Question

In △ABC, point D∈ AB with AD:DB=5:3, point E∈ BC so that BE:EC=1:4. If AABC=40 in2, find AADC, ABDC, and ACDE.

Answer 1

1. If AD : DB = 5 : 3, then AD = 5x in and DB = 3x in.

2. If BE : EC = 1 : 4, then BE = y and EC = 4y.

3. Consider ΔABc. The area of this triangle is

$A_{ABC}=\dfrac{1}{2}\cdot AB\cdot h_C,$

where $h_C$ is the heigth drawn to the side AB.

Since AB = AD + DB = 5x + 3x = 8x, you have that

$A_{ABC}=\dfrac{1}{2}\cdot 8y\cdot h_C=40\ in^2,\\ \\x\cdot h_C=10\ in^2.$

4. Consider ΔADC. The area of this triangle is

$A_{ADC}=\dfrac{1}{2}\cdot AD\cdot h_C=\dfrac{1}{2}\cdot 5x\cdot h_C=\dfrac{1}{2}\cdot 5\cdot 10=25\ in^2.$

5. Consider ΔBDC. The area of this triangle is

$A_{BDC}=\dfrac{1}{2}\cdot BD\cdot h_C=\dfrac{1}{2}\cdot 3x\cdot h_C=\dfrac{1}{2}\cdot 3\cdot 10=15\ in^2.$

On the other hand,

$A_{BDC}=\dfrac{1}{2}\cdot BC\cdot h_D=\dfrac{1}{2}\cdot (y+4y)\cdot h_D=15\ in^2,\\ \\y\cdot h_D=6\ in^2.$

6. Consider ΔCDe. The area of this triangle is

$A_{CDE}=\dfrac{1}{2}\cdot CE\cdot h_D=\dfrac{1}{2}\cdot 4y\cdot h_D=12\ in^2.$

Answer: $A_{ADC}=25\ in^2,\ A_{BDC}=15\ in^2,\ A_{CDE}=12\ in^2.$