Question

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

Answer 1

AM = 25, AC = 15, CM = 20

Explanation:

The given parameters are;

In ΔACM, ∠C = 90°,
`\overline{CP}` ⊥
`\overline{AM}`, AP = 9, and PM = 16

`\overline{AC}`² +
`\overline{CM}`² =
`\overline{AM}`²

`\overline{AM}` =
`\overline{AP}` + PM = 9 + 16 = 25

`\overline{AM}` = 25

`\overline{AC}`² =
`\overline{AP}`² +
`\overline{CP}`² = 9² +
`\overline{CP}`²

∴
`\overline{AC}`² = 9² +
`\overline{CP}`²

Similarly we get;

`\overline{CM}`² = 16² +
`\overline{CP}`²

Therefore, we get;

`\overline{AC}`² +
`\overline{CM}`² = 9² +
`\overline{CP}`² + 16² +
`\overline{CP}`² =
`\overline{AM}`² = 25²

2·
`\overline{CP}`² = 25² - (9² + 16²) = 288

`\overline{CP}`² = 288/2 = 144

`\overline{CP}` = √144 = 12

From
`\overline{AC}`² = 9² +
`\overline{CP}`², we get

`\overline{AC}` = √(9² + 12²) = 15

`\overline{AC}` = 15

From,
`\overline{CM}`² = 16² +
`\overline{CP}`², we get;

`\overline{CM}` = √(16² + 12²) = 20

`\overline{CM}` = 20.