From an outside point P, Two tangents, PA and PB, are drawn to a circle. A third tangent to the circle at a point Q intersects PA and PB at R and s respectively. If PA = 10 cm, What is the perimeter of $\delta Prs$?
From an outside point P, Two tangents, PA and PB, are drawn to a circle. A third tangent to the circle at a point Q intersects PA and PB at R and s respectively. If PA = 10 cm, What is the perimeter of $\triangle PRS$?
एक बाहरी बिंदु P से एक वृत्त पर दो स्पर्श रेखाएँ PA और PB खींची गई हैं। एक तीसरी स्पर्श रेखा वृत्त को बिंदु Q पर स्पर्श करती है और PA तथा PB को क्रमशः बिंदु R और S पर प्रतिच्छेद करती है। यदि PA = 10 सेमी है, तो $\triangle PRS$ का परिमाप क्या होगा?
Detailed Solution & Logic
20 cm/20 सेमी
Given:
PA = 10 cm
Property: From an external point, tangents to a circle are equal.
So, PA = PB = 10 cm
Now, the third tangent at $Q meets PA at R and PB at S.
Using tangent properties:
RQ = R A and SQ = SB
Also, the total lengths split such that:
PR + RA = PA = 10
PS + SB = PB = 10
Adding:
PR + PS + (RA + SB) = 20
But RA + SB = RS (along the tangent structure)
So, perimeter of $\triangle PRS$:
$PR + RS + PS = 20$
Final Answer:
20 cm
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