Question

find the area of the figure.
6
4\frac{1}{2}
1\frac{1}{2}
4\frac{1}{2}
\square units^2

Explanation:

Step1: Split the figure

The figure can be split into a rectangle and a triangle. The rectangle has length \( 4\frac{1}{2} \) (which is \( \frac{9}{2} \)) and width \( 4\frac{1}{2} \) ( \( \frac{9}{2} \) ). The triangle has a base of \( 6 - 4\frac{1}{2}=1\frac{1}{2}=\frac{3}{2} \) and height \( 4\frac{1}{2}=\frac{9}{2} \).

Step2: Calculate area of rectangle

Area of rectangle \( A_{rect} = length \times width=\frac{9}{2}\times\frac{9}{2}=\frac{81}{4} = 20.25 \)

Step3: Calculate area of triangle

Area of triangle \( A_{tri}=\frac{1}{2}\times base\times height=\frac{1}{2}\times\frac{3}{2}\times\frac{9}{2}=\frac{27}{8}=3.375 \)

Step4: Total area

Total area \( A = A_{rect}+A_{tri}=\frac{81}{4}+\frac{27}{8}=\frac{162 + 27}{8}=\frac{189}{8}=23.625 \) or we can also calculate by considering the figure as a trapezoid (since it can be seen as a trapezoid with bases \( 6 \) and \( 4\frac{1}{2} \) and height \( 4\frac{1}{2} \)). Wait, actually, the correct way: the figure is a trapezoid? Wait no, when we split, or maybe it's a rectangle plus a triangle. Wait, another approach: the figure has a rectangle part (length \( 4\frac{1}{2} \), height \( 4\frac{1}{2} \)) and a trapezoid? No, better to use the formula for the area by adding the rectangle and the triangle. Wait, alternatively, the figure can be considered as a rectangle with length \( 6 \) and height \( 4\frac{1}{2} \) minus a triangle? No, no. Wait, looking at the figure: the left side is \( 4\frac{1}{2} \), bottom is \( 4\frac{1}{2} \), top is \( 6 \), right side has a segment of \( 1\frac{1}{2} \). So the horizontal difference between top and bottom is \( 6 - 4\frac{1}{2}=1\frac{1}{2} \). So the figure is a rectangle ( \( 4\frac{1}{2}\times4\frac{1}{2} \)) plus a triangle with base \( 1\frac{1}{2} \) and height \( 4\frac{1}{2} \). Wait, no, the triangle's height is \( 4\frac{1}{2} \)? Wait, no, the vertical side is \( 4\frac{1}{2} \), so the triangle has base \( 1\frac{1}{2} \) and height \( 4\frac{1}{2} \)? Wait, no, the right side: the vertical part? Wait, maybe I made a mistake. Let's re - examine. The figure: the bottom is \( 4\frac{1}{2} \), left is \( 4\frac{1}{2} \), top is \( 6 \), and on the right, there is a slant side with a horizontal segment of \( 1\frac{1}{2} \) (since \( 6-4\frac{1}{2}=1\frac{1}{2} \)) and vertical segment? Wait, no, the height from the bottom to the top on the right: the vertical length is \( 4\frac{1}{2} \), and the horizontal overhang is \( 1\frac{1}{2} \). So the figure can be divided into a square (or rectangle) of \( 4\frac{1}{2}\times4\frac{1}{2} \) and a triangle with base \( 1\frac{1}{2} \) and height \( 4\frac{1}{2} \). Wait, but actually, the correct formula: if we consider the figure as a trapezoid with bases \( b_1 = 6 \), \( b_2=4\frac{1}{2} \) and height \( h = 4\frac{1}{2} \). The area of a trapezoid is \( \frac{(b_1 + b_2)}{2}\times h \). Let's check that. \( b_1 = 6=\frac{12}{2} \), \( b_2 = 4\frac{1}{2}=\frac{9}{2} \), \( h = 4\frac{1}{2}=\frac{9}{2} \). Then \( A=\frac{(\frac{12}{2}+\frac{9}{2})}{2}\times\frac{9}{2}=\frac{\frac{21}{2}}{2}\times\frac{9}{2}=\frac{21}{4}\times\frac{9}{2}=\frac{189}{8}=23.625 \), which matches the previous result. So that's a better way: using the trapezoid area formula.

So, \( A=\frac{(6 + 4\frac{1}{2})}{2}\times4\frac{1}{2} \)

First, convert mixed numbers to improper fractions: \( 6=\frac{12}{2} \), \( 4\frac{1}{2}=\frac{9}{2} \)

Sum of bases: \( \frac{12}{2}+\frac{9}{2}=\frac{21}{2} \)

Divide by 2: \( \frac{21}{2}\div2=\frac{21}{4} \)

Multiply by height ( \( \frac{9}{2} \)): \( \frac{21}{4}\times\frac{9}{2}=\frac{189}{8}=23\frac{5}{8} \) or \( 23.625 \)

Answer:

\( \frac{189}{8} \) (or \( 23\frac{5}{8} \) or \( 23.625 \))