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Find the Area of the figure below,
composed of a parallelogram and one
semicircle. Rounded to the nearest tenths
place
24

Find the Area of the figure below composed of a parallelogram and one semicircle Rounded to the nearest tenths place 24 class=

Respuesta :

Answer:

169.1 unit^2 to the nearest tenth.

Step-by-step explanation:

Area of the parallelogram = base * height

= 24 * 6 unit^2.

Area of the semicircle = 1/2 pi r^2.

Here r = 1/2 * 8 = 4.

So the area is 1/2 * pi * 4^2

= 8pi unit^2.

The area of the whole figure = 24*6 + 8pi

= 169.13 unit^2.

deepakrai138

The area of the figure is [tex]169.1[/tex] .

Given the figure attached has composed of a parallelogram of sides 8 and 24 and the semicircle has diameter of 8.

We have to calculate the area of the figure.

We know that Area of parallelogram will be,

[tex]A=Base\times height[/tex]

Here, Base= 24 and Height = 6

So, [tex]A=24\times 6[/tex]

[tex]A=144[/tex]

Now, Area of semicircle will be,

[tex]A=\dfrac{\pi r^{2}}{2}[/tex]

Here diameter of semicircle is 8 so its radius will be half of it, means radius is 4.

So, [tex]A=\dfrac{\pi (4)^{2} }{2} \\[/tex]

[tex]A=\dfrac{\pi \times 16}{2}[/tex]

[tex]A=\pi \times 8[/tex]

[tex]A=25.132[/tex]

Now, the area of the Figure will be  [tex](144+25.132=169.132)[/tex] ,on rounding it to nearest tenth place we get Area is [tex]169.1[/tex] .

Hence the area of the figure is [tex]169.1[/tex] .

For more details follow the link:

https://brainly.com/question/1238286

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