Question

大正方形ABCD内有一小正方形DEFG,对角线DF为6厘米,已知小正方形DEFG想东北方向平移3 厘米,就得到正方形D&

Answer 1

EG=6√2,

S=6×6+6√2×3=36+18√2.

或BD=6√2+3,

S正方形ABCD=1/2×(6√2+3)^2=1/2(81+36√2),

AE=3÷√2=3√2/2,

S=S正方形ABCD-SΔAEE‘-SΔCGG’

=1/2(81+36√2)-2×1/2×(3√2/2)^2

=36+18√2.