Isosceles trapezoid

Calculate the content of an isosceles trapezoid whose bases are at ratio 5:3, the arm is 6cm long and it is 4cm high.

Correct result:

S =  71.5542 cm²

Solution:

$$\begin{gathered} a:c=5:3 \\ b=6\text{ cm } \\ d=b=6=6\text{ cm } \\ h=4\text{ cm } \\ x=\sqrt{b^2-h^2}=\sqrt{6^2-4^2}=2\sqrt{5}\text{ cm}\doteq 4.4721\text{ cm } \\ a=x+c+x \\ 3a=5c \\ 3\cdot (c+2x)=5c \\ 3c+6x=5c \\ 2c=6x \\ c=3\cdot x=3\cdot 4.4721=6\sqrt{5}\text{ cm}\doteq 13.4164\text{ cm } \\ a=5\mathrm{/}3\cdot c=5\mathrm{/}3\cdot 13.4164=10\sqrt{5}\text{ cm}\doteq 22.3607\text{ cm } \\ S=(a+c)\cdot h\mathrm{/}2=(22.3607+13.4164)\cdot 4\mathrm{/}2=32\sqrt{5}\doteq 71.5542=71.5542{\text{cm}}^2 \\ \text{ Verifying Solution: } \\ {k}_1={\displaystyle \frac{a}{c}}={\displaystyle \frac{22.3607}{13.4164}}={\displaystyle \frac{5}{3}}\doteq 1.6667 \end{gathered}$$

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