Question

Given: △ACM, m∠C=90°, CP ⊥ AM . AC:CM=3:4, MP-AP=1. Find AM.

Answer 1

`AM=(25)/(7)`

Explanation:

It is given that AC:CM = 3:4.

Therefore, let AC = 3k and CM = 4k

It is given that Δ ACM is right angled.

Therefore,

`AM^(2) =AC^(2) +CM^(2)`

`=(3k)^(2) +(4k)^(2)`

`=9k^(2) +16k^(2)`

`=25k^(2)`

AM = 5k

But, AM = MP + AP

Therefore, MP + AP = 5k --- (1)

It is given that MP - AP = 1 --- (2)

Multiply (1) and (2), we get.

(MP + AP)(MP - AP) = 5k × 1 = 5k

`MP^(2) -AP^(2) =5k`

Add and subtract
`CP^(2)` on the left side.

`MP^(2) + CP^(2) - CP^(2) -AP^(2) =5k`

`(MP^(2) + CP^(2)) - (CP^(2) +AP^(2)) =5k` --- (3)

But, since CP ⊥ AM, Δ CMP and Δ CAP are right triangles. Therefore,

`MP^(2) +CP^(2) =CM^(2)` and

`CP^(2) +AP^(2) =AC^(2)`

Now, (3) becomes,

`CM^(2) -AC^(2) =5k`

`(4k)^(2) -(3k)^(2) =5k`

`7k^(2) =5k`

7k = 5 or

`k=(5)/(7)`

AM = 5k

`=5((5)/(7) )`

`=(25)/(7)`