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