Ratio in trapezium

The ratio of the height v and the base a, c in the trapezoid ABCD is 1:6:3. Its area is 324 square cm, and the peak angle B is 35 degrees. Determine the perimeter of the trapezoid.

Final Answer:

o =  106.9691 cm

Step-by-step explanation:

$$\begin{gathered} v:a:c=1:6:3 \\ S=324{\text{cm}}^2 \\ \beta =35{}^{\circ } \\ a=6v \\ c=3v \\ S=\frac{a+c}{2}\cdot v \\ S=\frac{6v+3v}{2}\cdot v \\ S=\frac{9}{2}\cdot v^2 \\ v=\sqrt{2\cdot S/9}=\sqrt{2\cdot 324/9}=6\sqrt{2}\text{cm}\doteq 8.4853\text{cm} \\ a=6\cdot v=6\cdot 8.4853=36\sqrt{2}\text{cm}\doteq 50.9117\text{cm} \\ c=3\cdot v=3\cdot 8.4853=18\sqrt{2}\text{cm}\doteq 25.4558\text{cm} \\ a=x_1+c+x_2 \\ \tan \beta =v:x_2 \\ {x}_2=v/\tan \beta =v/\tan 35°=8.4853/\tan 35°=8.4853/0.700208=12.11824\text{cm} \\ {x}_1=a-c-x_2=50.9117-25.4558-12.1182\doteq 13.3376\text{cm} \\ b=\sqrt{v^2+x_2^2}=\sqrt{8.4853^2+12.1182^2}\doteq 14.7936\text{cm} \\ d=\sqrt{v^2+x_1^2}=\sqrt{8.4853^2+13.3376^2}\doteq 15.808\text{cm} \\ o=a+b+c+d=50.9117+14.7936+25.4558+15.808\doteq 106.9691\text{cm} \\ \text{ Verifying Solution: } \\ {S}_1=\frac{a+c}{2}\cdot v=\frac{50.9117+25.4558}{2}\cdot 8.4853=324{\text{cm}}^2 \end{gathered}$$

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