Find the area of the following shape.

57 units^2

Step-by-step explanation:

First find the area of the triangle on the left

ABC

It has a base AC which is  9 units and a height of 3 units

A = 1/2 bh = 1/2 ( 9) *3 = 27/2 = 13.5

Then find the area of the triangle on the right

DE

It has a base AC which is  6 units and a height of 1 units

A = 1/2 bh = 1/2 ( 6) *1  = 3

Then find the area of the triangle on the top

It has a base AC which is  3 units and a height of 3 units

A = 1/2 bh = 1/2 ( 3) *3  = 9/2 = 4.5

Then find the area of the rectangular region

A = lw = 6*6 = 36

Add them together

13.5+3+4.5+36 =57 units^2

Total Area = 57 sq. units

Step-by-step explanation:

will make it simple and short

Total Area = A1 + A2 + A3

A1 = (7 + 6) * 6/2 = 39 sq. units  (area of a trapezoid)

A2 = 1/2 (9 * 3) = 13.5 sq. units (area of a triangle)

A3 = 1/2 (3 * 3) = 4.5 sq. units  (area of a triangle)

Total Area = 39 + 13.5 + 4.5 = 57 sq. units

just do 5.2 + 2 ( 7.2) and 1/3 x 3 (.9) and youv'e got your equation. 0.9( 9/10) + 7.2 ^2 = 16.4 hope i could !

step-by-step explanation:

idk

step-by-step explanation: i just don't