# 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?

Result

S =  64 cm2

#### Solution:

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

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