# in the given figure pq is a chord of length 8 cm of a circle of radius 5cm

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## PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents P and Q intersect at a point T . Find the length TP .

Click here👆to get an answer to your question ✍️ PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents P and Q intersect at a point T . Find the length TP .

Question

## PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents P and Q intersect at a point T. Find the length TP.

**A**

3 8 cm

**B**

3 20 cm

**C**

3 17 cm

**D**

3 11 cm Medium Open in App Solution Verified by Toppr

Correct option is B)

Joint OT.Let it meet PQ at the point R.

Then ΔTPQ is isosceles and TO is the angle bisector of ∠PTO.

[∵TP=TQ= Tangents from T upon the circle]

∴OT⊥PQ ∴OT bisects PQ. PR=RQ=4 cm Now, OR= OP 2 −PR 2 = 5 2 −4 2 =3 cm Now, ∠TPR+∠RPO=90 ∘ (∵TPO=90 ∘ ) =∠TPR+∠PTR(∵TRP=90 ∘ ) ∴∠RPO=∠PTR

∴ Right triangle TRP is similar to the right triangle

PRO. [By A-A Rule of similar triangles]

∴ PO TP = RO RP ⇒ 5 TP = 3 4 ⇒TP= 3 20 cm.

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## In the given figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP. [4 MARKS]

In the given figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP. [4 MARKS]

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In the given figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP. [4 MARKS]

Question

In the given figure, PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length TP.

Open in App Solution

Join OP and OT

Let OT intersect PQ at a point R.

Then, T P = T Q a n d ∠ P T R = ∠ Q T R . ∴ T R ⊥ P Q and TR bisects PQ. ∴ P R = R Q = 4 c m . Also O R = √ O P 2 − P R 2 = √ 5 2 − 4 2 cm = √ 25 − 16 c m = √ 9 c m = 3 c m . Let T P = x c m and T R = y c m From right Δ T R P , we get T P 2 = T R 2 + P R 2 ⇒ x 2 = y 2 + 16 ⇒ x 2 − y 2 = 16 … … ( i ) From right Δ O P T , we get T P 2 + O P 2 = O T 2 ⇒ x 2 + 5 2 = ( y + 3 ) 2 [ O T 2 = ( O R + R T ) 2 ] ⇒ x 2 − y 2 = 6 y − 16 … … ( i i )

From (i) and (ii) , we get

6 y − 16 = 16 ⇒ 6 y = 32 ⇒ y = 16 3 Putting y = 16 3 in (i) we get x 2 = 16 + ( 16 3 ) 2 = ( 256 9 + 16 ) = 400 9 ⇒ x = √ 400 9 = 20 3 Hence , lenght T P = x c m = 6.67 c m Suggest Corrections 104

SIMILAR QUESTIONS

**Q.**

P Q

is a chord of length

8 c m

of a circle of radius

5 c m . The tangents at P and Q

intersect at a point

T

(see figure). Find the length of the tangent

T P .

**Q.**

P Q

is a chord of length

8 c m

of a circle of radius

5 c m . The tangent at P and Q intersect at point T . Find the length T P .

**Q.**PQ is a chord of length S cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T. Find the length of TP.

**Q.**Point O is the centre of the circle suppose PQ is a chord of length 8 cm of a circle of radius 5 cm the tangent at point P and Q intersect at a point P find TP

**Q.**PQ is a chord of length

8

cm of a circle of radius

5

cm. The tangents at P and Q intersect at a point T. Find length TP.

## Example 3

Example 3(Method 1) PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see figure). Find the length ..

**Check sibling questions**

## Example 3 - Chapter 10 Class 10 Circles (Term 2)

Last updated at Feb. 25, 2017 by Teachoo

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### Transcript

Example 3(Method 1) PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see figure). Find the length TP. Join OT. Let OT intersect PQ at R From theorem 10.2, Lengths of tangents from external point are equal So, TP = TQ In ΔTPQ, TP = TQ, i.e. two sides are equal, So, Δ TPQ is an isosceles triangle Here, OT is bisector of ∠ PTQ, So OT ⊥ PQ Since OT ⊥ PQ So, PR = RQ So, PR = QR = 1/2PQ = 8/2 = 4 cm In right triangle ORP By Pythagoras theorem (Hypotenuse)2 = (Height)2 + (Base)2 (OP)2 = (PR)2 + (OR)2 52 = 42 + OR2 25 = 16 + OR2 25 – 16 = OR2 9 = OR2 OR2 = 9 OR = √9 = √(3^2 ) = 3 cm Let TP = x By Pythagoras theorem, (Hypotenuse)2 = (Height)2 + (Base)2 (OT)2 = (OP)2 + (TP)2 (OT)2 = 52 + x2 (OR + RT)2 = 52 + x2 (3 + RT)2 = 52 + x2 32 + RT2 + 2(3)RT= 52 + x2 9 + RT2 + 6RT= 25 + x2 9 + RT2 + 6RT= 25 + 16 + RT2 RT2 + 6RT – RT2 = 25 + 16 – 9 6RT = 32 RT = 32/6 RT = 16/3 From (1) x2 = 16 + RT2 Putting value of RT x2 = 16 + (16/3)^2 x2 = 16 + 256/9 x2 = (16 × 9 + 256)/9 x2 = (144 + 256)/9 x2 = 400/9 x = √(400/9) x = √((20)2/32) = √((20/3)^2 ) = 20/3 Hence, TP = x = 20/3 cm Example 3(Method 2) PQ is a chord of length 8 cm of a circle of radius 5 cm. The tangents at P and Q intersect at a point T (see figure). Find the length TP. Let TP = x Join OQ Since TP is a tangent, OP ⊥ TP ∴ ∠ OPT = 90° Similarly, Since TQ is a tangent, OQ ⊥ TQ ∴ ∠ OQT = 90° In Δ OPT & Δ OQT TP = TQ OT = OT OP = OQ So, Δ OPT ≅ Δ OQT So, ∠ POT = ∠ QOT In Δ OPR & Δ OQR OP = OQ ∠ POR = ∠ QOR OR = OR So, Δ OPR ≅ Δ OQR So, PR = QR So, PR = QR = 1/2PQ = 8/2 = 4 cm Also, Since Δ OPR ≅ Δ OQR So, ∠ PRO = ∠ QRO But, ∠ PRO + ∠ QRO = 180° ∠ PRO + ∠ PRO = 180° 2∠ PRO = 180° ∠ PRO = (180°)/2 ∠ PRO = 90° So, ∠ QRO = 90° Also, ∠ PRT = ∠ QRO = 90° By Pythagoras theorem, (Hypotenuse)2 = (Height)2 + (Base)2 In right triangle ORP By Pythagoras theorem (OP)2 = (PR)2 + (OR)2 52 = 42 + OR2 25 = 16 + OR2 25 – 16 = OR2 9 = OR2 OR2 = 9 OR = √9 OR = √(3^2 ) OR = 3 cm Pythagoras theorem, (Hypotenuse)2 = (Height)2 + (Base)2 9 + RT2 + 6RT= 25 + x2 9 + RT2 + 6RT= 25 + 16 + RT2 RT2 + 6RT – RT2 = 25 + 16 – 9 6RT = 32 RT = 32/6 RT = 16/3 From (3) x2 = 16 + RT2 Putting value of RT x2 = 16 + (16/3)^2 x2 = 16 + 256/9 x2 = (16 × 9 + 256)/9 x2 = (144 + 256)/9 x2 = 400/9 x = √(400/9) x = √((20)2/32) x = √((20/3)^2 ) x = 20/3 Hence, TP = x = 20/3 cm

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### Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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