The bases

The bases of the isosceles trapezoid ABCD have lengths of 10 cm and 6 cm. Its arms form an angle α = 50˚ with a longer base. Calculate the circumference and content of the ABCD trapezoid.

Correct result:

o =  22.2229 cm
S =  19.0681 cm²

Solution:

$$\begin{gathered} a=10\text{ cm } \\ c=6\text{ cm } \\ \alpha =50^{\circ } \\ x=(a-c)\mathrm{/}2=(10-6)\mathrm{/}2=2\text{ cm } \\ \cos \alpha =x:r \\ r=x\mathrm{/}\cos (\alpha )=2\mathrm{/}\cos (50^{\circ })=3.11145 \\ o=a+c+2\cdot r=10+6+2\cdot 3.1114=22.2229\text{ cm} \end{gathered}$$

$$\begin{gathered} \tan \alpha =h:x \\ h=x\cdot \tan (\alpha )=2\cdot \tan (50^{\circ })=2.38351 \\ S={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{10+6}{2}}\cdot 2.3835=19.0681{\text{cm}}^2 \end{gathered}$$

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