Question

Suppose that we have a right triangle $ABC$ with the right angle at $B$ such that $AC = \sqrt{61}$ and $AB = 5.$ A circle is drawn with its center on $AB$ such that the circle is tangent to $AC$ and $BC.$ If $P$ is the point where the circle and side $AC$ meet, then what is $CP$?

Answer 1

Answer:

  CP = 6

Step-by-step explanation:

The length of segment BC is given by the Pythagorean theorem:

  AC² = AB² +BC²

  (√61)² = 5² + BC² . . . . . fill in the given numbers

  61 -25 = BC² = 36 . . . . .subtract 25

  BC = 6 . . . . . . . . . . . . . . take the square root

Since the center of the circle is on AB and is tangent to BC, it must pass through point B. That is, segment BC of length 6 is one of the tangent lines from point C. The other one, to point P, must be the same length, so ...

  CP = 6