Answer:
AM = 25, AC = 15, CM = 20
Step-by-step explanation:
The given parameters are;
In ΔACM, ∠C = 90°, [tex]\overline{CP}[/tex] ⊥ [tex]\overline{AM}[/tex], AP = 9, and PM = 16
[tex]\overline{AC}[/tex]² + [tex]\overline{CM}[/tex]² = [tex]\overline{AM}[/tex]²
[tex]\overline{AM}[/tex] = [tex]\overline{AP}[/tex] + PM = 9 + 16 = 25
[tex]\overline{AM}[/tex] = 25
[tex]\overline{AC}[/tex]² = [tex]\overline{AP}[/tex]² + [tex]\overline{CP}[/tex]² = 9² + [tex]\overline{CP}[/tex]²
∴ [tex]\overline{AC}[/tex]² = 9² + [tex]\overline{CP}[/tex]²
Similarly we get;
[tex]\overline{CM}[/tex]² = 16² + [tex]\overline{CP}[/tex]²
Therefore, we get;
[tex]\overline{AC}[/tex]² + [tex]\overline{CM}[/tex]² = 9² + [tex]\overline{CP}[/tex]² + 16² + [tex]\overline{CP}[/tex]² = [tex]\overline{AM}[/tex]² = 25²
2·[tex]\overline{CP}[/tex]² = 25² - (9² + 16²) = 288
[tex]\overline{CP}[/tex]² = 288/2 = 144
[tex]\overline{CP}[/tex] = √144 = 12
From [tex]\overline{AC}[/tex]² = 9² + [tex]\overline{CP}[/tex]², we get
[tex]\overline{AC}[/tex] = √(9² + 12²) = 15
[tex]\overline{AC}[/tex] = 15
From, [tex]\overline{CM}[/tex]² = 16² + [tex]\overline{CP}[/tex]², we get;
[tex]\overline{CM}[/tex] = √(16² + 12²) = 20
[tex]\overline{CM}[/tex] = 20.
