Rectangular trapezoid

The ABCD rectangular trapezoid with the AB and CD bases is divided by the diagonal AC into two equilateral rectangular triangles. The length of the diagonal AC is 62cm. Calculate trapezium area in cm square and calculate how many differs perimeters of the ABC and ACD triangles in centimeters.

Correct result:

S =  2883 cm2
x =  62 cm

Solution:

u=62 cm a=u2+u2=622+62262 2 cm87.6812 cm  c2+c2=u2 c=u/2=62/231 2 cm43.8406 cm  S=a+c2 c=87.6812+43.84062 43.8406=2883 cm2u=62 \ \text{cm} \ \\ a=\sqrt{ u^2+u^2 }=\sqrt{ 62^2+62^2 } \doteq 62 \ \sqrt{ 2 } \ \text{cm} \doteq 87.6812 \ \text{cm} \ \\ \ \\ c^2 + c^2=u^2 \ \\ c=u / \sqrt{ 2 }=62 / \sqrt{ 2 } \doteq 31 \ \sqrt{ 2 } \ \text{cm} \doteq 43.8406 \ \text{cm} \ \\ \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ c=\dfrac{ 87.6812+43.8406 }{ 2 } \cdot \ 43.8406=2883 \ \text{cm}^2
o1=2 c+u=2 43.8406+62149.6812 cm o2=2 u+a=2 62+87.6812211.6812 cm  x=o2o1=211.6812149.6812=62 cmo_{1}=2 \cdot \ c + u=2 \cdot \ 43.8406 + 62 \doteq 149.6812 \ \text{cm} \ \\ o_{2}=2 \cdot \ u + a=2 \cdot \ 62 + 87.6812 \doteq 211.6812 \ \text{cm} \ \\ \ \\ x=o_{2}-o_{1}=211.6812-149.6812=62 \ \text{cm}



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