Answer:
[tex]\textsf{ Base of parallelogram = $\boxed{ 12 }$ in}[/tex]
[tex]\textsf{ Height of parallelogram = $\boxed{ 5 }$ in}[/tex]
[tex]\textsf{ Radius of half circles=$\boxed{ 6 }$ in}[/tex]
[tex]\textsf{ Area of Composite Figure =$\boxed{ 0.1}$\sf $m^2$}[/tex]
Step-by-step explanation:
By observation, we can say that:
Base of parallelogram = [tex]\sf 2\times (6)(radius ) = 12 [/tex] inches
Height of parallelogram = [tex]\sf 5 [/tex] inches
Radius of half circles = [tex]\sf 6 [/tex] inches
Now let's find an area of composite figures:
First calculate the area of the parallelogram:
[tex]\sf A_p = \textsf{Base} \times \textsf{Height} [/tex]
[tex]\sf A_p = 12 \times 5 [/tex]
[tex]\sf A_p = 60 \textsf{ square inches} [/tex]
Calculate the combined area of the two semicircles:
[tex]\sf A_c = \dfrac{1}{2} \pi r^2 [/tex]
[tex]\sf A_c = \dfrac{1}{2} \pi (6^2) [/tex]
[tex]\sf A_c = \dfrac{1}{2} \pi \times 36 [/tex]
[tex]\sf A_c = 18\textsf{ square inches} [/tex]
Since there are two semicircles, the total combined area is:
[tex]\sf 2 \times 18\pi = 36\pi [/tex] square inches.
Calculate the total area of the composite figure:
[tex]\sf A = A_p + A_c = 30 + 36\pi [/tex]
[tex]\sf A \approx 60 + 36 \times 3.1415926535897 [/tex]
[tex]\sf A \approx 60 + 113.09733552923 [/tex]
[tex]\sf A \approx 173.09733552923 [/tex] inches square
Since we need area in m², so we use scale factor.
1 in = 0.0254 m
Using this we get
1 in² = (0.0254)² m²
So,
173.09733552923 in² = 173.09733552923 × (0.0254)² m²/in²
= 0.1116754769900 m²
So,
[tex]\sf A \approx 0.1 \textsf{ square inches (in nearest tenth)} [/tex]
Summary:
[tex]\textsf{ Base of parallelogram = $\boxed{ 12 }$ in}[/tex]
[tex]\textsf{ Height of parallelogram = $\boxed{ 5 }$ in}[/tex]
[tex]\textsf{ Radius of half circles=$\boxed{ 6 }$ in}[/tex]
[tex]\textsf{ Area of Composite Figure =$\boxed{ 0.1}$\sf $m^2$}[/tex]
