Find the area of the shaded region. Round your answer to the nearest hundredth.

12=side length

The area of the shaded region is about — square units.

How do you find the area of the shaded region? Thank you!

Find The Area Of The Shaded Region. Round Your Answer To The Nearest Hundredth.12=side LengthThe Area

Answers

Answer 1

Answer:

Step-by-step explanation:

Divide 360 by the number of sides in a regular polygon to find θ

θ = 360/5 = 72°

Use law of cosines to find the radius of the circle which equals the common length sides of the isosceles triangle

12² = R² + R² - 2•R•Rcos72

12² = 2R² - 2R²cos72

12² =  2R²(1 - cos72)

R² = 12² / 2(1 - cos72)

R² = 104.1993788...

R = 10.2078...

The height of the isosceles triangle is

h = Rcos(θ/2) = 10.2078cos(72/2) = 8.258291...

The shaded area is the area of the circle minus the area of the pentagon.

A = π(104.1993) - 5(½(12)(8.258291))

A = 327.35200... - 247.74874...

A = 79.6032575...

A = 79.60 units²

Find The Area Of The Shaded Region. Round Your Answer To The Nearest Hundredth.12=side LengthThe Area
Answer 2

The Area of the shaded region is A = 79.60 units²

What is law of Cosine?

A planar triangle's side's square is equal to the sum of the squares of the other sides minus the product of those sides and the cosine of the angle that separates them by two.

Given:

length= 12

So, Divide 360 by the number of sides in a regular polygon

θ = 360/5

θ = 72°

Use law of cosines

12² = R² + R² - 2 x R x Rcos72

12² = 2R² - 2R² cos72

12² =  2R²(1 - cos72)

R² = 12² / 2(1 - cos72)

R² = 104.1993788...

R = 10.2078...

Thus, the height of the isosceles triangle is

h = R cos(θ/2)

   = 10.2078cos(72/2)

   = 8.258291...

and, the shaded area is

A = π(104.1993) - 5(½)(12)(8.258291))

A = 327.35200 - 247.74874

A = 79.6032575

A = 79.60 units²

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Answers

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Answers

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Answers

The angle is increased in approximately 0.095 radians (5.443°) for [tex]x = 0.45\,m[/tex].

Based on the statement we construct the geometric diagram, by definition of tangent we have expressions for the initial and final angles ([tex]\theta_{1}[/tex], [tex]\theta_{2}[/tex]), in radians, of the figure:

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[tex]\tan \theta_{1} = \frac{y_{1}}{x_{1}}[/tex] (1)

Final triangle

[tex]\tan \theta_{2} = \frac{y_{2}}{x_{2}}[/tex] (2)

By using (2), the equivalence [tex]\theta_{2} = \theta_{1}+\Delta \theta[/tex] and trigonometric identities we have the following expression:

[tex]\frac{y_{2}}{x_{2}} = \frac{\tan \theta_{1}+\tan \theta_{2}}{1-\tan \theta_{1}\cdot \tan \theta_{2}}[/tex] (3)

By (1), we simplify the expression:

[tex]\frac{y_{2}}{x_{2}} = \frac{\frac{y_{1}}{x_{1}} + \tan \Delta \theta}{1-\frac{y_{1}}{x_{1}}\cdot \tan \Delta \theta}[/tex] (3b)

If [tex]0 \le \Delta \theta \le \frac{\pi}{6}[/tex], then we can use the following approximation:

[tex]\tan \Delta\theta \approx \Delta \theta[/tex] (4)

Then, we reduce (3b) into an entirely algebraic expression:

[tex]\frac{y_{2}}{x_{2}} = \frac{\frac{y_{1}}{x_{1}}+\Delta \theta }{1-\frac{y_{1}}{x_{1}}\cdot \Delta \theta }[/tex] (3c)

Where [tex]y_{2} = \sqrt{r^{2}-x_{2}^{2}}[/tex].

Now we clear [tex]\Delta \theta[/tex] within the formula:

[tex]\Delta \theta = \frac{\frac{y_{2}}{x_{2}}-\frac{y_{1}}{x_{1}}}{\frac{y_{1}}{x_{1}}\cdot \left(1+\frac{y_{2}}{x_{2}} \right) }[/tex]  (5)

If we know that [tex]x_{1} = y_{1} = 0.5[/tex], [tex]r = 0.707[/tex] and [tex]x_{2} = 0.45[/tex], then we estimate the angle change:

[tex]y_{2} = \sqrt{0.707^{2}-0.45^{2}}[/tex]

[tex]y_{2} \approx 0.545[/tex]

[tex]\Delta \theta = \frac{\frac{0.545}{0.45}-1 }{1\cdot \left(1+\frac{0.545}{0.45} \right)}[/tex]

[tex]\Delta \theta = 0.095[/tex]

As [tex]\Delta \theta < \frac{\pi}{6}[/tex], then the result seems to be reasonable. The angle is increased in approximately 0.095 radians (5.443°) for [tex]x = 0.45\,m[/tex].

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