Math  /  Geometry

QuestionNella figura, QPQ P è tangente in TT alla semicirconferenza di centro OO e diametro 20.OB=6e20 . \overline{O B}=6 \mathrm{e} OP=503\overline{O P}=\frac{50}{3}. Trova PQ\overline{P Q}. [1256]\left[\frac{125}{6}\right]

Studdy Solution
Usa il teorema della tangente per trovare PQ \overline{PQ} :
Poiché QP QP è tangente alla semicirconferenza in T T , abbiamo:
OP2=OT2+PQ2 OP^2 = OT^2 + PQ^2
Dove OP=503 OP = \frac{50}{3} e OT=10 OT = 10 .
(503)2=102+PQ2 \left(\frac{50}{3}\right)^2 = 10^2 + PQ^2
25009=100+PQ2 \frac{2500}{9} = 100 + PQ^2
PQ2=25009100 PQ^2 = \frac{2500}{9} - 100
PQ2=250099009 PQ^2 = \frac{2500}{9} - \frac{900}{9}
PQ2=16009 PQ^2 = \frac{1600}{9}
PQ=16009 PQ = \sqrt{\frac{1600}{9}}
PQ=403 PQ = \frac{40}{3}
La lunghezza di PQ \overline{PQ} è:
403 \boxed{\frac{40}{3}}

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