SOLUTION: A rectangular lot 80 m x 40 m long is divided into two areas by an arc whose center is at the mid-point of the shorter side. If the radius of the arc is 30 m, what is the rat

Question 1182637: A rectangular lot 80 m x 40 m long is divided into two
areas by an arc whose center is at the mid-point of the
shorter side. If the radius of the arc is 30 m, what is the
ratio of the area of the smaller part to the area of the
bigger part?
Answer by greenestamps(13200) (Show Source): You can put this solution on YOUR website!

Here is a sketch....

Rectangle ABCD has side lengths 40 and 80; the circular arc is centered at E, the midpoint of AB. AE=20 and EG=30; the Pythagorean Theorem gives us AG=10*sqrt(5).

The area of the smaller portion of the lot is the area of the two congruent triangular regions EAG and EBF, plus the area of the circle sector EFG.

The areas of the triangles are easy to calculate.

For the area of the circle sector, you need to find what fraction of a circle the sector is, so you need to find the measure of angle FEG. Angle FEG is twice angle AGE; angle AGE is arcsin(20/30). Then the area of the sector is that fraction, times the area of the whole circle, which is (pi)r^2 = 900pi.

I leave it to you to do the calculations....

To check your calculations, here are my figures, slightly rounded:
Angle FEG: 83.62 degrees
Area of circle sector: 656.76 m^2
Area of smaller portion of lot (circle sector plus two triangles): 1103.97 m^2
Area of larger portion: 3200-1103.97 = 2096.03 m^2
Ratio of smaller to larger: 1104:2096

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