Answer:
Part 1) [tex]A=36(\pi-2)\ cm^2[/tex]
Part 2) [tex]P=6(\pi+2\sqrt{2})\ cm[/tex]
Step-by-step explanation:
Part 1) Find the area
we know that
The area of the shape is equal to the area of a quarter of circle minus the area of an isosceles right triangle
so
[tex]A=\frac{1}{4}\pi r^{2}-\frac{1}{2}(b)(h)[/tex]
we have that the base and the height of triangle is equal to the radius of the circle
[tex]r=12\ cm\\b=12\ cm\\h=12\ cm[/tex]
substitute
[tex]A=\frac{1}{4}\pi (12)^{2}-\frac{1}{2}(12)(12)\\A=(36\pi-72)\ cm^2[/tex]
simplify
Factor 36
[tex]A=36(\pi-2)\ cm^2[/tex]
Part 2) Find the perimeter
The perimeter of the figure is equal to the circumference of a quarter of circle plus the hypotenuse of the right triangle
The circumference of a quarter of circle is equal to
[tex]C=\frac{1}{4}(2\pi r)=\frac{1}{2}\pi r[/tex]
substitute the given values
[tex]C=\frac{1}{2}\pi (12)\\C=6\pi\ cm[/tex]
The hypotenuse of right triangle is equal to (applying the Pythagorean Theorem)
[tex]AC=\sqrt{12^2+12^2}\\AC=\sqrt{288}\ cm[/tex]
simplify
[tex]AC=12\sqrt{2}\ cm[/tex]
Find the perimeter
[tex]P=(6\pi+12\sqrt{2})\ cm[/tex]
simplify
Factor 6
[tex]P=6(\pi+2\sqrt{2})\ cm[/tex]
