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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.
- Prove that: ∆ABC ~ ∆AMP
- Find: AB and BC.
Solution
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
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