Question

hey guys please answer my question and I'll mark u brainliest

Answer 1

Answer:

OP is equals to a radius is equals to 5 let AE equals to BE be x.

in triangle OPT, OP²+OT²=PT²

Then, PT=12cm.

Now, AT=12-x and AE=x

And TE=13-5=8cm.

Now in triangle AET, AT²=AE²-ET²

x= 10/3cm.

AB=AE+ BE =2x=2×10/3=20/3cm.

Answer 2

Answer: 20 / 3 cm

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Here, PT and QT are tangents. ∴ ∠PTO = ∠QTO

In ΔOPT,

PT² = OT² - OP²

PT² = 13² - 5²

PT² = 169 - 25

PT² = 144

PT = √144

PT = 12

∠AEO = ∠BEO = 90° (Angles formed by tangent and radius)

∠AEO and ∠AET are linear pairs. ∴ ∠AEO = ∠AET = 90°

∠BEO and ∠BET are linear pairs. ∴ ∠BEO = ∠BET = 90°

Consider triangles ATE and BTE.

∠AET = ∠BET = 90°

∠ATE = ∠BTE

TE = TE (common)

∴ ΔATE ≅ ΔBTE

∴ AE = BE = 1/2 × AB

∴ AB = 2AE = 2BE

So we can find the length of AB by finding AE or BE and by taking its double.

Here, we know that ΔOPT ≅ ΔOQT.

(∵ OP = OQ,  PT = QT,  &  ∠P = ∠Q)

Consider triangles OPT and AET.

∠OPT = ∠AET = 90°

∠OTP = ∠ATE (common)

∴ ∠TOP = ∠TAE

∴ ΔOPT ~ ΔAET

∴ AE / OP  =  TE / TP

⇒ AE / 5 = 8 / 12

⇒ AE / 5 = 2 / 3

⇒ AE = (2 / 3) × 5

⇒ AE = 10 / 3

AB = 2AE = 2 × 10 / 3 = 20 / 3 cm

Hope this helps you. ^_^

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Thank you. :-))