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In Δ ABC and Δ PQR, BC = QR, ∠ A = ∠P, ∠Q = 90°, QR = 5 cm, PQ = 12 cm.Find the diameter of the circumcircle of Δ ABC. |
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Answer» QR = BC ∠A = ∠P PR Diameter of the circumcircle of Δ ABC Diameter = PR= \(\sqrt{12^2+5^2}\) = \(\sqrt{169}=\) 13cm. |
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