Question
Download Solution PDFIn the given fig. O is the centre of the circle and OP||QR, QR is tangent of the circle and OP = 6 cm, find the area of ∆OPR.
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFOP||QR (given)
⇒ Extend the point P to the point S such that QR = OS
Thus, rectangle QRSO is formed
⇒ OQ = SR = 6 cm (Height of the ∆OPR)
⇒ OP = 6 cm = Base of the ∆OPR
⇒ Area of ∆OPR = 1/2 × 6 × 6 = 18 cm2
Last updated on Jun 13, 2025
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