# Trapezoid MO

The rectangular trapezoid ABCD with right angle at point B, |AC| = 12, |CD| = 8, diagonals are perpendicular to each other.

Calculate the perimeter and area of ​​the trapezoid.

Result

p =  33.31
A =  69.25

#### Solution:

$|AC| = 12 \ \\ |CD| = 8 \ \\ \ \\ \sin \Theta = \dfrac{|BC|}{|AC|} \ \\ \cos \Theta = \dfrac{|BC|}{|BD|} \ \\ \ \\ \cos^2 \Theta + \dfrac{ |CD|}{|AC|}\cos \Theta - 1 =0 \ \\ x^2 + \dfrac{ |CD|}{|AC|}x - 1 =0 \ \\ \ \\ x^2 +0.667x -1 =0 \ \\ \ \\ a=1; b=0.667; c=-1 \ \\ D = b^2 - 4ac = 0.667^2 - 4\cdot 1 \cdot (-1) = 4.4444444444 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ -0.67 \pm \sqrt{ 4.44 } }{ 2 } \ \\ x_{1,2} = -0.33333333 \pm 1.0540925533895 \ \\ x_{1} = 0.72075922005613 \ \\ x_{2} = -1.3874258867228 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -0.72075922005613) (x +1.3874258867228) = 0 \ \\ \ \\ \Theta = 43^\circ 52'58" \ \\ \ \\ |BC| = |AC| \sin \Theta = 8.3182260804446 \ \\ |AB| = |AC| \cos \Theta = 8.6491106406735 \ \\ |AD| = \sqrt{ |BC|^2 + (|AB|-|CD|)^2} = 8.3435142325775 \ \\ \ \\ o = |AB|+|BC|+|CD| + |AD| = 33.31$
$A = \dfrac{(|AB|+|CD|)\cdot |BC|}{2}= 69.25$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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