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Core Connections Integrated II, 2015

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Core Connections Integrated II, 2015 View details

2. Section 9.2

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Exercise 71 Page 508

The height of the parallelogram is the hypotenuse of a 30^(∘)-60^(∘)-90^(∘) triangle.

231.28 square units

Practice makes perfect

To find the area of the figure we should calculate the area of the sector and parallelogram separately, then add the results. We can immediately calculate the area of the sector by calculating the area of a circle with the given radius and multiplying by the ratio of the sector's central angle to 360^(∘).

A_S=π r^2* θ/360^(∘)

r= 12, θ= 60^(∘)

A_S=π( 12)^2* 60^(∘)/360^(∘)

▼

Simplify right-hand side

a/b=.a /60^(∘)./.b /60^(∘).

A_S=π(12)^2* 1/6

Calculate power

A_S=π144* 1/6

CommutativePropMult

Commutative Property of Multiplication

A_S=144π* 1/6

MoveLeftFacToNumOne

a* 1/b= a/b

A_S=144π/6

a/b=.a /6./.b /6.

A_S=24π

The area of the sector is 24π. To calculate the area of the parallelogram we need to know its height. To find this dimension we will add some information to the diagram.

As we can see, the height of the parallelogram is also a side in a 30^(∘)-60^(∘)-90^(∘) triangle. In such a triangle the shorter leg is half the length of the hypotenuse and the longer leg is sqrt(3) times the shorter leg.

We can calculate the area of the parallelogram.

A_p=bh

b= 15, h= 6sqrt(3)

A_p= 15( 6sqrt(3))

Multiply

A_p=90sqrt(3)

Finally, we will add the two areas to determine the area of the figure. A=24π +90sqrt(3) ≈ 231.28 square units

Subchapter links

9.2.1 Quadratic Applications with Inequalities p.502-503

9.2.2 Solving Systems of Equations p.507-508