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प्रश्न
In the given figure ΔABC and ΔAMP are right angled at B and M respectively. Given AC = 10 cm, AP = 15 cm and PM = 12 cm.
1) Prove ΔABC ~ ΔAMP
2) Find AB and BC.
उत्तर
In ABC and AMP
∠ABC = ∠AMP (each 90°)
∠BAC = ∠PAM (common)
∴ΔABC ~ ΔAMP (By AA similarity)
Since the triangles are similar, we have
`(AB)/(AM) = (BC)/(MP) = (AC)/(AP)`
`=> (AB)/(AM) = (BC)/(MP) = (AC)/(AP)`
`=> (AB)/(AM) = (BC)/12 = 10/15`
Taking `(BC)/12 = 10/15 => BC = (12 xx 10)/15` = 8 cm
Now, using Pythagoras theorem in ΔABC
`(AB)^2 + (BC)^2 = (AC)^2`
`=> (AB)^2 = 10^2 - 8^2 = 36`
=> AB = 6 cm
Hence AB = 6 cm and BC = 8 cm
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