Let \(\Delta ABC\) be an isosceles triangle with \(AC=CB\), \(AB=7\) and height \(CD = 9\). Denote \(a,b,c,d,e,f,g,h\) and \(i\) to be parallel segments \(AB\) and divide \(CD\) into 9 equal segments.
Compute \(a+b+c+d+e+f+g+h+i\).
+128407 i = 7/2 * (9/9) = 3.5
h = (7/2) * (8/9)
g = (7/2) * ( 7/9)
f = (7/2) * (6/9)
e = (7/2)* (5/9)
d = (7/2) * ( 4/9)
c = (7/2) * (3/9)
b =(7/2) * (2/9)
a = (7/2) * (1/9)
Their sum =
(7/2) ( 9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) / 9 =
(7/18) * ( 45) =
(45/18) * 7 =
(5/2) * 7 =
35 / 2 =
17.5
+1639 The distance between the horizontal lines is 1
The tangent of angles ACD and BCD is 7/18 or 0.38888889 ==> t
a = 1 * t = 0.38888888
b = 2 * t = 0.77777777
c = 3 * t = 1.16666666
d = 4 * t = 1.55555555
e = 5 * t = 1.94444444
f = 6 * t = 2.33333333
g = 7 * t = 2.72222222
h = 8 * t = 3.111111111
i = 9 * t = 3.50000000
Add them up and you'll get 17.5 
