First consider triangle DUS (S is next to the 5 at the right angle)
Use pythagoras theorem to find US (The height) which will be 13^2-5^2=144=12 So the height is 12 units
Constract a line from C perpendicular on DK and let it be F
so CFK is another triangle, notice the height is just the same which is 12
so CF=12=US=12
We want to find FK
so 20^2-12^2=256 sqrt of that =16
so FK=16
Notice UC=15 so do SF=15
Now apply the formula to find the area
which is
1/2(b_1+b_2)*h
1/2(15+36)*12=306
The are is 306 units squared.
+128707 The triangle on the left is a 5 - 12 - 13 right triangle
The area = (1/2) product of the legs = (1/2)(5)(12) = 30 units^2
Draw a perpendicular from C to DK.....call the intersection point E
So....in the middle....we have a rectangle with a width of 15 and a height of 12
So....the area = 15 * 12 = 180 units^2
And triangle CEK is a right triangle with CE = 12 and CK = 20
So...
EK = sqrt [ CK^2 - CE^2 ] = sqrt [ 20^2 - 12^2] = sqrt [ 400 - 144] = sqrt [ 256] = 16
So....the area of CEK = (1/2)(16)(12) = 96 units^2
So....the area of the trapezoid =
30 + 180 + 96 =
306 units^2
