Exercise 55 Page 637

In our exercise, we are given that the perimeter of square 2 is 200 units, and the perimeter of square 1 is 150 units. We are asked to find the perimeter of square 3.

We will start with evaluating the side length of squares 1 and 2. Notice that a side length of a square can be evaluated by dividing its perimeter by 4. Let a and b represent the side lengths of squares 1 and 2. a=150/4=37.5 b=200/4=50 We can use these values to evaluate the side length of square 3, which will be represented by c. Since the triangle formed by these three squares is right, we can use the Pythagorean Theorem to find the value of c. a^2+ b^2=c^2 37.5^2+ 50^2=c^2 Let's solve the above equation. Remember that since c is a side length, it is always a positive number, so we do not need to consider two cases when taking a square root of c^2.

37.5^2+50^2=c^2

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Solve for c

CalcPow

Calculate power

1406.25+2500=c^2

AddTerms

Add terms

3906.25=c^2

RearrangeEqn

Rearrange equation

c^2=3906.25

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(c^2)=sqrt(3906.25)

SqrtPowToNumber

sqrt(a^2)=a

c=sqrt(3906.25)

CalcRoot

Calculate root

c=62.5

The side length of square 3 is 62.5 units. Therefore, we can find the perimeter of this square by multiplying this value by 4, as there are four congruent sides in each square. 62.5* 4=250 The perimeter of square 3 is 250 units.