Answer:
ABC shaded area = 36[tex]\pi[/tex] - 72 cm²
ABC shaded area perimeter = [tex]6\pi +12\sqrt{2}[/tex] cm
ABCD area = [tex]\dfrac52 \pi[/tex] cm²
ABCD perimeter = [tex]3\pi +2[/tex] cm
Step-by-step explanation:
Shape ABC
Assuming you want the area and perimeter of the shaded part of the shape only...
Area
Area of a sector = [tex]\dfrac12r^2\theta[/tex] (where r is the radius and [tex]\theta[/tex]
⇒ area of a sector = [tex]\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2[/tex]
Area of triangle = 1/2 x base x height
⇒ area of triangle = 1/2 x 12 x 12 = 72 cm²
Therefore, area of shaded area = area of sector - area of triangle
⇒ area = 36[tex]\pi[/tex] - 72 cm²
Perimeter
Arc length = [tex]r\theta[/tex] (where r is the radius and [tex]\theta[/tex]
⇒ arc length = [tex]12\times\dfrac12\pi =6\pi \ \textsf{cm}[/tex]
Hypotenuse of triangle = [tex]\sqrt{a^2+b^2}[/tex] (where a and b are the legs of the right triangle)
⇒ hypotenuse = [tex]\sqrt{12^2+12^2} =12\sqrt{2}[/tex] cm
Therefore, perimeter = arc length + hypotenuse
⇒ perimeter = [tex]6\pi +12\sqrt{2}[/tex] cm
Shape ABCD
Area
Area of a semicircle = [tex]\dfrac12 \pi r^2[/tex] (where r is the radius)
⇒ area of large semicircle ABC = [tex]\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2[/tex]
⇒ area of small semicircle AD = [tex]\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2[/tex]
⇒ area of shape ABCD = [tex]\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2[/tex]
Perimeter
1/2 circumference = [tex]\pi r[/tex]
⇒ perimeter = [tex]2\pi +2+\pi=3 \pi+2 \ \textsf{cm}[/tex]
