Find the area and perimeter of a segment of a circle

Question

Find the area and perimeter of the shaded region in the figure

My work

$$ \begin{align} \text{Area} &= \frac{r^2}{2} \theta - \frac12 r^2 \sin(\theta)\\ &= \frac{8^2}{2} \frac{37 \pi}{180}- \frac128^2\sin\left(\frac{37 \pi}{180}\right)\\ &= 41.3 \end{align} $$

Answer 1

Your work on the area is correct except the $37$'s should be $74$'s. There is no need to divide the angle by $2$. (Think through what the answer would be if the angle were $90$ instead of $74$. You'd be looking at a quarter circle with a right isosceles triangle removed, so the area would be $r^2({\pi\over4}-{1\over2})$.)

Answer 2

The solution is simple, if you consider that the perimeter of the shaded region is simply the sum of the arc length and the chord length. The arc length = Total Perimeter * (74/360) = 2*pi*8*74/360 = 10.332 m. The chord length = 2*r*sin(Angle/360) = 16*sin(74*pi/360) The sum of these two lengths is the perimeter of the shaded area.

Answer 3

For perimeter, use the sine law for the line and use angle ratio for the arc.

The bottom angle of the triangle is $53$ degree so

line $={8\over sin(53)}\cdot sin(74)=9.629$

arc $=2\pi8\cdot{74\over360}=10.332$

Hence the total is $19.961$.