In the given figure the radii of two concentric circles are 7 cm and 8 cm if PA 15cm then find PB

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Solution : <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/RSA_MATH_X_C08_S01_004_S01.png" width="80%"> <br> We have <br> `OAbotAP` and `OB bot BP` [`:'` the tangents at any point of a circle is perpendicular to the radius through the point of contact]. <br> Join `OP`. <br> In right `DeltaOAP`, we have <br> `OA=8cm`, `AP=15cm`. <br> `:.OP^(2)=OA^(2)+AP^(2)` <br> [by Pythogoras's theorem] <br> `impliesOP=sqrt(OA^(2)+AP^(2))=sqrt(8^(2)+15^(2))cm=sqrt(289)cm=17cm`. <br> In right `DeltaOBP`, we have <br> `OB=5cm`, `OP=17cm` <br> `:.OP^(2)=OB^(2)+BP^(2)`[ by Pythagoras' theorem] <br> `impliesBP=sqrt(OP^(2)-OB^(2))=sqrt(17^(2)-5^(2))cm=sqrt(264)cm` <br> Thus, the length of `BP=sqrt(264)cm=16.25cm` (approx).

Solution:

The chord of the larger circle is a tangent to the smaller circle as shown in the figure below.

PQ is a chord of a larger circle and a tangent of a smaller circle.

Tangent PQ is perpendicular to the radius at the point of contact S.

Therefore, ∠OSP = 90°

In ΔOSP (Right-angled triangle)

By the Pythagoras Theorem,

OP2 = OS2 + SP2

52 = 32 + SP2

SP2 = 25 - 9

SP2 = 16

SP = ± 4

SP is the length of the tangent and cannot be negative

Hence, SP = 4 cm.

QS = SP (Perpendicular from center bisects the chord considering QP to be the larger circle's chord)

Therefore, QS = SP = 4cm

Length of the chord PQ = QS + SP = 4 + 4

PQ = 8 cm

Therefore, the length of the chord of the larger circle is 8 cm.

☛ Check: NCERT Solutions Class 10 Maths Chapter 10

Video Solution:

Maths NCERT Solutions Class 10 Chapter 10 Exercise 10.2 Question 7

Summary:

If two concentric circles are of radii 5 cm and 3 cm, then the length of the chord of the larger circle which touches the smaller circle is 8 cm.

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Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig.3. If AP =15 cm, then find the length of BP.

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Tangents PA and PB are drawn from an external point P to two concentric circles with centre O and radii 8 cm and 5 cm respectively, as shown in Fig. 3. If AP = 15 cm, then find the length of BP.

To find: BP

Construction: Join OP.

`Now ,OA _|_AP`and `OB_|_BP`     `[therefore\text{Tangent to a circle is prependicular to the radius through the point of contact}]`

⇒ ∠OAP = ∠OBP = 90°

On applying Pythagoras theorem in ΔOAP, we obtain:

(OP)2 = (OA)2 + (AP)2

⇒ (OP)2 = (8)2 + (15)2

⇒ (OP)2 = 64 + 225

⇒ (OP)2 = 289

`rArr OP=sqrt289`

⇒ OP = 17

Thus, the length of OP is 17 cm.

On applying Pythagoras theorem in ΔOBP, we obtain:

(OP)2= (OB)2 + (BP)2

⇒ (17)2 = (5)2 + (BP)2

⇒ 289 = 25 + (BP)2

⇒ (BP)2 = 289 − 25

⇒ (BP)2 = 264

⇒ BP = 16.25 cm (approx.)

Hence, the length of BP is 16.25 cm.

Concept: Concept of Circle - Centre, Radius, Diameter, Arc, Sector, Chord, Segment, Semicircle, Circumference, Interior and Exterior, Concentric Circles

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