Exercise 11 Page 434

a Recall that two polygons are similar if and only if the two following conditions are met.

All corresponding angles are congruent.
All pairs of corresponding sides are proportional.

We are told to assume corresponding angles are congruent, so we only have to determine whether the second statement is true. Let's check if the ratio of the shorter sides of the $a i r$ $h o c k e y$ $t a b l e$ and the $h o c k e y$ $r i n k$ is equal to the ratio of their longer sides. Note that $85 feet = 1020$ inches and $200 feet = 2400$ inches.

Shorter Sides	Longer Sides
$\frac{4 0 . 8 in .}{1 0 2 0 in .}$	$\frac{9 6 in .}{2 4 0 0 in .}$
$\frac{1}{2 5}$	$\frac{1}{2 5}$

The ratios are equal. Since the air hockey table and the hockey rink are rectangles, we have enough information to conclude that the surfaces are similar.

b From Part A we know that the air hockey table and the hockey rink are similar. Therefore, we will find the ratio of their perimeters and the ratio of their areas. Let's start by determining the scale factor

k

of the

h o c k e y

r i n k

to the

a i r

h o c k e y

t a b l e .

k = \frac{4 0 . 8 in .}{8 5 ft} = \frac{4 0 . 8 in .}{1 0 2 0 in .} = \frac{1}{2 5}

We have that

k = \frac{1}{2 5} .

It means that the hockey rink is

25

times as large as the air hockey table. Recall that when two similar polygons have a scale factor of

k,

the ratio of their perimeters is

k

and the ratio of their areas is

k^{2} .

Ratio of Perimeters \frac{1}{2 5} Ratio of Areas (\frac{1}{2 5})^{2} = \frac{1}{6 2 5}

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Exercises p.434