Answer:
Step-by-step explanation:
Given that ACM is a right angled triangle.
Angle C =90 and also angle P =90
AC:CM=3:4
[tex]MP-AP=1[/tex] ... i
Let AC=3k and CM=4k
Then by pythagorean theorem
[tex]AM^2=(3k)^2+(4k)^2\\AM=5k[/tex]
From i, we get
[tex]AM=AP+MP = AP+AP+1=5k[/tex]
[tex]AP = \frac{5k-1}{2}[/tex]
Triangles APC and ACM are similar, since
angle A is common and one angle is right angle
AP/AC = AC/AM (Since triangles are similar)
[tex]AC^2 = 9k^2 = AP(AM) = (\frac{5k-1}{2} )AM\\=5k(\frac{5k-1}{2} )\\25k^2-5 = 18k^2\\7k^2=5\\k^2 =5/7\\k =\sqrt \frac{5}{7} \\AM=\sqrt \frac{5}{7}[/tex]
