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In figure, if AQ5cm, find the perimeter of triangle ABC |
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Answer» Let PQ touch the circle at the point R. We know that tangents drawn from an external point to a circle are equal in length. AB = AC = 5cm AP +BP = AQ+ QC = 5cm AP + PR = AQ + QR = 5cm…………(1) [ BP = PR & QC = QR] Now, perimeter of ∆ APQ = AP +PQ+AQ = AP +RP+QR+AQ = 5 + 5 [ from eq 1] Perimeter of ∆ APQ= 10 cm |
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