Exercise 34 Page 624

a The given function expresses the perimeter

P

of a square in terms of the area

A

of the square.

P = 4 A

We want to graph this square root function. To do so, let's first make a table of values. Note that the area cannot be negative.

Now, we will plot these ordered pairs on a coordinate plane and draw a smooth curve that connects them.

b We will determine the perimeter of a square with an area of

225

square meter using the given function. To do this, let's substitute

A = 225

into the equation and then solve it for

P .

P = 4 A

$A = 225$

P = 4225

Calculate root

P = 4 (15)

P = 60

c Finally, we will find when the perimeter and the area will have the same value. To find this, let's first consider a square with sides of length

a .

In this case, the

perimeter

of the square is

4 a

while its

area

a^{2} .

Now, let's assume that these values are equal.

4 a = a^{2}

We will solve this equation for

a .

4 a = a^{2}

$LHS - a^{2} = RHS - a^{2}$

4 a - a^{2} = 0

Factor out $a$

a (4 - a) = 0

a (4 - a) = 0

Use the Zero Product Property

a = 0 4 - a = 0 (I) (II)

$(II):$ $LHS + a = RHS + a$

a = 0 4 = a

$(II):$ Rearrange equation

a = 0 a = 4

The solutions to the equation are

a = 0

and

a = 4 .

However, since

a

represents a length, it should be greater than

0 .

. Therefore, the perimeter and the area of a square will be the same when the square has a side length of

4 .

Check Your Understanding p.624

Practice and Problem Solving p.624-625

H.O.T. Problems p.625

Standardized Test Practice p.625

Spiral Review p.625

Skills Review p.625