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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

4. Parallel Lines and Proportional Parts

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Exercise 56 Page 581

Recall the formula for the area of a circle.

A

Practice makes perfect

Let's begin with recalling the formula for the area of a circle. A=π r^2In this formula r is the radius of this circle. Since we are given that the area of a circle is 16 square meters, we can substitute this value for A in the above formula and solve for r.

A=π r^2

A= 16

16=π r^2

.LHS /π.=.RHS /π.

16/π=r^2

Rearrange equation

r^2=16/π

Next, we will take a square root of both sides of the equation. Remember that since r is always a positive number, we do not need to consider two cases when evaluating the square root of r^2.

r^2=16/π

sqrt(LHS)=sqrt(RHS)

sqrt(r^2)=sqrt(16/π)

SqrtPowToNumber

sqrt(a^2)=a

r=sqrt(16/π)

▼

Simplify right-hand side

sqrt(a/b)=sqrt(a)/sqrt(b)

r=sqrt(16)/sqrt(π)

Calculate root

r=4/sqrt(π)

Identity Property of Multiplication

r=4/sqrt(π)*1

Rewrite 1 as sqrt(π)/sqrt(π)

r=4/sqrt(π)*sqrt(π)/sqrt(π)

Multiply fractions

r=4sqrt(π)/(sqrt(π))^2

( sqrt(a) )^2 = a

r=4sqrt(π)/π

This corresponds with answer A.

Subchapter links

Exercises p.577-581