The area of the shape is 24π and the perimeter is 12π+8.
Given to us:
Talking about the complete shape which is 3 semi-circles connected to each other such that the first semi-circle AHCB is connected to HGDC which is connected to GFED the last semi-circle, each facing opposite to the last one.
Height of the shape(semi-circles), h = 4 cm;
Diameter of the shape(semi-circles), d = 4 cm;
Side of the square, s = 4 cm;
Radius of the semi-circle, r = d/2 = 2 cm;
Now,
The perimeter can be calculated by,
[tex]\begin{aligned}Perimeter_{shape}= &(Perimeter\ of\ semicircle\times no.\ of\ sides)\\&+(2\times\ height\ of\ the\ shape)\end{aligned}[/tex]
As there are 2 sides of each semi-circle, therefore, a total of 6 sides(AH, BC, HG, CD, GF, DE are there.
Also, there are 2 height sides (AB and EF), therefore we multiply the height by 2.
Solving the equation,
[tex]\begin{aligned}Perimeter_{shape}= &(Perimeter\ of\ semicircle\times no.\ of\ sides)\\&+(2\times\ height\ of\ the\ shape)\\=&(\pi \times r\times 6)+(2\times 4)\\=&(\pi \times 2\times 6)+(2\times 4)\\=&12\pi+8\end{aligned}[/tex]
For the Area of the shape,
[tex]\begin{aligned}\\Area_{shape} &=surface\ area\ of\ AHCB+surface\ area\ of\ HGDC\\ &+surface\ area\ of\ AHCB \\\end{aligned}[/tex]
[tex]\begin{aligned}\\Area_{shape}&= (\pi rl)_{AHCB}+ (\pi rl)_{HGDC}+ (\pi rl)_{GFED}\\\end{aligned}[/tex]
As radius for all the semi-circles is the same which is 2 cm.
[tex]\begin{aligned}\\Area_{shape}&= (\pi rl)_{AHCB}+ (\pi rl)_{HGDC}+ (\pi rl)_{GFED}\\&= 3\times (\pi rl)\\&= 3\times (\pi \times 2\times 4)\\&=3\times (8\times \pi)\\&= 24\pi\end{aligned}[/tex]
Therefore, the area of the shape is [tex]\bold{24\pi}[/tex] and the perimeter is [tex]\bold{12\pi+8}[/tex].
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