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In rectangle $ABCD$, points $E$ and $F$ lie on segments $AB$ and $CD$, respectively, such that $AE = \frac{AB}{8}$ and $CF = \frac{CD}{5}$. 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.

 Jun 4, 2024
 #1
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Let D= (0,0)   Let  B= (40,10)

Slope of line  through DB = ( 10 - 0 ) / (40 -0)  =  1/4

Equation of  this line

y =(1/4)x

 

Let E = (5,10)  Let F = (8,0)

Slope of line through EF  = (0 -10)/ (8-5)  = -10/3

Equation of this line

y = (-103)(x -8)

y = (-10/3)x + 80/3

 

To find the x coordinate of P

(1/4)x = (-10/3)x + 80/3

(1/4 + 10/3)X = 80/3

(3/12 + 40/12)x = 80/3

(43/12)x = 80/3

x = (80/3) (12/43)  =  320/43

y coordinate of P  =  (1/4) (320/43)  = 80/43

 

Height of triangle EBP  =  10 - 80/43 =  350/43

Base of EBP = 40 - 5 = 35

Area of EBP  = (1/2) (35) (350/43)  = 6125/43

 

[EBP ] / [ ABCD]  =  (6125/43) / [400]  = 245 / 688

 

cool cool cool

 Jun 4, 2024

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