+532 In rectangle ABCD, points E and F lie on segments AB and CD, respectively, such that AE=AB/3 and CF=CD/2. Segment BD intersects segment EF at P. What fraction of the area of rectangle ABCD lies in triangle EBP? Express your answer as a common fraction.
+129899 Not a particularly "geometric" solution....but see the following image
Let A = (0,4) B = (6, 4) C = (6,0) D = ( 0, 0) E = (2, 4) F = (3,0)
The segment connecting BD has the slope 2/3
And its equation is y = 2/3x
The segment connecting EF has a slope of [ 4 - 0 ] / [ 2 - 3 ] = -4
So....its equation is y = -4(x - 3) ⇒ y = -4x + 12
The x coordinate for P is
(2/3)x = -4x + 12
(2/3)x + 4x = 12
(14/3)x = 12
x = 36/14 = 18/7
And its y coordinate is y = (2/3)(18 7) = 36/21 = 12/7
So.....the base of EBP = 4
And its height is 4 - 12/7 = 16/7
So...its area = (1/2)(4)(16/7) = 32/7
And the area of the rectangle = 6 * 4 = 24
So....the fractional area of EBP to the area of the rectangle is
(32/7) / 24 = 32 / 168 = 4 / 21
+129899 Thanks, asdf...!!!!
Here's a more general solution :
Angle EBP = Angle FDP
And angle BPE = angle DPF
So...by AA congruency.....triangle EBP is similar to triangle FDP
Note that EB is 2/3 of AB = 2/3 of CD
And FD is 1/2 of CD
So.... EB /FD = (2/3)CD / (1/2)CD = 4/3
So....EB is (4/3) FD
So....the height of EBP must be (4/3) the height of FDP
So.....there are 7 equal parts of AD.....and the height of EBP must be 4 of these
So.....height of EBP = (4/7)AD
So.....the area of EBP = (1/2)*(2/3)AB * (4/7)AD = (1/3)(4/7)(AB)(AD) = (4/21)(AB)(AD) =
(4/21) area of ABCD
