# a is a point at a distance 13 cm from the centre o of a circle of radius 5 cm. ap and aq are the tangents to the circle at p and q. if a tangent bc is drawn at a point r lying on the minor arc pq to intersect ap at b and aq at c, find the perimeter of the ∆abc.

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get a is a point at a distance 13 cm from the centre o of a circle of radius 5 cm. ap and aq are the tangents to the circle at p and q. if a tangent bc is drawn at a point r lying on the minor arc pq to intersect ap at b and aq at c, find the perimeter of the ∆abc. from screen.

## Question 14 A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm . AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ABC.

Question 14 A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm . AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ABC.

Byju's Answer Standard X Mathematics

Radius and Tangent to a Circle Are Perpendicular

Question 14 A... Question Question 14

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q . If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the ∆ABC.

Open in App Solution

Given two tangents are drawn from an external point A to the circle with centre O.

OA = 13 cm

Tangent BC is drawn at a point R radius of circle 5 cm

To find perimeter of

Δ ABC Proof ∠ O P A = 90 ∘

[ tangent at any point of circle is perpendicular to the radius through the point of contact]

O A 2 = O P 2 + P A 2

[ by Pythagoras theorem]

( 13 ) 2 = 5 2 + P A 2 ⇒ P A 2 = 144 = 12 2 ⇒ P A = 12 c m Now, perimeter of Δ ABC= AB+BC+CA = (AB+BR) + (RC+CA) =AB+BP+CQ+CA [ ∵

BR = BP , RC = CQ tangents from internal point to a circle are equal]

= AP + AQ = 2 AP = 2 (12) = 24 cm

[ AP = AQ tangent from internal point to a circle are equal]

Hence the perimeter of

Δ ABC = 24 cm. Suggest Corrections 3 Video Solution

EXEMP - Grade 10 - Mathematics - Circles - Q44

MATHEMATICS

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SIMILAR QUESTIONS

**Q.**A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm . AP and AQ are the tangents to the circle at P and Q . If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C , find the perimeter of the

∆ ABC .

**Q.**Two tangents are drawn to a circle from an external point

A

, touching the circle at the points

P and Q

. A third tangent intersects segment

A P at B and segment A Q at C

and touches the circle at

R . If A Q = 10

units, then the perimeter of

Δ A B C is

**Q.**From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at a point E and PA = 14 cm, find the perimeter of Δ PCD.

**Q.**

From an external point P, tangents PA and PB are drawn to a circle with centre O. At one point E on the circle tangent is drawn, which intersects PA and PB at C and D respectively. If

P A = 14 c m

, find the perimeter of

Δ P C D .

**Q.**From an external point P, tangents PA and PB are drawn to a circle with centre O. If CD is the tangent to the circle at the point E and PA = 14 cm, find the perimeter of ∆PCD.

Figure View More EXPLORE MORE

Radius and Tangent to a Circle Are Perpendicular

Standard X Mathematics

## A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C ,find the perimeter of the ΔABC .

Click here👆to get an answer to your question ✍️ A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C ,find the perimeter of the ΔABC .

Question

## A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C ,find the perimeter of the ΔABC.

Medium Open in App Solution Verified by Toppr

Correct option is A)

BP=BR and CR=CQ

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

Perimeter of △ABC=AB+BR+RC+CA=AB+BP+QC+CA

=AP+QA(AP=QA) =2AP

In △APO, using pythagaras theorem

AO 2 =AP 2 +PO 2 ⇒13 2 =AP 2 +5 2 ⇒AP= 169−25 AP= 144 =12

∴ Perimeter of △ABC=2AP=2×12=24cm

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## A is a Point at a Distance 13 Cm from the Centre O of a Circle of Radius 5 Cm . Ap and Aq Are the Tangents to the Circle at P and Q . If a Tangent Bc is Drawn at a Point R Lying on the Minor Arc Pq

A is a Point at a Distance 13 Cm from the Centre O of a Circle of Radius 5 Cm . Ap and Aq Are the Tangents to the Circle at P and Q . If a Tangent Bc is Drawn at a Point R Lying on the Minor Arc Pq

Advertisement Remove all ads Short Note

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm . AP and AQ are the tangents to the circle at P and Q . If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C , find the perimeter of the

∆ ABC . Advertisement Remove all ads

### SOLUTION

A is a point 13 cm from centre O. AP and AQ are the tangents to the circle with centre O.

AP = AQ In ∆APO, AP2+OP2=AO2 ⇒AP2=AO2−OP2 ⇒AP2=132−52 ⇒AP2=169−25=144 ⇒AP=12cm Now In ∆ABC,

Perimeter = AB + BC + AC

= AB + BR + RC + AC

= AB + BP + CQ + AC ( BR = BP, RC = CQ)

= AP + AQ = 12 + 12 = 24 cm

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

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Chapter 8: Circles - Exercise 8.2 [Page 35]

Q 24 Q 23 Q 25

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RD Sharma Class 10 Maths

Chapter 8 Circles

Exercise 8.2 | Q 24 | Page 35

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