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:

$$\begin{gathered} u=62\text{ cm } \\ a=\sqrt{u^2+u^2}=\sqrt{62^2+62^2}=62\sqrt{2}\text{ cm}\doteq 87.6812\text{ cm } \\ {c}^2+c^2=u^2 \\ c=u\mathrm{/}\sqrt{2}=62\mathrm{/}\sqrt{2}=31\sqrt{2}\text{ cm}\doteq 43.8406\text{ cm } \\ S={\displaystyle \frac{a+c}{2}}\cdot c={\displaystyle \frac{87.6812+43.8406}{2}}\cdot 43.8406=2883{\text{cm}}^2 \end{gathered}$$

$$\begin{gathered} o_1=2\cdot c+u=2\cdot 43.8406+62\doteq 149.6812\text{ cm } \\ {o}_2=2\cdot u+a=2\cdot 62+87.6812\doteq 211.6812\text{ cm } \\ x=o_2-o_1=211.6812-149.6812=62\text{ cm} \end{gathered}$$

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