trapezoid ABCD a = 35 m, b=28 m c = 11 m and d = 14 m. How to calculate its area?


S =  321.949 cm2


a=35 b=28 d=14 c=11 h>0  d2=x2+h2 b2=y2+h2 c=axy  142=x2+h2 282=y2+h2 c=35xy  x=1/4=14=0.25 y=97/4=974=24.25 c=axy=35(0.25)24.25=11  h=d2x2=142(0.25)213.9978 cm S=(a+c) h/2=(35+11) 13.9978/2321.9487321.949 cm2a=35 \ \\ b=28 \ \\ d=14 \ \\ c=11 \ \\ h>0 \ \\ \ \\ d^2=x^2+h^2 \ \\ b^2=y^2+h^2 \ \\ c=a-x-y \ \\ \ \\ 14^2=x^2+h^2 \ \\ 28^2=y^2+h^2 \ \\ c=35-x-y \ \\ \ \\ x=-1/4=- \dfrac{ 1 }{ 4 }=-0.25 \ \\ y=97/4=\dfrac{ 97 }{ 4 }=24.25 \ \\ c=a-x-y=35-(-0.25)-24.25=11 \ \\ \ \\ h=\sqrt{ d^2-x^2 }=\sqrt{ 14^2-(-0.25)^2 } \doteq 13.9978 \ \text{cm} \ \\ S=(a+c) \cdot \ h/2=(35+11) \cdot \ 13.9978/2 \doteq 321.9487 \doteq 321.949 \ \text{cm}^2

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How did you come up with x = -0.25 and y = 24.25?


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