Question
Download Solution PDFवर्तुळाची त्रिज्या 10 सेमी आहे. PQ आणि PR या वर्तुळाच्या दोन स्पर्शिका आहेत आणि ∠QOR = 120° आहे, मग (PQ + PR + QR) चे मूल्य किती आहे?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFदिल्याप्रमाणे, ∠QOR = 120°
म्हणून, ∠QOP = ∠ROP = 60°
∠OQP = ∠ORP = 90°
⇒ ∠QPO = ∠RPO = 30°
OQ = OR = 10 सेमी
∴
⇒
⇒ 1/2 = 10/OP
⇒ OP = 20 सेमी
∴
⇒
⇒
∴
आपल्याला ΔOQP चे क्षेत्रफळ माहीत आहे =
⇒
⇒
⇒
∴ QR = 2 × KQ = 2 × 5√3 = 10√3 सेमी
प्रश्नानुसार,
(PQ + PR + QR) चे मूल्य =
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