Question

find the area of the figure. (sides meet at right angles.) 9 yd 4 yd 5 yd 5 yd 9 yd \boxed{} yd²

Explanation:

Step1: Divide the figure

We can divide the given figure into two rectangles. One rectangle has dimensions \(9\) yd (length) and \(9 - 5=4\) yd (width). The other rectangle has dimensions \(5\) yd (length) and \(5\) yd (width) (Wait, actually, let's re - check. Alternatively, we can consider the large square of side \(9\) yd and subtract the missing rectangle. The missing rectangle has length \(4\) yd and width \(5\) yd.

Step2: Calculate the area of the large square

The area of a square with side \(s\) is \(A = s\times s\). For the large square with side \(s = 9\) yd, the area \(A_{1}=9\times9 = 81\) square yards.

Step3: Calculate the area of the missing rectangle

The missing rectangle has length \(l = 4\) yd and width \(w = 5\) yd. The area of a rectangle is \(A=l\times w\), so \(A_{2}=4\times5=20\) square yards.

Step4: Calculate the area of the figure

The area of the given figure \(A = A_{1}-A_{2}\). Substituting the values, we get \(A = 81 - 20=61\) square yards.

Alternatively, we can divide the figure into two rectangles:

• Rectangle 1: Length \(9\) yd, width \(9 - 5 = 4\) yd. Area \(A_{1}=9\times4 = 36\) square yards.
• Rectangle 2: Length \(5\) yd, width \(5\) yd. Area \(A_{2}=5\times5 = 25\) square yards. Wait, no, that's incorrect. Wait, another way: The vertical side of the left - part: \(9 - 5=4\) yd. The horizontal side of the bottom - part: \(9 - 4 = 5\) yd. So the figure can be divided into a rectangle of \(9\times4\) and a rectangle of \(5\times5\)? No, wait, let's look at the correct division.

Wait, the correct way: The figure can be split into a rectangle with length \(9\) yd and width \(4\) yd (top part) and a rectangle with length \(5\) yd and width \(5\) yd (bottom right part)? No, that's not right. Wait, the total height is \(9\) yd, the bottom rectangle has height \(5\) yd and width \(5\) yd. The top rectangle has height \(9 - 5=4\) yd and width \(9\) yd? No, because there is a notch. Wait, the horizontal length of the top part: the total length is \(9\) yd, and the notch has a horizontal length of \(4\) yd? Wait, maybe my first method is better. The large square (if it were a square) has area \(9\times9 = 81\). The missing part is a rectangle with length \(4\) and width \(5\), area \(20\). So \(81-20 = 61\).

Let's verify with another division:

• Rectangle 1: Length \(5\) yd, width \(5\) yd (bottom right). Area \(25\).
• Rectangle 2: Length \(9\) yd, width \(4\) yd (top). Area \(36\).
• Wait, but \(25 + 36=61\), which matches. Yes, because the top rectangle has width \(9\) yd and height \(4\) yd (since \(9 - 5 = 4\)), and the bottom rectangle has width \(5\) yd and height \(5\) yd (since the bottom part's width is \(5\) yd and height is \(5\) yd). So \(9\times4+5\times5=36 + 25 = 61\).

Answer:

\(61\)