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Dovator [93]
3 years ago
11

A rope of length 18 feet is arranged in the shape of a sector of a circle with central angle O radians, as shown in the

Mathematics
1 answer:
creativ13 [48]3 years ago
6 0

Answer:

A(\theta)=\frac{162 \theta}{(\theta+2)^2}

Step-by-step explanation:

The picture of the question in the attached figure

step 1

Let

r ---> the radius of the sector

s ---> the arc length of sector

Find the radius r

we know that

2r+s=18

s=r \theta

2r+r \theta=18

solve for r

r=\frac{18}{2+\theta}

step 2

Find the value of s

s=r \theta

substitute the value of r

s=\frac{18}{2+\theta}\theta

step 3

we know that

The area of complete circle is equal to

A=\pi r^{2}

The complete circle subtends a central angle of 2π radians

so

using proportion find the area of the sector by a central angle of angle theta

Let

A ---> the area of sector with central angle theta

\frac{\pi r^{2} }{2\pi}=\frac{A}{\theta} \\\\A=\frac{r^2\theta}{2}

substitute the value of r

A=\frac{(\frac{18}{2+\theta})^2\theta}{2}

A=\frac{162 \theta}{(\theta+2)^2}

Convert to function notation

A(\theta)=\frac{162 \theta}{(\theta+2)^2}

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