Answer:
Step-by-step explanation:
It is given that in △ACM, m∠C=90°, CP ⊥ AM
, AC:CM=3:4, MP-AP=1.
Let AC=3x and CM=4x, then from △ACM, we get
[tex](AM)^2=(AC)^2+(CM)^2[/tex]
[tex](AM)^2=(3x)^2+(4x)^2[/tex]
[tex](AM)^2=25x^2[/tex]
[tex]AM=5x[/tex]
Now, AM=MP+AP⇒5x=MP+AP (1)
It is also given that MP-AP=1 (2)
Therefore, from (1) and (2),
(MP+AP)(MP-AP)=5x
[tex](MP)^2-(AP)^2=5x[/tex]
Add and subtract [tex](CP)^2[/tex] on the left side,
[tex](MP)^2+(CP)^2-(CP)^2-(AP)^2=5x[/tex]
[tex]((MP)^2+(CP)^2)-((CP)^2+(AP)^2)=5x[/tex] (3)
But, since CP ⊥ AM, Δ CMP and Δ CAP are right triangles. Therefore,
[tex](MP)^2+(CP)^2=(CM)^2[/tex] and
[tex](CP)^2+(AP)^2=(AC)^2[/tex]
Thus, Equation (3) becomes,
[tex](CM)^2-(AC)^2=5x[/tex]
[tex](4x)^2-(3x)^2=5x[/tex]
[tex]7x^2=5x[/tex]
[tex]x=\frac{5}{7}[/tex]
Therefore, AM=5x
[tex]AM=5(\frac{5}{7})=\frac{25}{7}[/tex].
