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In trapezoid ABCD, \overline{AB} \parallel \overline{CD}. Find the area of the trapezoid.

 

 Jun 13, 2024
 #1
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Draw AE and BF  perpemdicular to BC

Let DE  = x

Let CF = [ 104 - ( 78 + x ) ] = 26 - x 

We have rwo right triangles ADE and BCF

Let the height of the  trapezoid  = h

 

So

h = sqrt [  10^2 - x^2]

h = sqrt [ 24^2 - (26  -x)^2 ]

 

So

 

sqrt [ 10^2 -x^2]  = sqrt [ 24^2 - (26 -x)^2 ]      square both sides

 

10^2-x^2  = 24^2 - (26 -x)^2

 

100 - x^2  = 576 - x^2 + 52x - 676  simplify

 

52x - 200 = 0

 

13x - 50 = 0

 

x = 50/13

 

h = sqrt [ 10^2 - (50/13)^2]  = 120/13

 

Area of  trapezoid   =  (1/2) (120/13) (78 + 104)   =  840

 

cool cool cool

 Jun 13, 2024

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