Answer:
ar ΔBDC=6 [tex]in^2[/tex]
ar ΔABD=8 [tex]in^2[/tex]
ar ΔABC=14 [tex]in^2[/tex]
Explanation:
Please take a look of attach figure.
We are given area of triangle ar ΔCDE=5 [tex]in^2[/tex]
ar ΔCDE=[tex]\frac{1}{2}\times DN\times EC[/tex]------------(1)
ar ΔBDE=[tex]\frac{1}{2}\times DN\times BE[/tex]------------(2)
Divide eq(2) by eq(1) and we get,
[tex]\frac{ar\ ΔBDE}{ar\ ΔCDE}=\frac{BE}{EC}[/tex]
[tex]\frac{ar\ ΔBDE}{5}=\frac{1}{5}[/tex]
So, ar ΔBDE=1 [tex]in^2[/tex]
ar ΔBDC=ar ΔBDE+ ar ΔCDE
ar ΔBDC=1+5 = 6 [tex]in^2[/tex]
Similarly,
[tex]\frac{ar\ ΔABD}{ar\ ΔBDC}=\frac{AD}{DC}[/tex]
[tex]\frac{ar\ ΔABD}{6}=\frac{4}{3}[/tex]
ar ΔABD=8 [tex]in^2[/tex]
ar ΔABC=ar ΔABD + ar ΔBDC
ar ΔABC=8+6 = 14 [tex]in^2[/tex]
