Answer:
Step-by-step 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
[tex](AM)^{2}=(AC)^{2}+(CM)^{2}[/tex]
[tex](AM)^{2}=(3k)^{2}+(4k)^{2}[/tex]
[tex](AM)^{2}=25k^{2}[/tex]
[tex]AM=5k[/tex]
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
[tex](MP)^{2}-(AP)^{2}=5k[/tex]
Add and subtract [tex](CP)^{2}[/tex] on left side
[tex](MP)^{2}+(CP)^{2}-(CP)^{2}-(AP)^{2}=5k[/tex]
[tex]((MP)^{2}+(CP)^{2})-((CP)^{2}-(AP)^{2})=5k[/tex] (3)
But, since CP ⊥ AM, Δ CMP and Δ CAP are right triangles. Therefore,
[tex](MP)^{2}+(CP)^{2}=(CM)^{2}[/tex]
[tex](CP)^{2}+(AP)^{2}=(AC)^{2}[/tex]
Thus, Equation (3) becomes,
[tex](CM)^{2}-(AC)^2=5k[/tex]
[tex](4k)^{2}-(3k)^2=5k[/tex]
[tex]7k^{2}=5k[/tex]
[tex]k=\frac{5}{7}[/tex]
Thus, AM=5k
⇒AM=[tex]\frac{25}{7}[/tex]
