Question

hey gyues

solve this pls
​

Answer 1

Answer:

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 2

$\huge\sf\pink{Answer}$

☞ PT is equal to $\sf \dfrac{20}{3} \ cm$

$\rule{110}1$

$\huge\sf\blue{Given}$

✭ PQ is a chord of Length 8 cm

✭ Radius of circle = 5 cm

✭ The tangents at p and q intersect at t

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ The Length of TP?

$\rule{110}1$

$\huge\sf\purple{Steps}$

We know that,

OT ⊥ PQ,

∴ PM = MQ = 4 cm (Perpendicular draw from the centre of the circle to a chord bisect the chord)

In right ΔOPM,

➳ $\sf {OP{}^{2} = PM{}^{2}+ OM{}^{2}}$

➳ $\sf {52=42+OM{}^{2}}$

➳ $\sf{ OM^2=25-16}$

➳ $\sf {OM =\sqrt{9}}$

➳ $\sf\red{OM = 3}$

In right ΔPTM,

≫ $\sf{ PT{}^{}2=TM{}^{2}+PM{}^{2}\qquad -eq(1)}$

≫ $\sf \angle OPT = 90^{\circ}$(Radius is perpendicular to tangent at point of contact])

In right ΔOPT,

≫ $\sf {OT^2 = PT^2 + OP^2\qquad-eq(2)}$

From equations eq(1) and eq(2)

➝ $\sf{OT^2=(TM^2+PM^2)+OP^2}$

➝ $\sf{(TM+OM)^2=(TM^2+PM^2)+OP^2}$

➝ $\sf{TM^2+OM^2+2×TM×OM=TM^2+PM^2+OP^2}$

➝ $\sf{OM^2+2×TM×OM=PM^2+OP^2}$

➝ $\sf{32+2×TM×3=42+52}$

➝ $\sf{9 + 6TM = 16 + 25}$

➝ $\sf{6TM=32}$

➝ $\sf{TM=\dfrac{32}{6}}$

➝ $\sf\green{TM =\dfrac{16}{3}}$

Substituting the value of TM in eq(1)

➢ $\sf{PT^2=TM^2+PM^2}$

➢ $\sf {\dfrac{\dfrac{16}{3}}{2} + 42}$

➢ $\sf \dfrac{256}{9} + 16 = \dfrac{256 + 144}{9}$

➢ $\sf \dfrac{400}{9} = \dfrac{20}{3} \times 2$

➢ $\sf\orange{PT = \dfrac{20}{3}}$

$\rule{170}3$