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An isosceles trapezoid has legs of length 30 cm each, two diagonals of length 40 cm each and the longer base is 50 cm. What is the trapezoid's area in sq cm?

RektTheNoob  Aug 9, 2017
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Probably several ways to do this....but here's a method using Heron's Formula

 

Let 50 cm  = the bottom  base

 

We can find the area of one of the triangles that comprise the area of the trapezoid.....its sides are 30, 40 and 50  cm

 

Let s be the semi-perimeter of one of the triangle  = [ 50 + 40 + 30 ] / 2  = 60 cm

 

And the area of  of this triangle  =  sqrt  [ s ( s - a) (s - b) ( s -c) ]     where a,b, c are the sides

 

So...we have    sqrt [ 60 ( 60 - 30) (60 - 40) (60 - 50) ]  = sqrt [ 60 * 30 * 20 * 10 ] =

 

sqrt [ 1800 * 200]   =   sqrt [ 3600 * 100]  = 60 * 10  = 600 cm ^2

 

Now the base of this triangle = 50 and the height can be found as follows :

 

Area =  (1/2) 50 * height

 

600 = 25 * height

 

height  =  24 cm   and this is the height of the trapezoid

 

Now....the top base can be found as   50  -  2 sqrt [ 30^2 - 24^2 ]  = 14 cm

 

So....the area of the trapezoid  = (1/2) height ( sum of the bases)  =  (1/2) (24) ( 50 + 14)  = 12 ( 64)  = 768 cm'^2

 

Here's a pic :

 

 

 

cool cool cool

CPhill  Aug 9, 2017
edited by CPhill  Aug 9, 2017

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