Find the area and perimeter of shaded region.

Question

(a) Find the area of the shaded region. (Round your answer to one decimal place.)

(b) Find the perimeter of the shaded region. (Round your answer to one decimal place.)

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Answer 1

For area Use formula of area of sector-area of ttiangle which is given by $\frac{\theta}{360}πr^2-\frac{1}{2}r^2sin\theta$ plug in the values. For length or perimeter we can uae length of sector +length of third side . $\frac{\theta}{360}2πr+length of third side$ . For length use components ie $l=rsin(54.5)=10sin(54.5)$ hope now im clear on it.

Answer 2

The area of the circle is $\pi r^2$ so the area of a sector of $1$ degree is $\frac{\pi r^2}{360}$ hence your sector has an area equal to $71\cdot\frac{\pi r^2}{360}\cdot $. You have now to rest the area $\frac{100 \sin (71)}{2}$ of the triangle in your figure.

Answer 3

The area is given by the area of the circular sector (⌔): $$\frac{\theta}{2} \times r^2$$ minus the are of triangle (△): $$\frac{b \times h}{2}=\frac{r \times r sin(\theta)}{2}.$$ For both equations, $\theta$ is the absolute value of the angle, given by $$\theta= \frac{71^{\circ}}{360^{\circ}}\times 2 \pi.$$ Thus, the required area is $$\frac{\theta}{2} \times r^2 - \frac{r \times r sin(\theta)}{2} = \frac{\theta-\sin(\theta)}{2} \times r^2 = \frac{\frac{71^{\circ}}{360^{\circ}}\times 2 \pi - \sin(\frac{71^{\circ}}{360^{\circ}}\times 2 \pi)}{2} \times (10[m])^2 = 14.7 \,m^2.$$

For the perimeter of this region, we just have to add the arc (⌒) lengh (fraction of the the total perimeter of the circumference) and the straight line length (given by the Law of Cosines): $$\theta \times r + \sqrt{r^2 + r^2 - 2 \times r \times r\times \cos(\theta)} = 24.0 \,m.$$