In the figure, two circles with centres P and Q touch externally at R. Tangents AT and BT meet the common tangent TR at T. If AP = 6 cm and PT = 10 cm, then BT =?
8 cm
In right-angled ΔPAT, 90 at A.
PT 2= AP 2+ AT 2
⇒ 10 2= 6 2+ AT 2
⇒ AT 2= 100 - 36 = 64
⇒ AT = √64
⇒ AT = 8 cm
As we know,
AT = TR = 8 cm [common tangent]
⇒ TR = TB = 8 cm [common tangent]Similar Questions
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