Diagonals at right angle

In the trapezoid ABCD this is given:
AB=12cm
CD=4cm
And diagonals crossed under a right angle. What is the area of this trapezoid ABCD?

Correct result:

S =  64 cm2

Solution:

a=12 cm c=4 cm  a:c=h1:h2  h=h1+h2  h1=a/2=12/2=6 cm h2=c/2=4/2=2 cm  h=h1+h2=6+2=8 cm  S=a+c2 h=12+42 8=64 cm2a=12 \ \text{cm} \ \\ c=4 \ \text{cm} \ \\ \ \\ a:c=h_{1}:h_{2} \ \\ \ \\ h=h_{1}+h_{2} \ \\ \ \\ h_{1}=a/2=12/2=6 \ \text{cm} \ \\ h_{2}=c/2=4/2=2 \ \text{cm} \ \\ \ \\ h=h_{1}+h_{2}=6+2=8 \ \text{cm} \ \\ \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ 12+4 }{ 2 } \cdot \ 8=64 \ \text{cm}^2



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