Question

payton cut out two shapes, as shown, that she will later put together to resemble a house. what is the total area of the two shapes? 22 square units 21 square units 23 square units 20 square units

Explanation:

Step1: Calculate the area of the first shape (triangle - like part)

Use the formula for the area of a polygon. Count the number of full - squares and half - squares. For the first part, we can also use the formula \(A=\frac{1}{2}(b_1 + b_2)h\) (treating it as a trapezoid in a way of decomposition). But a simpler way is to use the formula for the area of a polygon with coordinates (Pick's theorem: \(A = I+\frac{B}{2}-1\), where \(I\) is the number of interior points and \(B\) is the number of boundary points). However, by counting:
The first shape (upper - left part):
We can divide it into two triangles. One triangle with base \(4\) (from \(x = 1\) to \(x=5\)) and height \(3\) (\(y\) from \(y = 4\) to \(y = 7\)) has area \(A_1=\frac{1}{2}\times4\times3 = 6\). Another triangle with base \(4\) (from \(x = 5\) to \(x = 9\)) and height \(4\) (\(y\) from \(y = 4\) to \(y = 8\)) has area \(A_2=\frac{1}{2}\times4\times4=8\).

Step2: Calculate the area of the second shape (lower - part)

The second shape (lower - part) is a quadrilateral. We can divide it into two triangles. One triangle with base \(3\) (from \(x = 1\) to \(x = 4\)) and height \(3\) (\(y\) from \(y = 1\) to \(y = 4\)) has area \(A_3=\frac{1}{2}\times3\times3=\frac{9}{2}\). Another triangle with base \(5\) (from \(x = 4\) to \(x = 9\)) and height \(3\) (\(y\) from \(y = 1\) to \(y = 4\)) has area \(A_4=\frac{1}{2}\times5\times3=\frac{15}{2}\). The sum of the areas of the two triangles in the second shape is \(A_{lower}=\frac{9 + 15}{2}=12\).

Step3: Calculate the total area

The total area \(A=A_1 + A_2+A_{lower}=6 + 8+8=22\).

Answer:

22 square units