Question

find the area of the figure.
9
4 1/2
7 1/2
1 1/2
units²

Explanation:

Step1: Split the figure

The figure can be split into a rectangle and a trapezoid (or a rectangle and a triangle, but here we'll use rectangle and a smaller rectangle or a trapezoid). Alternatively, we can consider it as a rectangle with length \(7\frac{1}{2}\) and height \(4\frac{1}{2}\), plus a trapezoid on the right. Wait, actually, another way: the top length is 9, bottom is \(7\frac{1}{2}\), so the difference in length is \(9 - 7\frac{1}{2}=1\frac{1}{2}\). The height of the right - hand part (the triangle - like or trapezoid - like) can be found from the vertical sides. The vertical side on the left is \(4\frac{1}{2}\), and on the right, the vertical part of the slanted side: the total vertical length is \(4\frac{1}{2}\), and the small vertical segment on the right is \(1\frac{1}{2}\), so the height of the right - hand trapezoid (or triangle) is \(4\frac{1}{2}-1\frac{1}{2} = 3\).

But a better way: The figure can be considered as a rectangle with length \(7\frac{1}{2}\) and width \(4\frac{1}{2}\) plus a trapezoid with bases \(1\frac{1}{2}\) and \(4\frac{1}{2}\) and height \(9 - 7\frac{1}{2}=1\frac{1}{2}\).

First, convert mixed numbers to improper fractions:
\(7\frac{1}{2}=\frac{15}{2}\), \(4\frac{1}{2}=\frac{9}{2}\), \(1\frac{1}{2}=\frac{3}{2}\), \(9=\frac{18}{2}\)

Step2: Area of the rectangle

The area of the rectangle \(A_{1}\) with length \(l = \frac{15}{2}\) and width \(w=\frac{9}{2}\) is \(A_{1}=l\times w=\frac{15}{2}\times\frac{9}{2}=\frac{135}{4}\)

Step3: Area of the trapezoid

The formula for the area of a trapezoid is \(A_{2}=\frac{(a + b)h}{2}\), where \(a=\frac{3}{2}\), \(b = \frac{9}{2}\), and \(h=\frac{3}{2}\) (since \(9 - 7\frac{1}{2}=\frac{18}{2}-\frac{15}{2}=\frac{3}{2}\))

\(A_{2}=\frac{(\frac{3}{2}+\frac{9}{2})\times\frac{3}{2}}{2}=\frac{\frac{12}{2}\times\frac{3}{2}}{2}=\frac{6\times\frac{3}{2}}{2}=\frac{9}{2}\)

Step4: Total area

Now, add the area of the rectangle and the trapezoid:
\(A = A_{1}+A_{2}=\frac{135}{4}+\frac{9}{2}=\frac{135 + 18}{4}=\frac{153}{4}=38\frac{1}{4}\)

Wait, another approach: The figure can be seen as a large rectangle with length 9 and width \(4\frac{1}{2}\) minus a triangle (or a trapezoid) on the bottom right. The base of the triangle (the horizontal segment) is \(9 - 7\frac{1}{2}=1\frac{1}{2}\), and the height (vertical segment) is \(4\frac{1}{2}-1\frac{1}{2}=3\).

Area of the large rectangle: \(A_{large}=9\times4\frac{1}{2}=9\times\frac{9}{2}=\frac{81}{2}\)

Area of the triangle: \(A_{triangle}=\frac{1}{2}\times1\frac{1}{2}\times3=\frac{1}{2}\times\frac{3}{2}\times3=\frac{9}{4}\)

Then the area of the figure \(A = A_{large}-A_{triangle}=\frac{81}{2}-\frac{9}{4}=\frac{162 - 9}{4}=\frac{153}{4}=38\frac{1}{4}\)

Or, more simply, we can use the formula for the area of a trapezoid - like figure by considering the two parallel sides (the top and bottom) and the height. Wait, the top length is 9, the bottom length is \(7\frac{1}{2}\), and the average of the two parallel sides multiplied by the height (the vertical height). Wait, no, the vertical sides: the left side is \(4\frac{1}{2}\), and the right side has a vertical component. Wait, actually, the figure is a trapezoid with the two parallel sides being the left and right vertical sides? No, the top and bottom are horizontal.

Wait, the correct way: The figure can be divided into a rectangle and a right triangle. The rectangle has length \(7\frac{1}{2}\) and height \(4\frac{1}{2}\). The triangle has base \(9 - 7\frac{1}{2}=1\frac{1}{2}\) and height \(4\frac{1}{2}-1\frac{1}{2}=3\)? No, that's not right. Wait, let's look at the vertical lengths: the left side is \(4\frac{1}{2}\), the right side has a vertical segment of \(1\frac{1}{2}\) and then a slanted part. The horizontal lengths: top is 9, bottom is \(7\frac{1}{2}\). So the horizontal difference is \(1\frac{1}{2}\), and the vertical difference between the left and right vertical segments is \(4\frac{1}{2}-1\frac{1}{2}=3\). So the area is the area of the rectangle (\(7\frac{1}{2}\times4\frac{1}{2}\)) plus the area of the trapezoid with bases \(1\frac{1}{2}\) and \(4\frac{1}{2}\) and height \(1\frac{1}{2}\) (the horizontal difference).

\(7\frac{1}{2}\times4\frac{1}{2}=\frac{15}{2}\times\frac{9}{2}=\frac{135}{4}\)

Trapezoid area: \(\frac{(1\frac{1}{2}+4\frac{1}{2})\times1\frac{1}{2}}{2}=\frac{( \frac{3}{2}+\frac{9}{2})\times\frac{3}{2}}{2}=\frac{6\times\frac{3}{2}}{2}=\frac{9}{2}\)

Total area: \(\frac{135}{4}+\frac{9}{2}=\frac{135 + 18}{4}=\frac{153}{4}=38\frac{1}{4}\)

Answer:

\(38\frac{1}{4}\) (or \(\frac{153}{4}\))