Question

In △ABC, point D∈ AC with AD:DC=4:3, point E∈ BC so that BE:EC=1:5. If ACDE=5 in2, find ABDC, AABD, and AABC.

Answer 1

Answer:

ar ΔBDC=6 $in^2$

ar ΔABD=8 $in^2$

ar ΔABC=14 $in^2$

Explanation:

Please take a look of attach figure.

We are given area of triangle ar ΔCDE=5 $in^2$

ar ΔCDE=$\frac{1}{2}\times DN\times EC$------------(1)

ar ΔBDE=$\frac{1}{2}\times DN\times BE$------------(2)

Divide eq(2) by eq(1) and we get,

$\frac{ar\ ΔBDE}{ar\ ΔCDE}=\frac{BE}{EC}$

$\frac{ar\ ΔBDE}{5}=\frac{1}{5}$

So, ar ΔBDE=1 $in^2$

ar ΔBDC=ar ΔBDE+ ar ΔCDE

ar ΔBDC=1+5 = 6 $in^2$

Similarly,

$\frac{ar\ ΔABD}{ar\ ΔBDC}=\frac{AD}{DC}$

$\frac{ar\ ΔABD}{6}=\frac{4}{3}$

ar ΔABD=8 $in^2$

ar ΔABC=ar ΔABD + ar ΔBDC

ar ΔABC=8+6 = 14 $in^2$