Area of the shaded region [tex]$=36(\pi -2)[/tex] square cm
Perimeter of the shaded region [tex]=6 (\pi + 2\sqrt 2)[/tex] cm
Solution:
Radius of the quarter of circle = 12 cm
Area of the shaded region = Area of quarter of circle – Area of the triangle
[tex]$=\frac{1}{4} \pi r^2 - \frac{1}{2} bh[/tex]
[tex]$=\frac{1}{4} \pi \times 12^2 - \frac{1}{2} \times 12 \times 12[/tex]
[tex]$=36\pi -72[/tex]
[tex]$=36(\pi -2)[/tex] square cm.
Area of the shaded region [tex]$=36(\pi -2)[/tex] square cm
Using Pythagoras theorem,
[tex]AC^2=AB^2+BC^2[/tex]
[tex]AC^2=12^2+12^2[/tex]
[tex]AC^2=288[/tex]
Taking square root on both sides of the equation, we get
[tex]AC= 12\sqrt 2[/tex] cm
Perimeter of the quadrant of a circle = [tex]\frac{1}{4} \times 2\pi r[/tex]
[tex]$=\frac{1}{4} \times 2 \times \pi \times 12[/tex]
[tex]$=6 \pi[/tex] cm
Perimeter of the shaded region = [tex]6 \pi + 12\sqrt 2[/tex] cm
[tex]=6 (\pi + 2\sqrt 2)[/tex] cm
Hence area of the shaded region [tex]$=36(\pi -2)[/tex] square cm
Perimeter of the shaded region [tex]=6 (\pi + 2\sqrt 2)[/tex] cm