#1**0 **

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**

geno3141 Jun 23, 2020