Answer:
Part 1) The area of the shaded region is [tex]576\ cm^{2}[/tex]
Part 2) The area of the shaded region is [tex]A=(18+4.5\pi)\ cm^{2}[/tex]
Step-by-step explanation:
Part 1) Figure N 1
I assume that the figure ABCD is a square
we know that
The area of the shaded region is equal to the area of the square ABCD minus the area of semicircle BC plus the area of semicircle AD
therefore
The area of the shaded region is equal to the area of the square ABCD
The area of the square is
[tex]A=24^{2}=576\ cm^{2}[/tex]
Part 2) Figure N 2
I assume that the triangle ABC is a right isosceles triangle
so
AB=BC
AB ⊥ BC
The area of the shaded region is equal to the area of triangle plus the area of semicircle
A) Find the area of the triangle ABC
The area of triangle is
[tex]A=\frac{1}{2}(AB)(BC)[/tex]
substitute
[tex]A=\frac{1}{2}(6)(6)[/tex]
[tex]A=18\ cm^{2}[/tex]
B) Find the area of semicircle
The area of semicircle is equal to
[tex]A=\frac{1}{2}\pi r^{2}[/tex]
we have
[tex]r=BC/2=6/2=3\ cm[/tex] -----> the radius is half the diameter
substitute
[tex]A=\frac{1}{2}\pi (3)^{2}[/tex]
[tex]A=4.5\pi\ cm^{2}[/tex]
therefore
The area of the figure is equal to
[tex]A=(18+4.5\pi)\ cm^{2}[/tex]
