Question

Please find the perimeter and area of these shapes

Answer 1

Answer:

ABC shaded area = 36$\pi$ - 72   cm²

ABC shaded area perimeter = $6\pi +12\sqrt{2}$    cm

ABCD area = $\dfrac52 \pi$  cm²

ABCD perimeter = $3\pi +2$   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 = $\dfrac12r^2\theta$ (where r is the radius and $\theta$

⇒ area of a sector = $\dfrac12 \times 12^2\times \dfrac{\pi}{2} =36\pi \ \textsf{cm}^2$

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$\pi$ - 72 cm²

Perimeter

Arc length = $r\theta$ (where r is the radius and $\theta$

⇒ arc length = $12\times\dfrac12\pi =6\pi \ \textsf{cm}$

Hypotenuse of triangle = $\sqrt{a^2+b^2}$ (where a and b are the legs of the right triangle)

⇒ hypotenuse = $\sqrt{12^2+12^2} =12\sqrt{2}$ cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter = $6\pi +12\sqrt{2}$  cm

Shape ABCD

Area

Area of a semicircle = $\dfrac12 \pi r^2$ (where r is the radius)

⇒ area of large semicircle ABC = $\dfrac12 \times \pi \times 2^2=2\pi \ \textsf{cm}^2$

⇒ area of small semicircle AD = $\dfrac12 \times \pi \times 1^2=\dfrac12\pi \ \textsf{cm}^2$

⇒ area of shape ABCD = $\dfrac12 \pi + 2 \pi=\dfrac52 \pi \ \textsf{cm}^2$

Perimeter

1/2 circumference = $\pi r$

⇒ perimeter = $2\pi +2+\pi=3 \pi+2 \ \textsf{cm}$