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Core Connections Integrated II, 2015

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Core Connections Integrated II, 2015 View details

4. Section 8.4

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Exercise 95 Page 467

Practice makes perfect

a From the exercise, we know that the greatest area that can be enclosed with a given perimeter is a circle. Therefore, if we have a string that is 24 cm long, the greatest area that can be enclosed is a circle with a perimeter of 24 cm.

To calculate the area of this circle we must know its radius. We can find the radius by substituting the circumference in the formula C=2rπ and solving for r.

C=2rπ

C= 24

24=2rπ

▼

Solve for r

.LHS /2π.=.RHS /2π.

24/2π=r

a/b=.a /2./.b /2.

12/π=r

Rearrange equation

r=12/π

When we know the radius we can determine the area of the circle.

A=π r^2

r= 12/π

A=π( 12/π)^2

▼

Simplify right-hand side

(a/b)^m=a^m/b^m

A=π144/π^2

MoveLeftFacToNum

a*b/c= a* b/c

A=144π/π^2

a/b=.a /π./.b /π.

A=144/π

b Similar to Part A, we have to first determine the radius of a circle with a circumference of 18π cm.

C=2rπ

C= 18π

18π=2rπ

▼

Solve for r

.LHS /2π.=.RHS /2π.

18π/2π=r

a/b=.a /2π./.b /2π.

9=r

Rearrange equation

r=9

When we know the radius we can determine the area of the circle.

A=π r^2

r= 9

A=π(9)^2

▼

Simplify right-hand side

Calculate power

A=π * 81

CommutativePropMult

Commutative Property of Multiplication

A=81π

A=81π

Subchapter links

8.4.1 A Special Ratio p.460-462

8.4.2 Arcs and Sectors p.466-468

8.4.3 Circles in Context p.471-473