

(Angles in the alternate segments)

9. In figure 8-58, PT is a tangent to the circle. Find the length of AB. Sol: Clearly PAB is a secant and PT is the tangent (Given)

Therefore PT 2 = PA × PB or 82 = 6 × PB or PB =

64 32 = cm 6 3

Now AB = PB – PA =

32 32 − 18 14 −6 = = cm 3 3 3

10. In figure 8-59, AB, AC and PQ are tangents. If AB= 8 cm. Find the perimeter of Triangle APQ Sol: Since AB and AC are the tangents from the same point A.

Therefore AB=AP=8 cm Similarly BP = PX and XQ= QC Now, perimeter of Triangle APQ = AP + AQ + PQ = AP + AQ + ( PX +XQ) = (AP + BP) + (AQ + QC) = AB + AC = 8 + 8 = 16 cm

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