SOLUTION: Let $ABCD$ be a rectangle having an area of 290. Let $E$ be on $\overline{BC}$ such that $BE:EC=3:2$. Let $F$ be on $\overline{CD}$ such that $CF:FD=3:1$. If $G$ is the intersectio

Question 1202377: Let $ABCD$ be a rectangle having an area of 290. Let $E$ be on $\overline{BC}$ such that $BE:EC=3:2$. Let $F$ be on $\overline{CD}$ such that $CF:FD=3:1$. If $G$ is the intersection of $\overline{AE}$ and $\overline{BF}$, compute the area of $\triangle{BEG}$.

Found 2 solutions by josgarithmetic, math_tutor2020:
Answer by josgarithmetic(39617) (Show Source): You can put this solution on YOUR website!
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Answer by math_tutor2020(3817) (Show Source): You can put this solution on YOUR website!

Start by drawing an xy axis

Then add rectangle ABCD, along with points E,F,G as well.
I'll have point A placed at the origin.

Let m and n represent the lengths of segment FD and segment CE respectively.
m = FD
n = CE

BE:EC = 3:2
BE/EC = 3/2
BE/EC = 1.5
BE = 1.5EC
BE = 1.5n

CF:FD = 3:1
CF/FD = 3/1
CF/FD = 3
CF = 3*FD
CF = 3m

Segment Lengths:
DF = m
FC = 3m
AB = DC = DF+FC = m+3m = 4m
CE = n
EB = 1.5n
AD = CB = CE+EB = n+1.5n = 2.5n

Point locations
A = (0,0)
B = (4m,0)
C = (4m,2.5n)
D = (0,2.5n)
E = (4m,1.5n)
F = (m,2.5n)
G = unknown for now

Focus on these two points
A = (0,0)
E = (4m,1.5n)

Determine the equation of line AE.
I'll skip the steps.
You should get the result: y = (3nx)/(8m)

Now focus on these points
B = (4m,0)
F = (m,2.5n)
The equation of line BF is
y = (-5nx)/(6m) + 10n/3

We need to solve this system of equations

to determine the (x,y) location where they cross.
Technically we only need the x coordinate.

y = (3nx)/(8m)
(-5nx)/(6m) + 10n/3 = (3nx)/(8m)
48m * [ (-5nx)/(6m) + 10n/3 ] = 48m*(3nx)/(8m)
-40nx + 160mn = 18nx
160mn = 18nx + 40nx
160mn = 58nx
x = (160mn)/(58n)
x = 80m/29
This is the x coordinate of point G.

The horizontal distance from G to B is:
xB - xG = 4m - 80m/29 = 36m/29

This expression 36m/29 represents the height of triangle BEG if we had side EB as the base.
It might help to rotate the diagram so EB is horizontal.

area of triangle BEG = (1/2)*base*height
area of triangle BEG = (1/2)*(3n/2)*(36m/29)
area of triangle BEG = (27/29)*mn
We'll pause things to take a slight detour.

Rectangle ABCD has:
base = AB = 4m
height = BC = 2.5n
so,
area of rectangle ABCD = base*height
area of rectangle ABCD = AB*BC
area of rectangle ABCD = 4m*2.5n
area of rectangle ABCD = 10mn

We're told the rectangle has an area of 290 square units.
area = 10mn = 290
10mn = 290
mn = 290/10
mn = 29

We can now return to where we paused earlier.
area of triangle BEG = (27/29)*mn
area of triangle BEG = (27/29)*29
area of triangle BEG = 27

Answer: 27 square units

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