Question

In the given figure, ∆ABC and ∆AMP are right angled at B and M respectively.

Given AC = 10 cm, AP = 15 cm and PM = 12 cm.

1. Prove that: ∆ABC ~ ∆AMP
2. Find: AB and BC.

Answer 1

i. In ∆ABC and ∆AMP,

∠BAC = ∠PAM   ...[Common]

∠ABC = ∠PMA  ...[Each = 90°]

∴ ΔABC ∼ ΔAMP  ...(By AA similarity)

ii.

`AM = sqrt(AP^2 - PM^2)`

= `sqrt(15^2 - 12^2)`

= 9

Since ∆ABC ∼ ∆AMP,

`(AB)/(AM) = (BC)/(PM) = (AC)/(AP)`

`=> (AB)/(AM) = (BC)/(PM) = (AC)/(AP)`

`=> (AB)/9 = (BC)/12 = (10)/(15)`

From this we can write,

`(AB)/9 = 10/15`

`=> AB = (10 xx 9)/15 = 6`

`(BC)/12 = 10/15`

`=>` BC = 8 cm