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Find the area. Round to the nearest hundredth when necessary. 16 in Area in? 2

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We have the next figure

Because we have an irregular figure we divide the figure into a rectangle and in a triangle

For the area of the rectangle


A=l* w

l=length

w= width

In our case

l=19

w=16

we substitute the values


A=19*16=304in^2

for the area of the triangle


A=(b* h)/(2)

b= base

h=height

in our case

b=16in

h=30-19=11in

we substitute the values


\begin{gathered} A=(16*11)/(2) \\ A=88in^2 \end{gathered}

Then we sum the areas of the figures and we obtain the total are


\begin{gathered} A=304+88 \\ A=392in^2 \end{gathered}

Find the area. Round to the nearest hundredth when necessary. 16 in Area in? 2-example-1
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