PQ is a chord of length 8 cm of a circle of radius 5 cm the tangents at P and Q intersect at point P find the length of TP
Answers
Answer:
∴ the length of tangent PT is (20/3) cm
Step-by-step explanation:
Given radius, OP = OQ = 5 cm
Length of chord, PQ = 4 cm
OT ⊥ PQ,
∴ PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right ΔOPM,
OP² = PM² + OM²
⇒ 52 = 42 + OM²
⇒ OM² = 25 – 16 = 9
Hence OM = 3cm
In right ΔPTM,
PT² = TM² + PM² → (1)
∠OPT = 90º [Radius is perpendicular to tangent at point of contact]
In right ΔOPT,
OT2² = PT² + OP² → (2)
From equations (1) and (2), we get
OT² = (TM² + PM²) + OP2²
⇒ (TM + OM)² = (TM² + PM²) + OP²
⇒ TM² + OM² + 2 × TM × OM = TM² + PM² + OP²
⇒ OM² + 2 × TM × OM = PM2 + OP²
⇒ 32 + 2 × TM × 3 = 42 + 52
⇒ 9 + 6TM = 16 + 25
⇒ 6TM = 32
⇒ TM = 32/6 = 16/3
Equation (1) becomes,
PT² = TM² + PM²
= (16/3)2 + 42
= (256/9) + 16 = (256 + 144)/9
= (400/9) = (20/3)2
⇒ PT = 20/3
∴ the length of tangent PT is (20/3) cm
Answer: TP = 6.66 cm.
Step-by-step explanation:Given radius, OP = OQ = 5 cm
Length of chord, PQ = 4 cm
OT ⊥ PQ,
∴ PM = MQ = 4 cm [Perpendicular draw from the centre of the circle to a chord bisect the chord]
In right ΔOPM,
OP² = PM² + OM²
⇒ 52 = 42 + OM²
⇒ OM² = 25 – 16 = 9
Hence OM = 3cm
In right ΔPTM,
PT² = TM² + PM² → (1)
∠OPT = 90º [Radius is perpendicular to tangent at point of contact]
In right ΔOPT,
OT2² = PT² + OP² → (2)
From equations (1) and (2), we get
OT² = (TM² + PM²) + OP2²
⇒ (TM + OM)² = (TM² + PM²) + OP²
⇒ TM² + OM² + 2 × TM × OM = TM² + PM² + OP²
⇒ OM² + 2 × TM × OM = PM2 + OP²
⇒ 32 + 2 × TM × 3 = 42 + 52
⇒ 9 + 6TM = 16 + 25
⇒ 6TM = 32
⇒ TM = 32/6 = 16/3
Equation (1) becomes,
PT² = TM² + PM²
= (16/3)2 + 42
= (256/9) + 16 = (256 + 144)/9
= (400/9) = (20/3)2
⇒ PT = 20/3 cm
PT = 6.66 cm
∴ the length of tangent PT is 6.66 cm.