# 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:

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

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