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?
+128475 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 !!!!
+1639 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?
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I'll borrow Phill's diagram.
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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
