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Please help me out please

Please help me out please-example-1

1 Answer

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The purple area is the sum of the circle sector and the triangle:


A = A_s+A_t

Let's compute them one at the time:

The sector is identified by an angle of 260°, because it is the remainder of a 100° angle.

We can build this simple proportion


\text{total area}/\text{total angle}=\text{sector area}/\text{sector angle}

The area of the circle is
\pi r^2, so we have


\pi 8.35^2/ 360= A_s/ 260

Solving for the sector area, we have


A_s = (\pi\cdot8.35^2\cdot260)/(360) = (8.35^2\cdot 13\pi)/(18)

The triangle is an isosceles triangle, because two of the sides are radii. This means that the height is also a bisector, so we can cut the triangle in two 90-50-40 triangles.

Using the law of sines, we can deduce that the height is


8.35\sin(40)

And half the base is


8.35\sin(50)

So, the area of the triangle is


A_t = 8.35^2\sin(40)\sin(50)

So, the purple area is


A = A_s+A_t =\\(8.35^2\cdot 13\pi)/(18)+8.35^2\sin(40)\sin(50) = 8.35^2\left((13)/(18)\pi+\sin(40)\sin(50)\right)\approx 192.5

User Yuriy Vasylenko
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