in a rectangle ABCD,AB=6,BC=3.point EbetweenB and C,and fbetween E and C are such thatBE=EF=FC intersect BD at P and Q respectively.the ratio of bp:pqqd can be written as r:s:t where the greatest common factor of r,s,t is 1.find r+s+t=?
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see diagram.
Let B be the origin.
Equation of st line BD is : y = 6/3 x or, y = 2 x
Equation of st line AE is : (y-0)/(x-1) = (6-0)/(0-1) or y + 6 x = 6
Intersection of BD and AE is : P(3/4, 3/2)
Equation of st line AF is : (y-0)/(x-2) = (6-0)/(0-2) or y + 3 x = 6
Intersection of BD and AF is: Q(6/5, 12/5)
r = BP = 3√5 / 4 cm
s = PQ = √[(6/5-3/4)² + (12/5 - 3/2)²] = √(81/400 + 81/100) = 9√5 / 20 cm
t = QD = BD - r - s = 3√5 - 3√5 /4 - 9 √5 / 20 cm
=> t = 9√5 / 5 cm
r : s : t = 3/4 : 9/20 : 9/5
= 5 : 3 : 12
r+s+t = 20
Let B be the origin.
Equation of st line BD is : y = 6/3 x or, y = 2 x
Equation of st line AE is : (y-0)/(x-1) = (6-0)/(0-1) or y + 6 x = 6
Intersection of BD and AE is : P(3/4, 3/2)
Equation of st line AF is : (y-0)/(x-2) = (6-0)/(0-2) or y + 3 x = 6
Intersection of BD and AF is: Q(6/5, 12/5)
r = BP = 3√5 / 4 cm
s = PQ = √[(6/5-3/4)² + (12/5 - 3/2)²] = √(81/400 + 81/100) = 9√5 / 20 cm
t = QD = BD - r - s = 3√5 - 3√5 /4 - 9 √5 / 20 cm
=> t = 9√5 / 5 cm
r : s : t = 3/4 : 9/20 : 9/5
= 5 : 3 : 12
r+s+t = 20
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