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Triangles help plz

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Find the area of the trapezoid D U C K below

Oct 30, 2019

#1
+2

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.

Oct 30, 2019
#2
+109202
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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

Oct 30, 2019