Question

sheet 1
7 in
4 in
9 in
5 in
area = ____________

Explanation:

Step1: Find the width of the lower rectangle

The total height is 9 in and the height of the right - side part is 5 in, so the height of the lower left rectangle is \(9 - 5=4\) in. The width of the lower left rectangle: the top rectangle has a width of 7 in and the right - side part has a width of 4 in, so the width of the lower left rectangle is \(7 - 4 = 3\) in.

Step2: Calculate the area of the top rectangle

The top rectangle has length \(l_1 = 7\) in and height \(h_1=9 - 5 = 4\) in (wait, no, actually, from the figure, the top rectangle's height: since the lower right part is 5 in, and the total height is 9 in, the height of the top rectangle is \(9 - 5=4\) in? Wait, maybe a better way: the top rectangle: length = 7 in, and its height: let's see, the lower part has a height of 5 in? Wait, maybe I made a mistake. Let's re - examine.

Alternative approach: The figure can be divided into two rectangles. The top rectangle: length = 7 in, and its height: let's find the height of the top rectangle. The lower right rectangle has height 5 in and width 4 in. The left - hand side has a total height of 9 in. So the height of the top rectangle is \(9 - 5 = 4\) in. The area of the top rectangle \(A_1=7\times4 = 28\) square inches.

The lower part: we have two rectangles? Wait, no. Wait, the left - hand rectangle: width is \(7 - 4=3\) in, height is 9 in? No, that's not right. Wait, maybe the figure is composed of a top rectangle and a left rectangle and a right rectangle? No, let's look at the dimensions again.

Wait, the total height is 9 in. The right - hand rectangle has height 5 in and width 4 in. The left - hand rectangle: width is \(7 - 4 = 3\) in, and height is 9 in? No, that can't be. Wait, maybe the correct division is: the top rectangle (length 7 in, height \(9 - 5=4\) in) and a lower rectangle which is composed of a left rectangle (width \(7 - 4 = 3\) in, height 9 in) and a right rectangle (width 4 in, height 5 in)? No, that would be over - counting.

Wait, another way: The area of the figure is equal to the area of the large rectangle (if we consider the outer dimensions) minus the area of the missing part. But maybe it's easier to divide into two rectangles:

1. Top rectangle: length = 7 in, height = \(9 - 5=4\) in. Area \(A_1 = 7\times4=28\) square inches.

2. Lower part: we have a rectangle with width 4 in and height 5 in, and a rectangle with width \(7 - 4 = 3\) in and height 9 in? No, that's not right. Wait, no, the lower left rectangle: width is \(7 - 4 = 3\) in, height is 9 in? No, that would make the area too big.

Wait, I think I messed up the division. Let's start over.

Looking at the figure:

- The top rectangle: length = 7 in, and its height: let's see, the vertical side on the left is 9 in. The right - hand side has a vertical segment of 5 in. So the height of the top rectangle is \(9 - 5 = 4\) in. So \(A_1=7\times4 = 28\).

- The lower part: we have a rectangle with width 4 in and height 5 in, and a rectangle with width \(7 - 4=3\) in and height 9 in? No, that's incorrect. Wait, no, the left - hand rectangle: width is \(7 - 4 = 3\) in, and height is 9 in? No, because the top rectangle is already covering part of it.

Wait, maybe the correct division is:

- Rectangle 1: top, length = 7 in, height = \(9 - 5 = 4\) in. Area \(A_1=7\times4 = 28\).

- Rectangle 2: left, width = \(7 - 4=3\) in, height = 9 in. Area \(A_2=3\times9 = 27\).

- Rectangle 3: right, width = 4 in, height = 5 in. Area \(A_3=4\times5 = 20\).

But that's adding three rectangles, but we are over - counting the overlapping parts.

Wait, I think I made a mistake in the initial analysis. Let's look at the figure again. The figure has a top rectangle (7 in long) and below it, on the left, a rectangle and on the right, a rectangle. The total height is 9 in. The right - hand rectangle (below the top) has height 5 in and width 4 in. The left - hand rectangle (below the top) has height 9 in and width \(7 - 4 = 3\) in. But the top rectangle and the lower rectangles: the top rectangle's bottom side is at height \(9 - 4=5\) in (since its height is 4 in). Wait, no, if the top rectangle has height 4 in, then its bottom edge is at 4 in from the top, and the lower rectangles have height 5 in (from 4 in to 9 in). So the top rectangle: length 7 in, height 4 in (area \(7\times4 = 28\)). The lower part: left rectangle (width \(7 - 4=3\) in, height 5 in) and right rectangle (width 4 in, height 5 in). So area of lower part \(A_2=(3 + 4)\times5=7\times5 = 35\)? No, \(3\times5+4\times5=(3 + 4)\times5 = 35\). Then total area \(A=28 + 35=63\) square inches? Wait, no, \(7\times4=28\), \(3\times5 = 15\), \(4\times5 = 20\), \(28+15 + 20=63\).

Wait, let's check with another method. The total area if we consider a rectangle with length 7 in and height 9 in is \(7\times9 = 63\) square inches. Wait, that's interesting. Because if we consider the figure as a 7 in by 9 in rectangle, is there any missing part? Wait, the right - hand side has a 4 in by 5 in rectangle, but no, maybe the figure is just a 7 in by 9 in rectangle? But that can't be, because there is a 4 in by 5 in rectangle? Wait, no, maybe the figure is a 7 in by 9 in rectangle, and the way it's drawn is with a 4 in by 5 in rectangle, but actually, the area is \(7\times9=63\) square inches? Wait, that seems too easy. Wait, let's check the dimensions again. The length is 7 in, the total height is 9 in. So if we multiply 7 and 9, we get 63. Maybe the figure is a rectangle with length 7 in and height 9 in, and the internal lines are just for division, but actually, the area is \(7\times9 = 63\) square inches.

Wait, let's verify with the division method. If we divide the figure into two rectangles:

- Top rectangle: length 7 in, height \(9 - 5 = 4\) in. Area \(7\times4 = 28\).

- Lower rectangle: length 7 in, height 5 in. Area \(7\times5 = 35\).

Then total area \(28+35 = 63\) square inches. Ah, that's the correct division! I made a mistake earlier by thinking the lower part was divided into two rectangles, but actually, the lower part is a single rectangle with length 7 in and height 5 in. Because the width of the lower part is the same as the top part (7 in), and the height is 5 in. The top part has height \(9 - 5=4\) in. So:

Step1: Calculate the area of the top rectangle

The top rectangle has length \(l_1 = 7\) in and height \(h_1=9 - 5 = 4\) in. The area formula for a rectangle is \(A = l\times h\). So \(A_1=7\times4 = 28\) square inches.

Step2: Calculate the area of the bottom rectangle

The bottom rectangle has length \(l_2 = 7\) in and height \(h_2 = 5\) in. Using the area formula for a rectangle, \(A_2=7\times5=35\) square inches.

Step3: Calculate the total area

The total area of the figure is the sum of the area of the top rectangle and the area of the bottom rectangle. So \(A = A_1+A_2=28 + 35=63\) square inches.

Answer:

\(63\) square inches