Question

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.

Answer 1

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