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Given ACM, angle C=90º. AP=9, PM=12. Find AC, CM, AM.

Given ACM Angle C90º AP9 PM12 Find AC CM AM class=

Sagot :

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.

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