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# new question

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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.

Jan 13, 2019

#1
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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

Jan 13, 2019
edited by CPhill  Jan 13, 2019
#2
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good job cphill!

Jan 13, 2019
#3
+111437
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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

(4/21) area of ABCD

Jan 13, 2019
edited by CPhill  Jan 13, 2019
edited by CPhill  Jan 13, 2019
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