If ΔABC and ΔPQR are two similar triangles shown in the figure. AM and PN are the medians on ΔABC and ΔPQRrespectively. The ratio of areas of ΔABC and ΔPQR is 9:25. If AM = PO = 5 cm. Find the value of 3(ON) in cm ___
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We are given two triangles i.e. ΔABC and ΔPQR such that ΔABC∼ΔPQR so, the corresponding ratio of sides and corresponding angles should be equal. ⇒ABPQ=BCQR=ACPR Also, 12×BC12×QR=BMQN Since both the triangles i.e. ΔABC and ΔPQR are similar, there angles will be equal i.e. ∠ A = ∠ P, ∠ B = ∠ Q and ∠ C = ∠ R. In ΔABM and ΔPQN, ABPQ=BMQN ∠ B = ∠ Q Therefore ΔABM∼ΔPQN [ SAS similarity] Given AM = PO = 5 cm ⇒AreaofΔABCAreaofΔPQR=AB2PQ2=925⇒ABPQ=35 ⇒ABPQ=AMPN=35⇒55+ON=35 Then 25=15+3(ON)ON=1033(ON)=10cm Hence the length of 3(ON) is 10 cm.