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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 \(a+b+c+d+e+f+g+h+i\).

 

 Jan 4, 2021
 #1
avatar+114090 
+2

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

 

 

cool cool cool

 Jan 4, 2021
 #2
avatar+824 
+1

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 smiley

 Jan 4, 2021

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