Question

question
find the area of the figure below, composed of a rectangle and two semicircles. round to the nearest tenths place.
answer attempt 1 out of 2
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Explanation:

Step1: Analyze the figure's components

The figure is composed of a rectangle and two semicircles (which form a full circle). The length of the rectangle is \( 12 \), the width of the rectangle (and the diameter of the circle) is \( 10 \), so the radius \( r=\frac{10}{2} = 5 \).

Step2: Calculate the area of the rectangle

The area of a rectangle is \( A_{rectangle}=length\times width \). Substituting the values, \( A_{rectangle}=12\times10 = 120 \).

Step3: Calculate the area of the circle (from two semicircles)

The area of a circle is \( A_{circle}=\pi r^{2} \). Substituting \( r = 5 \), we get \( A_{circle}=\pi\times5^{2}=25\pi\approx25\times3.1416 = 78.54 \).

Step4: Calculate the total area

The total area of the figure is the sum of the area of the rectangle and the area of the circle. So \( A_{total}=A_{rectangle}+A_{circle}=120 + 25\pi\approx120+78.54 = 198.54 \). Rounding to the nearest tenths place, we get \( 198.5 \) (wait, actually \( 25\pi\approx78.5398 \), so \( 120 + 78.5398=198.5398\approx198.5 \)? Wait, no, \( 25\pi\) is approximately \( 78.54 \), so \( 120+78.54 = 198.54 \), which rounds to \( 198.5 \) when rounded to the nearest tenth? Wait, no, \( 198.54 \) rounded to the nearest tenth is \( 198.5 \)? Wait, no, the tenths place is the first decimal, so \( 198.54 \), the digit in the hundredths place is \( 4 \), which is less than \( 5 \), so we round down? Wait, no, \( 198.54 \) to the nearest tenth: the tenths digit is \( 5 \), the hundredths digit is \( 4 \), so we keep the tenths digit as is, so \( 198.5 \)? Wait, but let's recalculate: \( 25\pi=78.53981634 \), so \( 120 + 78.53981634=198.53981634 \). Rounding to the nearest tenth: look at the hundredths place, which is \( 3 \)? Wait, no, \( 198.53981634 \): the number is \( 198.5\underline{3}981634 \), so the tenths place is \( 5 \), hundredths is \( 3 \), so we round to \( 198.5 \)? Wait, no, I think I made a mistake in the width. Wait, the diameter of the semicircle is equal to the width of the rectangle? Wait, the figure: the top and bottom are semicircles, so the diameter of each semicircle is equal to the width of the rectangle? Wait, the length of the rectangle is \( 12 \), and the distance between the two parallel sides (the width) is \( 10 \), so the diameter of the semicircles is \( 10 \), so radius \( 5 \). So the two semicircles make a full circle. So area of rectangle: \( 12\times10 = 120 \). Area of circle: \( \pi r^{2}=\pi\times5^{2}=25\pi\approx78.54 \). Total area: \( 120 + 78.54 = 198.54 \), which rounds to \( 198.5 \) when rounded to the nearest tenth? Wait, no, \( 198.54 \) to the nearest tenth: the tenths digit is \( 5 \), the hundredths digit is \( 4 \), so we round down, so \( 198.5 \). But let's check again. Wait, maybe I messed up the dimensions. Wait, the rectangle: length is \( 12 \), width is \( 10 \)? Wait, the vertical side is \( 12 \), the horizontal dashed line is \( 10 \), so the rectangle has length \( 12 \) and width \( 10 \), and the two semicircles have diameter \( 10 \), so radius \( 5 \). So that's correct. So total area is \( 120 + 25\pi\approx198.5 \) (rounded to the nearest tenth). Wait, but let's compute \( 25\pi\) more accurately: \( \pi\approx3.141592654 \), so \( 25\times3.141592654 = 78.53981635 \). Then \( 120+78.53981635 = 198.53981635 \). Rounding to the nearest tenth: the tenths place is \( 5 \), the hundredths place is \( 3 \), so we round to \( 198.5 \). Wait, but maybe I made a mistake in the rectangle's dimensions. Wait, is the length of the rectangle \( 12 \) and the width \( 10 \), or is the length \( 12 \) and the diameter of the semicircle \( 10 \), so the width of the rectangle is equal to the diameter of the semicircle, which is \( 10 \). So that's correct. So the total area is approximately \( 198.5 \) when rounded to the nearest tenth. Wait, but let's check with another approach. Wait, maybe the figure is a rectangle with length \( 12 \) and width \( 10 \), and two semicircles (top and bottom) with diameter \( 10 \), so together they make a circle. So yes, the calculation is correct.

Answer:

\( 198.5 \)