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 dra
wn 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 $CP$?
1 answer:
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
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