Exercise 21 Page 348

Practice makes perfect

a Let's start by connecting the towns with segments that form a triangle. The triangle that connects the towns has vertices at B(3,9), W(3,3), and O(9,3) and is a right triangle.

We need the strength of the signal to be the same for each town. Therefore, we need to find a point that is equidistant from the three towns, which are the vertices of △ BWO. The point that satisfies this condition is the circumcenter, which is the concurrency point of the perpendicular bisectors.

From the figure above, we have that the circumcenter is the point C(7,6). Hence, the cell phone tower must be located at (7,6).

b Since the tower is placed at the circumcenter of △ BWO, it is equidistant from the three towns. Therefore, it is enough to find the distance between the tower and one town only. Let's compute the distance between the tower and Bayville using the Distance Formula.

d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2) We can substitute the coordinates of C and B into the equation above.

d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)

SubstitutePoints

Substitute ( 7,6) & ( 3,9)

d = sqrt(( 7- 3)^2 + ( 6- 9)^2)

▼

Simplify right-hand side

SubTerms

Subtract terms

d=sqrt(4^2+(- 3)^2)

CalcPow

Calculate power

d=sqrt(16+9)

AddTerms

Add terms

d=sqrt(25)

CalcRoot

Calculate root

d=5

The distance from the tower to Bayville, and therefore also to Westfield and Oxboro, is 5 miles.

c To determine if other towns benefit from the tower, we can draw a circle centered at point C and with a radius of 5 (the distance from the tower to the three towns). This circle represents the region reached by the signal.

We can see that Seabury and Medfield are included in the circle. This means that both Seabury and Medfield benefit from the cell phone tower.

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Exercises p.346-348

Exercise 21 Page 348

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