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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

#1**+2 **

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 :

CPhill Aug 9, 2017