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Help to this explanation , this is a review practice.

Help to this explanation , this is a review practice.-example-1
User Thomasb
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To solve this problem we will use the Pythagorean theorem twice:


c^2=a^2+b^2,

where c is the length of the hypotenuse and, a, and b are the lengths of the legs.

Now, from the triangle OPD, we get ( we will omit the units to simplify the calculations):


OP^2=1.2^2+OD^2.

Notice that OD is a radius of the circle, therefore OD=1, and OP=OC+CP=1+CP, then we can rewrite the above equation as:


\begin{gathered} 1+CP=√(1.2^2+1^2), \\ CP=\sqrt{(11)/(5)}-1\approx0.56. \end{gathered}

Analogously, we get that:


\begin{gathered} AP^2=AD^2+1.2^2=2^2+1.2^2, \\ AB+BP=1.7+BP=√(4+1.2^2), \\ BP=√(4+1.2^2)-1.7\approx0.63. \end{gathered}

Therefore,


\begin{gathered} Distance(P,C)\approx0.56\text{ }in, \\ Distance(P,B)\approx0.63\text{ in.} \end{gathered}

Answer:


Point\text{ C.}

User SSteve
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