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The figure is shown in a rectangle topped with a semicircle. The base of the rectangle is one-third its height. If $h$ is the height of the rectangle, what is the area of the figure?

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(\frac{24 + π}{72}) h^{2}

{units}^{2}

(\frac{24}{72}) h^{2}

{units}^{2}

(\frac{34 + π}{72}) h^{2}

{units}^{2}

None

answer is A.

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Detailed Solution

Given that the height of the rectangle is equal to

h .

The base of the rectangle

= \frac{h}{3}

.
The base of the rectangle

=

Diameter of the semicircle.

∴

Radius of the circle

= \frac{h}{6}

.
Area of a rectangle is (length)×(breadth).
Area of a semicircle is

\frac{π}{2} r^{2}

.
Now, area of the figure

=

Area of the rectangular part

+

Area of the semicircular part.

\Rightarrow

Area of the figure

= h \times \frac{h}{3} + \frac{1}{2} \times π \times {(\frac{h}{6})}^{2}

\Rightarrow

Area of the figure

= \frac{h^{2}}{3} + \frac{π h^{2}}{72}

\Rightarrow

Area of the figure

= (\frac{24 + π}{72}) h^{2}

{units}^{2} .

Hence, option

1

is correct.

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The figure is shown in a rectangle topped with a semicircle. The base of the rectangle is one-third its height. If h is the height of the rectangle, what is the area of the figure?

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The figure is shown in a rectangle topped with a semicircle. The base of the rectangle is one-third its height. If $h$ is the height of the rectangle, what is the area of the figure?