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# Help can anyone solve this? Im sure that there are some ingeniousness people here.

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What is the area of the trapezoid shown?

Dec 4, 2017
edited by Guest  Dec 4, 2017
edited by Guest  Dec 4, 2017
edited by Guest  Dec 4, 2017

#1
+7352
+2

First let's draw two lines perpendicular to the base, forming two right triangles, like this...

Now we can find  h  and  x  .

sin 60°  =  h / 2          so          h   =   2 sin 60°   =   √3

cos 60°  =  x / 2          so          x   =   2 cos 60°   =   1

So...

the height of the trapezoid   =   √3   ,

the first base   =   2  ,   and

the second base   =   2 + 1 + 1  =  4

area of trapezoid  =  (first base + second base)/2  *  height

area of trapezoid  =  (2 + 4)/2  *  √3

area of trapezoid  =  3√3    sq. units

Dec 4, 2017

#1
+7352
+2

First let's draw two lines perpendicular to the base, forming two right triangles, like this...

Now we can find  h  and  x  .

sin 60°  =  h / 2          so          h   =   2 sin 60°   =   √3

cos 60°  =  x / 2          so          x   =   2 cos 60°   =   1

So...

the height of the trapezoid   =   √3   ,

the first base   =   2  ,   and

the second base   =   2 + 1 + 1  =  4

area of trapezoid  =  (first base + second base)/2  *  height

area of trapezoid  =  (2 + 4)/2  *  √3

area of trapezoid  =  3√3    sq. units

hectictar Dec 4, 2017