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In trapezoid $ABCD,$ $\overline{AB} \parallel \overline{CD}$. Find the area of the trapezoid.

 

 Jan 15, 2024
 #1
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Angle OCD = 45

Angle COD  = 90

So....Angle ODC = 45

So triangle DCO  is 45  - 45 - 90

So  OD  = 10/sqrt 2  = OC  =  5sqrt 2

 

Area of triangle DCO   = (1/2) (5sqrt 2)(5sqrt 2)   = 25

 

Height of triangle DCO can be found as

 

25  = (1/2)CD * height

 

50 = 10 * height

 

50/10 =  height   = 5

 

Triangle ABO  similar to Triangle DCO

 

And the scale factor from triangle ABO  to Triangle DCO   = 4/10 =  2/5

 

So the height of triangle ABO  = (2/5) (height of tragle DCO)  = (2/5) (5)  = 2

 

So the trapezoid has a height of  5 + 2 =  7

 

And the bases are 10 and 4

 

Area =   (1/2) (7) (10 + 4)  =  49

 

cool cool cool

 Jan 16, 2024

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