McGraw Hill Integrated II, 2012
MH
McGraw Hill Integrated II, 2012 View details
5. The Triangle Inequality
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Exercise 48 Page 451

Practice makes perfect
a

Since the three locations are not collinear, we can draw the triangle formed by them.

By applying the Triangle Inequality Theorem, we get the following three inequalities. HP + PS &> HS HS + PS &> HP HS + HP &> PS Next, let's substitute the measures of the sides and solve each inequality for HS. 3/4 + 1.5 > HS & ⇒ 2.25 > HS HS + 1.5 > 3/4 & ⇒ HS > -0.75 HS + 3/4 > 1.5 & ⇒ HS > 0.75 From the first and third inequalities we get two bounds for the distance between our house and the shopping center. 0.75 mi < HS < 2.25 mi In conclusion, the distance from our house to the shopping center must be greater than 0.75 miles and shorter than 2.25 miles.

b

Since the three places are collinear, there are two possible options for the location of our house, namely whether or not it could be between the park and the shopping center. Below, we draw both situations.

By applying the Segment Addition Postulate to the first case, we get the following relation. HS = HP + PS ⇒ HS &= 3/4 + 1.5 ⇒ HS &= 2.25 mi In the first case, the distance from our house to the shopping center is 2.25 miles. Let's use the same reasoning for the second case. PS = HS + HP ⇒ HS &= PS - HP ⇒ HS &= 0.75 mi In the second case, the distance from our house to the shopping center is 0.75 miles.