Trapezium

The lengths of parallel sides of a trapezium are (2x+3) and (x+8), and the distance between them is (x+4). if the area of the trapezium is 590, find the value of x.

Correct result:

x =  -23.6667

Solution:

$$\begin{gathered} S=590 \\ a=2x+3 \\ c=x+8 \\ h=x+4 \\ S={\displaystyle \frac{a+c}{2}}\cdot h \\ 590=((2x+3)+(x+8))\mathrm{/}2\ast (x+4) \\ 590=((2x+3)+(x+8))\mathrm{/}2\cdot (x+4) \\ -1.5x^2-11.5x+568=0 \\ 1.5x^2+11.5x-568=0 \\ a=1.5;b=11.5;c=-568 \\ D=b^2-4ac=11.5^2-4\cdot 1.5\cdot (-568)=3540.25 \\ D>0 \\ {x}_{1,2}={\displaystyle \frac{-b\pm \sqrt{D}}{2a}}={\displaystyle \frac{-11.5\pm \sqrt{3540.25}}{3}} \\ {x}_{1,2}=-3.83333333\pm 19.8333333333 \\ {x}_1=16 \\ {x}_2=-23.6666666667 \\ \text{ Factored form of the equation: } \\ 1.5(x-16)(x+23.6666666667)=0 \\ x>0 \\ x=x_2=(-23.6667)=-{\displaystyle \frac{71}{3}}\doteq -23.6667=-23.6667 \\ a=2\cdot x+3=2\cdot (-23.6667)+3=-{\displaystyle \frac{133}{3}}\doteq -44.3333 \\ c=x+8=(-23.6667)+8=-{\displaystyle \frac{47}{3}}\doteq -15.6667 \\ h=x+4=(-23.6667)+4=-{\displaystyle \frac{59}{3}}\doteq -19.6667 \\ {S}_2={\displaystyle \frac{a+c}{2}}\cdot h={\displaystyle \frac{(-44.3333)+(-15.6667)}{2}}\cdot (-19.6667)=590 \end{gathered}$$

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