Question

Please find the perimeter and area of these shapes

Answer 1

ABC shaded area = 36
`\pi` - 72 cm²

ABC shaded area perimeter =
`6\pi +12√(2)` cm

ABCD area =
`\frac52 \pi` cm²

ABCD perimeter =
`3\pi +2` cm

Explanation:

Shape ABC

Assuming you want the area and perimeter of the shaded part of the shape only...

Area

Area of a sector =
`\frac12r^2\theta` (where r is the radius and
`\theta`

⇒ area of a sector =
`\frac12 * 12^2* (\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*\frac12\pi =6\pi \ \textsf{cm}`

Hypotenuse of triangle =
`√(a^2+b^2)` (where a and b are the legs of the right triangle)

⇒ hypotenuse =
`√(12^2+12^2) =12√(2)` cm

Therefore, perimeter = arc length + hypotenuse

⇒ perimeter =
`6\pi +12√(2)` cm

Shape ABCD

Area

Area of a semicircle =
`\frac12 \pi r^2` (where r is the radius)

⇒ area of large semicircle ABC =
`\frac12 * \pi * 2^2=2\pi \ \textsf{cm}^2`

⇒ area of small semicircle AD =
`\frac12 * \pi * 1^2=\frac12\pi \ \textsf{cm}^2`

⇒ area of shape ABCD =
`\frac12 \pi + 2 \pi=\frac52 \pi \ \textsf{cm}^2`

Perimeter

1/2 circumference =
`\pi r`

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