Question

find the area of the shape.
(sides meet at right angles.)
5 in
2 in
2 in
5 in
3 in
3 in
square inches

Explanation:

Step1: Divide the shape into two rectangles

We can split the L - shaped figure into two rectangles. One rectangle has dimensions \(5\) in (length) and \(3\) in (width), and the other rectangle has dimensions \(2\) in (length) and \((5 - 3)\) in (width) or we can also split it as a \(5\times5\) square minus a \(2\times3\) rectangle. Let's use the method of splitting into two rectangles. The first rectangle: length \(l_1 = 5\) in, width \(w_1=3\) in. The second rectangle: length \(l_2 = 2\) in, width \(w_2=2\) in (wait, no, let's re - examine. Wait, the vertical side of the upper part: the total height is 5 in, the lower part has height 3 in, so the upper part has height \(5 - 3=2\) in, and the horizontal length of the upper part is \(5\) in? No, wait, the figure: the top horizontal side is 5 in, the left vertical side is 2 in, then a horizontal side of 2 in, then a vertical side of 3 in, then a horizontal side of 3 in, and a vertical side of 5 in. So another way: split into a rectangle of \(5\times2\) and a rectangle of \(3\times3\)? Wait, no. Let's do it correctly. The area of a rectangle is \(A = l\times w\).

Alternative approach: The large square (if we consider the outer square) has side 5 in, area \(5\times5 = 25\) square inches. Then the missing part is a rectangle with length 2 in and width 3 in (since \(5 - 3 = 2\) for the vertical side? Wait, no. Wait, the indentation: the horizontal length of the indentation is \(5 - 3=2\) in? No, the horizontal segment is 2 in, and the vertical segment is 3 in. So the area of the missing rectangle is \(2\times3 = 6\) square inches. Then the area of the shape is \(5\times5-2\times3\).

Wait, let's calculate using the two - rectangle method. First rectangle: top part: length 5 in, height 2 in. Area \(A_1=5\times2 = 10\) square inches. Second rectangle: bottom part: length 3 in, height 3 in? No, wait, the bottom part: the horizontal length is 3 in (since \(5 - 2 = 3\) in), and the vertical length is 3 in? No, the vertical length of the bottom part is 3 in, and the horizontal length is 3 in? Wait, no. Let's look at the coordinates. Let's place the figure on a coordinate system. The bottom - right corner is at \((5,0)\), the top - right corner is at \((5,5)\), the top - left corner is at \((0,5)\), then a point at \((0,3)\), then a point at \((2,3)\), then a point at \((2,0)\), then back to \((5,0)\). So the figure can be divided into two rectangles:

Rectangle 1: from \((0,3)\) to \((5,5)\): length \(l = 5\) in, width \(w=5 - 3 = 2\) in. Area \(A_1=5\times2=10\) square inches.

Rectangle 2: from \((2,0)\) to \((5,3)\): length \(l = 5 - 2=3\) in, width \(w = 3\) in. Area \(A_2=3\times3 = 9\) square inches.

Then total area \(A=A_1 + A_2=10 + 9=19\) square inches? Wait, no, that's wrong. Wait, \(5\times2=10\), \(3\times3 = 9\), \(10 + 9 = 19\)? But \(5\times5=25\), and \(2\times3 = 6\), \(25-6 = 19\). Yes, that's correct. Because the large square (5x5) has an area of 25, and the missing rectangle (2x3) has an area of 6, so \(25 - 6=19\).

Wait, let's check again. The missing part: the horizontal length is 2 in (the segment from x = 0 to x = 2 at y from 0 to 3? No, wait, the indentation is a rectangle with length 2 in (horizontal) and height 3 in (vertical). So area of missing rectangle is \(2\times3 = 6\). Area of the 5x5 square is \(25\). So \(25-6 = 19\).

Step2: Calculate the area of the square and the missing rectangle

- Area of the square with side \(s = 5\) in: \(A_{square}=s\times s=5\times5 = 25\) square inches.
- Area of the missing rectangle: length \(l = 2\) in, width \(w = 3\) in. So \(A_{rectangle}=l\times w=2\times3 = 6\) square inches.
- Area of the shape \(A = A_{square}-A_{rectangle}=25 - 6=19\) square inches.

Answer:

19