Question
Download Solution PDFवृत्त की त्रिज्या 10 सेंटीमीटर है। PQ और PR वृत्त के दो स्पर्शरेखा हैं और ∠QOR = 120° है, तो, (PQ + PR + QR) का मान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिया गया हैं, ∠QOR = 120°
इसलिए, ∠QOP = ∠ROP = 60°
∠OQP = ∠ORP = 90°
⇒ ∠QPO = ∠RPO = 30°
OQ = OR = 10 सेंटीमीटर
∴
⇒
⇒ 1/2 = 10/OP
⇒ OP = 20 सेंटीमीटर
∴
⇒
⇒
∴
हम क्षेत्रफल ΔOQP को जानते हैं =
⇒
⇒
⇒
∴ QR = 2 × KQ = 2 × 5√3 = 10√3 सेंटीमीटर
प्रश्न के अनुसार,
(PQ + PR + QR) का मान =
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