Exercise 35 Page 647

To find the area of the shaded region, we will subtract the area of the small hexagon from the area of the big hexagon. To do so, we will use the area formula for a regular polygon. A= 1/2ap We will first find the apothem and perimeters of the hexagons, then their areas.

Finding Apothems and Perimeters

We will start by finding the central angle of a regular hexagon. The formula for the measure of the central angle of a n-sided regular polygon is 360n. Let's find the central angle of a hexagon. 360/n ⇒ 360/6 = 60 The measure of the central angle of a regular hexagon is 60^(∘). Since the apothem bisects this angle, the measure of the angle between the apothem and the radius of the hexagon is 30^(∘). Let's show the apothem of each hexagon.

Using the trigonometric ratios of 30^(∘), we can find the apothems, a and b, and half the side lengths, x and y.

	Hexagon 1	Hexagon 2
Cosine Ratio	cos 30^(∘)=a/6 ⇕ a= 6 cos30 ^(∘)	cos 30^(∘)=b/12 ⇕ b= 12cos30 ^(∘)
Sine Ratio	sin 30^(∘)=x/6 ⇕ x=6 sin30 ^(∘)	sin 30^(∘)=y/12 ⇕ y=12sin 30^(∘)

Next, we will find the perimeters of the hexagons. Since x and y represent half the side lengths, we need to multiply them by 12 to find the perimeters of the hexagons. rc Hexagon1: & 12 * x = 12* 6sin30^(∘) [0.5em] & = 72 sin30^(∘) [0.5em] [-0.6em] Hexagon2: & 12 * y =12* 12sin30^(∘) [0.5em] & = 144 sin30^(∘)

Areas of Hexagons

Now, we can find the area A_1 of Hexagon 1 by substituting 6cos 30^(∘) for a and 72sin30^(∘) for p. Recall trigonometric ratios of 30^(∘), cos30^(∘) = sqrt(3)2 and sin30^(∘)= 12.

A_1=1/2ap

SubstituteII

a= 6cos 30^(∘), p= 72sin30^(∘)

A_1=1/2 * 6cos 30^(∘) * 72sin30^(∘)

▼

Evaluate right-hand side

Multiply

A_1=1/2 * 432 * cos30^(∘) * sin30^(∘)

MoveRightFacToNumOne

1/b* a = a/b

A_1=432/2 * cos30^(∘) * sin30^(∘)

\ifnumequal{30}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{30}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{30}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{30}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{30}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{30}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{30}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{30}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{30}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{30}{360}{\cos\left(360^\circ\right)=1}{}

A_1=432/2 * sqrt(3)/2 * sin30^(∘)

\ifnumequal{30}{0}{\sin\left(0^\circ\right)=0}{}\ifnumequal{30}{30}{\sin\left(30^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{45}{\sin\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\sin\left(60^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{90}{\sin\left(90^\circ\right)=1}{}\ifnumequal{30}{120}{\sin\left(120^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{135}{\sin\left(135^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\sin\left(150^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{180}{\sin\left(180^\circ\right)=0}{}\ifnumequal{30}{210}{\sin\left(210^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{30}{225}{\sin\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\sin\left(240^\circ\right)=\text{-} \dfrac {\sqrt 3}2}{}\ifnumequal{30}{270}{\sin\left(270^\circ\right)=\text{-}1}{}\ifnumequal{30}{300}{\sin\left(300^\circ\right)=\text{-}\dfrac {\sqrt 3}2}{}\ifnumequal{30}{315}{\sin\left(315^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\sin\left(330^\circ\right)=\text{-} \dfrac 1 2}{}\ifnumequal{30}{360}{\sin\left(360^\circ\right)=0}{}

A_1=432/2 * sqrt(3)/2 * 1/2

MultFrac

Multiply fractions

A_1=432 sqrt(3)/8

CalcQuot

Calculate quotient

A_1 = 54 sqrt(3)

Similarly, the area A_2 of Hexagon 2 is half the product of its apothem b and perimeter p.

A_2=1/2bp

SubstituteII

b= 12cos 30^(∘), p= 144sin30^(∘)

A_2=1/2 * 12cos 30^(∘) * 144sin30^(∘)