Trapezoid MO

Right-angled trapezoid ABCD has a right angle at vertex B. Given that |AC| = 12, |CD| = 8, and the diagonals are perpendicular to each other.

Calculate the perimeter and area of the trapezoid.

Final Answer:

o =  33.31
S =  69.25

Step-by-step explanation:

$$\begin{gathered} AC=12 \\ CD=8 \\ \sin \Phi =\frac{|BC|}{|AC|} \\ \cos \Phi =\frac{|BC|}{|BD|} \\ {\cos }^2\Phi +\frac{|CD|}{|AC|}\cos \Phi -1=0 \\ {x}^2+CD/AC\cdot x-1=0 \\ {x}^2+8/12\cdot x-1=0 \\ {x}^2+0.667x-1=0 \\ a=1;b=0.667;c=-1 \\ D=b^2-4ac=0.667^2-4\cdot 1\cdot (-1)=4.4444444444 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-0.67\pm \sqrt{4.44}}{2} \\ {x}_{1,2}=-0.333333\pm 1.054093 \\ {x}_1=0.72075922 \\ {x}_2=-1.387425887 \\ x=x_1=0.7208\doteq 0.7208 \\ \Phi =\arccos x=\arccos 0.7208\doteq 0.7659\text{rad} \\ {\Phi }_2=\Phi \to \text{°}=\Phi \cdot \frac{180}{\pi }\text{°}=0.7659\cdot \frac{180}{\pi }\text{°}=43.8828\text{°} \\ BC=AC\cdot \sin (\Phi )=12\cdot \sin 0.7659\doteq 8.3182 \\ AB=AC\cdot \cos (\Phi )=12\cdot \cos 0.7659\doteq 8.6491 \\ AD=\sqrt{B{C}^2+(AB-CD{)}^2}=\sqrt{8.3182^2+(8.6491-8{)}^2}\doteq 8.3435 \\ o=AB+BC+CD+AD=8.6491+8.3182+8+8.3435=33.31 \end{gathered}$$

Our quadratic equation calculator calculates it.

$S=\frac{(AB+CD)\cdot BC}{2}=\frac{(8.6491+8)\cdot 8.3182}{2}=69.25$

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