a)
Since DE // AC, then
Triangles BDE and BAC are similar
Then the ratio between their areas is the square of the ratio between their corresponding side
Since BD = 2 cm
Since DA = 4 cm, then
BA = 2 + 4 = 6 cm
The side BD in triangle BDE = 2
The side BA in triangle BAC = 6
The ratio between them is
[tex]\begin{gathered} \frac{BD}{BA}=\frac{2}{6} \\ \frac{BD}{BA}=\frac{\frac{2}{2}}{\frac{6}{2}} \\ \frac{BD}{BA}=\frac{1}{3} \end{gathered}[/tex]
We will use this ratio to find the ratio between their areas
[tex]\frac{A_{BDE}}{A_{BAC}}=(\frac{BD}{BA})^2[/tex]
Since the area of the triangle, BDE is 2 cm^2, then
[tex]\begin{gathered} \frac{2}{A_{BAC}}=(\frac{1}{3})^2 \\ \frac{2}{A_{BAC}}=\frac{1}{9} \end{gathered}[/tex]
By using the cross-multiplication
[tex]\begin{gathered} A_{BAC}\times1=2\times9 \\ A_{BAC}=18cm^2 \end{gathered}[/tex]
To find area DECA subtract area triangle BDE from area triangle BAC
[tex]\begin{gathered} A_{DECA}=18-2 \\ A_{DECA}=16cm^2 \end{gathered}[/tex]
The missing is 16
