+1262 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
+118608 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.
