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Consider a square ABCD with side length 2. Let E be the midpoint of AB, F the midpoint of BC, and P and Q the points at which line segment AF intersects DE and DB, respectively.  What is the area of EBQP?

 Jan 21, 2021
 #1
avatar+116126 
+1

See  the  following :

 

 

Segment   DB   has a slope of    (2-0) / (0-2)  =2/-2  = -1

And the equation of  line containing DB  is    y = -x  + 2      (1)

 

Segment AF  has a slope of  ( 1-0) /(2 - 0)   =1/2

And the equation of the  line  containing  AF  is

y = (1/2)x     (2)

 

Find the x  coordinate of  Q  ⇒    (1)  = (2)

 

-x + 2 =  1/2x

2 =  3/2x

x = 4/3

 

And the y coordinate  is   (1/2)(4/3)  =  4/6 = 2/3

Thie is the height  of    triangle  AQB....and the  base  is  AB  = 2

So  the area of tis triangle is   (1/2) (2/3) (2)  =  2/3        (3)

 

And the  slope of the segment containing DE  is  (2 - 0) ( 0-1)  =  -2

And the equation of the line comtaining this segment is   y = -2x + 2     (4)

 

Find the  x coordinate  of P    ( 2)  = (4)

 

1/2x =   -2x + 2

5/2x  = 2

x  =   4/5

And the  y coordinate of   P  =  (1/2)( 4/5)  =  4/10  =  2/5

And this is the  height  of triangle APE    and the base = AE =  1

And its area is  (1/2) ( 1) (2/5)  = 1/5     (   5)

 

The area of   EBQP   =  ( 3)  - ( 5)  =   2/3   -  1/5  =      7/ 15 

 

EDIT TO CORRECT A  PRIOR ERROR  !!!!

 

 

cool cool cool

 Jan 21, 2021
edited by CPhill  Jan 21, 2021
 #2
avatar+1035 
+3

Consider a square ABCD with side length 2. Let E be the midpoint of AB, F the midpoint of BC, and P and Q the points at which line segment AF intersects DE and DB, respectively.  What is the area of EBQP?

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

I'll borrow Phill's diagram.

~~~~~~~~~~~~~~~~~~~~~~~~

Area of Δ ABD = 2

Since DE is a median of Δ ABD, the area of Δ BED = 1

∠ADE = tan-1(1/2)

DP = cos-1(∠ADE) * 2

∠EDB = 45 - ∠ADE

PQ = tan(∠EDB) * DP

Area of Δ DPQ = 1/2 (DP * PQ)

[EBQP] = [EDB] - [DPQ] = 7/15     or     0.466666667 square units

 Jan 21, 2021

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