McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. Parts of Similar Triangles
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Exercise 24 Page 588

Right fielder, see solution.

Practice makes perfect
Let's begin with recalling the Triangle Angle Bisector Theorem. An angle bisector in a triangle separates the opposite side into two segments that are proportional to the lengths of the other two sides. Now let's look at the given picture. Notice that the lengths of CH and RH are the distances that the center fielder and right fielder have to the ball.

Since we know that BH is an angle bisector of ∠ B, we can write a proportion. CH/RH=BC/BR Next, we can evaluate the fraction on the right-hand side by substituting 202 for BC and 197 for BR. CH/RH=202/197 ⇓ CH/RH≈1.03 Since this ratio is greater than 1, the length of CH is greater than the length of RH. Therefore, the right fielder is closer to the ball than the center fielder.