Question

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

Find AM.

Answer 1

25/7

Explanation:

Answer 2

Explanation:

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

Solution: Let AC=3k and CM=4k, then from △ACM, we get

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

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

`(AM)^(2)=25k^(2)`

`AM=5k`

Now, AM=MP+AP⇒5k=MP+AP (1)

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

Therefore, from (1) and (2),

(MP+AP)(MP-AP)=5k

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

Add and subtract
`(CP)^(2)` on 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)`

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

Thus, Equation (3) becomes,

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

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

`7k^(2)=5k`

`k=(5)/(7)`

Thus, AM=5k

⇒AM=
`(25)/(7)`