Exercise 33 Page 496

In order to find the length of the side of the square, we need to focus on where to place the diameter, how the square's sides relate to the circle, and the numerical relationship between the square's sides and the circle's diameter.

Diameter

The critical points are the places where the objects intersect. In this case, we can draw our diameter so that it meets where the circle and square touch.

We can see that the diagonal of the square is the diameter of the circle.

Sides of the Square

The square has sides length x. We can see that the diagonal of the square forms two right triangles, where the square's adjacent sides are perpendicular and become the legs of the triangles.

Relationship between Square's Sides and Circle's Diameter

Since we have a right triangle, we know that the sum of the squares of the square's legs is equal to the square of the hypotenuse. Let's use the Pythagorean Theorem to find the side lengths of the square.

a^2+b^2=c^2

SubstituteValues

Substitute values

x^2 + x^2 = 6^2

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Solve for x

CalcPow

Calculate power

x^2 + x^2 = 36

AddTerms

Add terms

2x^2 = 36

DivEqn

.LHS /2.=.RHS /2.

x^2 = 18

SqrtEqn

sqrt(LHS)=sqrt(RHS)

x = ± sqrt(18)

SplitIntoFactors

Split into factors

x = ± sqrt(9*2)

CalcRoot

Calculate root

x = ± 3sqrt(2)

UseCalc

Use a calculator

x ≈ ± 4.2

We are working with length, so we can disregard the negative solution. The square's side is sqrt(2) times half the length of the diameter. The largest value of x that fits into the hoop is about 4.2 inches.