Answer:
Part 1)
a) The area is [tex]A=36(\pi-2)\ cm^2[/tex]
b) The perimeter is [tex]P=6(\pi+2\sqrt{2})\ cm[/tex]
Part 2)
a) The area is [tex]A=576\ cm^2[/tex]
b) The perimeter is [tex]P=24(\pi+2)\ cm[/tex]
Step-by-step explanation:
see the attached figure to better understand the problem
Part 1)
a) Find the area of the figure N 1
we know that
The area of the figure N 1 is equal to the area of a quarter circle minus the area of the triangle
Find the area of the quarter circle
The area of quarter circle is equal to
[tex]A=\frac{1}{4}\pi r^{2}[/tex]
we have
[tex]r=12\ cm[/tex]
substitute
[tex]A=\frac{1}{4}\pi (12)^{2}[/tex]
[tex]A=36\pi\ cm^2[/tex]
Find the area of the triangle
The area of triangle is equal to
[tex]A=\frac{1}{2}(b)(h)[/tex]
we have
[tex]b=12\ cm\\h=12\ cm[/tex]
substitute
[tex]A=\frac{1}{2}(12)(12)[/tex]
[tex]A=72\ cm^2[/tex]
therefore
The area of the figure is
[tex]A=(36\pi-72)\ cm^2[/tex]
Simplify
[tex]A=36(\pi-2)\ cm^2[/tex]
b) Find the perimeter of figure N 1
we know that
The perimeter of the figure N 1 is equal to the circumference of a quarter circle plus the side AC of triangle
Find the circumference of a quarter circle
The perimeter of a quarter of circle is equal to
[tex]C=\frac{1}{4}2\pi r[/tex]
simplify
[tex]C=\frac{1}{2}\pi r[/tex]
we have
[tex]r=12\ cm[/tex]
substitute
[tex]C=\frac{1}{2}\pi (12)[/tex]
[tex]C=6\pi\ cm[/tex]
Find the length side AC
Applying the Pythagorean Theorem
[tex]AC^2=12^2+12^2[/tex]
[tex]AC^2=288[/tex]
[tex]AC=\sqrt{288}\ cm[/tex]
simplify
[tex]AC=12\sqrt{2}\ cm[/tex]
The perimeter of the figure is
[tex]P=(6\pi+12\sqrt{2})\ cm[/tex]
simplify
[tex]P=6(\pi+2\sqrt{2})\ cm[/tex]
Part 2)
a) Find the area of the figure N 2
we know that
The area of the figure N 2 is equal to the area of a semicircle plus the area of a square minus the area of semicircle
so
The area of the figure is equal to the area of the square
[tex]A=24^2\\A=576\ cm^2[/tex]
b) Find the perimeter of the figure N 2
we know that
The perimeter of the figure N 2 is equal to the length side AB plus the length side DC plus the circumference of two semicircles
so
The perimeter of the figure N 2 is equal to two times the length side AB plus the circumference of one circle
[tex]P=2(AB)+\pi D[/tex]
[tex]P=2(24)+\pi (24)[/tex]
[tex]P=(48+24\pi)\ cm[/tex]
simplify
[tex]P=24(\pi+2)\ cm[/tex]
