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frika
The upper semicircle ABCD has radius R=2 and area [tex]A_u= \frac{1}{2} \pi R^2= \frac{1}{2} \pi\cdot 2^2=2\pi[/tex] and the lower semicircle AD has radius r=1 and area [tex]A_l= \frac{1}{2} \pi r^2= \frac{1}{2} \pi\cdot 1^2= \frac{1}{2}\pi [/tex]. The whole fugure area is [tex]A=A_u+A_l=2\pi + \frac{1}{2}\pi= \frac{5}{2} \pi[/tex].
The perimeter consists of upper semicircle  length, lower semicircle length and CD length.
[tex]L_u= \frac{1}{2} \cdot(2\pi R)=2\pi[/tex]
[tex]L_l= \frac{1}{2} \cdot(2\pi r)=\pi[/tex]
[tex]L_{CD}=2[/tex]
and the perimeter [tex]P=L_u+L_l+L_{CD}=2\pi+\pi+2=3\pi+2[/tex]
The area is 5π/2 in² and the perimeter is 3π+2 in.

Explanation:
The area of a circle is given by the formula A=
π*r². The top part of the figure is half a circle that has a radius of 2; this means the area would be
A=(1/2)
π*2² = (1/2)π*4 = 2π.

The bottom part of the figure is half a circle with a diameter of 2; since the diameter is twice as long as the radius, this means the radius is 2/2 = 1, and the area would be A=(1/2)
π*1² = 1/2 π.

Adding these, we have 2 1/2 
π, which can be written as 5π/2.

The perimeter of a circle is the circumference, which is given by the formula C=
π*d. The top part has a radius of 2, so the diameter is 2*2=4; this makes the circumference C=1/2(π)(4) = 2π.

The bottom part has a diameter of 2, which makes the circumference C=1/2(
π)(2) = 1π.

Together this gives us 3
π.

We also must add the length of CD to this; CD=2, which gives us 3
π+2.