Answer:
The answer is below
Step-by-step explanation:
The shape of the figure attached is the shape of a sector with an angle of 90° (quarter of a circle).
From the sector, AB = AC = radius of the sector (r) = 12. Therefore:
[tex]Area\ of\ sector=\frac{\theta}{360}*\pi r^2 = \frac{90}{360}*\pi * (12)^2 = 0.25 * \pi * 144=36\pi\\\\Area\ of\ triangle\ ABC=\frac{1}{2}*base *height= \frac{1}{2}*AB*BC=\frac{1}{2}*12*12=72\\\\Area \of\ shaded\ region = Area\ of\ sector-Area\ of\ triangle=36\pi-72\\\\Area \of\ shaded\ region =36\pi-72[/tex]
From the triangle: AC² = AB² + BC²
AC² = 12² + 12² = 144 + 144
AC² = 288
AC=√288 = 12√2
[tex]Perimeter\ of\ sector=\frac{\theta}{360}*2\pi r = \frac{90}{360}*2\pi * (12) = 0.25 * 2\pi * 12=6\pi\\\\Perimeter \of\ shaded\ region = Perimeter\ of\ sector+AC=6\pi + 12\sqrt{2}[/tex]
