+0  
 
0
66
4
avatar+891 

AB and CD are parallel, and AB < CD

. .

 Jul 29, 2020
 #1
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0

By Heron's formulas, the area of the trapezoid is 27*sqrt(3).

 Jul 29, 2020
 #2
avatar+1130 
0

we have (9+8+7)/2=12 and (sqrt(12(12-9)(12-8)(12-7)))/8 to find the altitude and we find that the altitude is sqrt720/8=sqrt(11.25) then we can use the pythagorean's theorem to have sqrt(49-11.25)=sqrt37.75 and (((8-2sqrt37.75)+8)*sqrt11.25)/2=8sqrt11.25-sqrt424.6875

 Jul 29, 2020
 #3
avatar+891 
0

What?

qwertyzz  Jul 29, 2020
 #4
avatar+110713 
-1

Use Heron's formula to find the area of triangle ACD

12sqrt5   units squared

 

Use the normal area of a triangle formula to get the height

12sqrt5=1/2 * 8 *h

h= 3sqrt5

 

units

 

Let X be the point such that AX is a height

Let Y be the point such that BY is a height

(SBYX is a rectangle)

since ABCD is a isosceles trapezium, DX = YC

 

Find YC

YC^2 = 49-45=4

YC=DX=2


XY=AB=4

 

\(Area=\frac{4+8}{2}*3\sqrt5\\ area=18\sqrt5 units^2\)

 

 

 

You need to draw the pic and check it.

I largely did it in my head so there could easily be errors.

 Jul 30, 2020

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