I'm reviewing for finals but this geometry question got me.
The figure above is an isosceles right triangle with a semicircle attached
a) Show that the area of the figure is about 5.1 square centimeters when L = 2 cm
b) Show that L is about 3.9 cm when the area of the figure is 20 cm
c) Express the distance L as a function of the area of the figure
This is my work so far, but I didn't get full points on it on the test we had way back. Can anyone point out what I did wrong?
A) r = sq. root (2^2 + 2^2) = sq. root 8 = 2(sq. root 2)
Area of triangle = 1/2 (2 sq. root 2)(sq. root 2) = 2
Area of semicircle = [pi(2 sq. root 2)^2]/2 = 12.5
Total area = 14.5 square cm
B) 20 = 1/2(sq. root 2l^2)(sq. root 2l^2/2) + [pi(sq. root 2l^2)2]/2
20 = 1/2 (2l^2/2) + [pi(2l^2)]/2
80 = 2l^2 + 2pi(2l^2)
2l^2 = 10.9
L = 2.3 cm
C) A(L) = 2l^2/4 + pi(2l^2)/2