What is the area of the shaded region, knowing that arc AC is 1/4 of the circumference centered on D
I can't finish??
$GB=\frac{DB}{2} = \frac{a\sqrt2}{2}\\ \triangle ABG:(\frac{a\sqrt2}{2})^2=a^2+(a\sqrt2)^2-2a^2\sqrt2cos\theta \implies cos \theta =\frac{5\sqrt2}{8}\\ DE = acos\theta =\frac{5a\sqrt2}{8}\\\triangle DEG: a^2=EG^2+DE^2 \therefore EG = \frac{a}{4}\sqrt{\frac{7}{2}}\\ S\triangle DGH = 2S\triangle DGE = 2. \frac{1}{2}.\frac{5a\sqrt2}{8}.\frac{a}{4}.\sqrt{\frac{7}{2}}=\frac{5a^2\sqrt7}{32}\\ S(ACB):S_{ABCD}-S_{sect DAC}=a^2-\frac{\pi a^2}{4}=\frac{a^2(4-\pi)}{4}$
