In the given figure, if PQ = 6, QR = 10, PS = 8 find TS.

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Solution

Using the theorem of chords intersecting outside the circle, we have

PT × PS = PR × PQ

⇒ (TS + PS) × PS = (QR + PQ) × PQ

⇒ (TS + 8) × 8 = (10 + 6) × 6

⇒ 8TS + 64 = 16 × 6 = 96

⇒ 8TS = 96 − 64 = 32

⇒ TS = $\frac{32}{8}$ = 4 units

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Similar questions

P Q = 6

Q R = 10

P S = 8

T S

P S = Q R, P Q = S R

∠ Q P R = ∠ P S R, P Q = 6 c m, P S = 4 c m

R S = 2 c m,

{PR}^{2} = {PS}^{2} + QR \times ST + {(\frac{QR}{2})}^{2}

{PQ}^{2} = {PS}^{2} - QR \times ST + {(\frac{QR}{2})}^{2}

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