Solution:
The area of the triangle on the left will be calculated using the image below
Concept:
The area of the triangle is given below
[tex]\begin{gathered} A_{\text{traingle}}=\frac{1}{2}\times base\times height \\ \text{base}=2in \\ \text{height}=9in \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} A_{\text{traingle}}=\frac{1}{2}\times base\times height \\ A_{\text{traingle}}=\frac{1}{2}\times2in\times9in \\ A_{\text{traingle}}=\frac{18in^2}{2} \\ A_{\text{traingle}}=9in^2 \end{gathered}[/tex]
Hence,
The area of the triangle on the left = 9in²
Step 2:
The image below will be used to calculate the area of triangle on the right
The area of the triangle is given below
[tex]\begin{gathered} A_{\text{traingle}}=\frac{1}{2}\times base\times height \\ \text{base}=2in \\ \text{height}=9in \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} A_{\text{traingle}}=\frac{1}{2}\times base\times height \\ A_{\text{traingle}}=\frac{1}{2}\times2in\times9in \\ A_{\text{traingle}}=\frac{18in^2}{2} \\ A_{\text{traingle}}=9in^2 \end{gathered}[/tex]
Hence,
The area of the triangle on the right = 9in²
Step 3:
The image below represents the rectangle
The area of the rectangle is calculated below as
[tex]\begin{gathered} A_{\text{rectangle}}=\text{length}\times breadth \\ \text{where,} \\ \text{length}=12in \\ \text{breadth}=9in \end{gathered}[/tex]
By substituting the values, we will have
[tex]\begin{gathered} A_{\text{rectangle}}=\text{length}\times breadth \\ A_{\text{rectangle}}=12in\times9in \\ A_{\text{rectangle}}=108in^2 \end{gathered}[/tex]
Hence,
The area of the rectangle is =108 in²
The area of the trapezoid will be calculated using the formula below
[tex]\begin{gathered} A_{\text{trapezoid}}=A_{\text{traingle}}+A_{\text{triangle}}+A_{\text{recatngle}} \\ A_{\text{trapezoid}}=9in^2+9in^2+108in^2 \\ A_{\text{trapezoid}}=126in^2 \end{gathered}[/tex]
Hence,
The area of the trapezoid is = 126 in²