body{margin:0;} noscript{z-index: 10000; position: fixed; display: block; height: 100%; width: 100%; background: #fff;} noscript .fa{font-size: 40px; display:block; color: #daa520;} noscript div{margin-top: 6%; padding-left: 20px; padding-right: 20px; text-align: center;} ! You must have JavaScript enabled to use this site.

Dashboard

McGraw Hill Integrated II, 2012 View details

4. Parallel Lines and Proportional Parts

Finish the assignment

Continue to next subchapter

Exercise 56 Page 581

Recall the formula for the area of a circle.

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

Substitute

A= 16

16=π r^2

DivEqn

.LHS /π.=.RHS /π.

16/π=r^2

RearrangeEqn

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

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(r^2)=sqrt(16/π)

SqrtPowToNumber

sqrt(a^2)=a

r=sqrt(16/π)

▼

Simplify right-hand side

SqrtQuot

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

r=sqrt(16)/sqrt(π)