Question
Download Solution PDFஒரு வட்டத்தின் ஆரம் 10 செ.மீ மற்றும் PQ மற்றும் PR ஆகியவை வட்டத்தின் இரு தொடுகோடுகள் மேலும் ∠QOR = 120° எனில், (PQ + PR + QR) இன் மதிப்பு என்ன?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFகொடுக்கப்பட்டவை, ∠QOR = 120°
இவ்வாறு, ∠QOP = ∠ROP = 60°
∠OQP = ∠ORP = 90°
⇒ ∠QPO = ∠RPO = 30°
OQ = OR = 10 செ.மீ
∴
⇒
⇒ 1/2 = 10/OP
⇒ OP = 20 செ,மீ
∴
⇒
⇒
∴
நாம் அறிவோம், ΔOQP இன் பரப்பளவு =
⇒
⇒
⇒
∴ QR = 2 × KQ = 2 × 5√3 = 10√3 செ.மீ
கேள்வியின் படி,
The value of (PQ + PR + QR) இன் மதிப்பு =
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