# 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.

Correct result:

p =  33.31
A =  69.25

#### Solution:

$|AC| = 12 \ \\ |CD| = 8 \ \\ \ \\ \sin \Theta = \dfrac{|BC|}{|AC|} \ \\ \cos \Theta = \dfrac{|BC|}{|BD|} \ \\ \ \\ \cos^2 \Theta + \dfrac{ |CD|}{|AC|}\cos \Theta - 1 =0 \ \\ x^2 + \dfrac{ |CD|}{|AC|}x - 1 =0 \ \\ \ \\ x^2 +0.667x -1 =0 \ \\ \ \\ a=1; b=0.667; c=-1 \ \\ D = b^2 - 4ac = 0.667^2 - 4\cdot 1 \cdot (-1) = 4.4444444444 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ -0.67 \pm \sqrt{ 4.44 } }{ 2 } \ \\ x_{1,2} = -0.33333333 \pm 1.05409255339 \ \\ x_{1} = 0.720759220056 \ \\ x_{2} = -1.38742588672 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -0.720759220056) (x +1.38742588672) = 0 \ \\ \ \\ \ \\ \Theta = 43^\circ 52'58" \ \\ \ \\ |BC| = |AC| \sin \Theta = 8.31822608044 \ \\ |AB| = |AC| \cos \Theta = 8.64911064067 \ \\ |AD| = \sqrt{ |BC|^2 + (|AB|-|CD|)^2} = 8.34351423258 \ \\ \ \\ o = |AB|+|BC|+|CD| + |AD| = 33.31$
$A=\frac{\left(\mathrm{\mid }AB\mathrm{\mid }+\mathrm{\mid }CD\mathrm{\mid }\right)\cdot \mathrm{\mid }BC\mathrm{\mid }}{2}=69.25$

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