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# help

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The figure below shows a trapezium ABCD, where AB : DC = 2 : 3 is the mid-point of AB. Straight lines DE and DB meet AC at F and G respectively. Find the radio of AF : FG : GC. Nov 12, 2019

#1
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Probably  a way better "geometric" way to do this

Let

A = (3,4)

B = (7,4)

C = (6,0)

D = (0,0)

E = (5,4)

The equation of the line containing DE  is   y = (4/5)x   (1)

The equation of the line contining DB  is  y = (4/7)x      (2)

The slope of the line containing AC  is [ 4 - 0] / [ 3 - 6 ]  =  -4/3

And the equation of this line is  y = (-4/3)(x - 6)  = (-4/3)x + 8   (3)

The intersection of (1)  and (3)  will give the x coordnate  of F

So

(4/5)x  = (-4/3)x + 8

(4/5)x + (4/3)x  = 8

([ 12 + 20] / 15)x  = 8

([ 32] / 15 )x =   8

x = 15 * 8  / 32

x = 15/4  = 3.75

And y  = (4/5) (15/4)  = 3

So  F = ( 3.75, 3)

And the intersection of (2) and (3) will give the x coordinate of G

So

(4/7)x  = (-4/3)x + 8

(4/7)x + (4/3)x  = 8

([ 12 + 28]  / 12) x  = 8

( [ 40 ]/ 21) x  = 8

x =  8 * 21 / 40

x =  21/ 5  =  4.2

And y = (4/7)(21/5)  = (4/5)* 3  = 12/5  = 2.4

So  G  = ( 4.2, 2.4)

And AF =  √[(3 - 3.75)^2  + ( 4- 3)^2 ] =  √[.75^2 + 1] =  1.25

And FG  = √[ (3.75- 4.2)^2 + ( 3 - 2.4)^2 ]=  √[[.45^2 + .6^2] = √.5625  = .75

And GC = √[(4.2 - 6)^2  + 2.4^2 ]  = √ [ 1.8^2 + 2.4^2 ] = √9  = 3

So

AF : FG : GC  =

1.25  : .75 :  3   =

5 (.25) : 3 (.25) :  12 (.25)  =

5   :  3     :   12   Nov 13, 2019