Respuesta :

Answer:

21.6 units² (nearest tenth)

Step-by-step explanation:

[tex]\textsf{Area of a trapezoid}=\dfrac{1}{2}(a+b)h \quad \textsf{(where a and b are the bases and h is the height)}[/tex][tex]\textsf{Area of a semicircle}=\dfrac12 \pi r^2 \quad \textsf{(where r is the radius)}[/tex]

Given:

  • a = 2
  • b = 8
  • h = 4
  • r = 2 ÷ 2 = 1

[tex]\begin{aligned}\textsf{Area of figure} & =\textsf{area of trapezoid + area of semicircle}\\& = \dfrac{1}{2}(a+b)h+\dfrac{1}{2} \pi r^2\\& = \dfrac{1}{2}(2+8)(4)+\dfrac{1}{2} \pi (1)^2\\& = 20+\dfrac{1}{2} \pi\\& = 21.6\: \sf units^2 \:(nearest\:tenth)\end{aligned}[/tex]

Area of the trapezium

  • 1/2(Sum of parallel sides)×Height
  • 1/2(2+8)(4)
  • 2(10)
  • 20units²

Area of semicircle

  • π(2/2)²/2
  • π/2
  • 1.57units²

Total area

  • 20+1.57
  • 21.6units²(Rounded)