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In the diagram below, RT:TS=1:2 and SR = PQ = 20. Find UV.

 

I tried using similar triangles, but I couldn't work it out.

 Jun 22, 2020
edited by tdmdfever  Jun 22, 2020
 #1
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I can do this; but I use coordinate geometry.

 

Place this diagram on a coordinate axis:  Place Q at the origin, QR on the positive x-axis and QP on the

positive y-axis.

Q = (0,0)    P = (0,20)    R = (a,0)             [use a variable like 'a' because we don't know exactly where it is]

 

Since  RT : TS  =  1 : 2,  RT will be one-third of SR.

Since  SR = 20,  RT = 6 2/3.

Coordinates of T = (a, 6 2/3)     S = (a,20)

 

Equation of QS:  slope  =  (20 - 0) / (a - 0)  =  20/a     y-intercept = 0     --->   y =  (20/a)·x

 

Equation of PT:  slope  =  ( 6 2/3 - 20) / (a - 0)

            [multiplying the numerator and the denominator of the slope by 3]

                           slope  =  (20 - 60) / (3a)   =  -40/(3a)     y-intercept = 20

     --->     y  =  [ -40/(3a) ]·x + 20

 

Finding where these two lines intersect:  (20/a)·x  =  [ -40/(3a) ]·x + 20

   [multiply each term by 3a]                            60x  =  -40x + 60a

                                                                       100x  =  60a

                                                                             x  =  (3/5)a

 

Substituting this value of x into the equation:  y =  (20/a)·x     --->     y  =  (20/a)·(3/5)a  =  12

 Jun 23, 2020
 #2
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Oh, ok, thank you so much!

tdmdfever  Jun 23, 2020

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