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 answer:

x =  -23.6667

Step-by-step explanation:

$$\begin{gathered} S=590 \\ a=2x+3 \\ c=x+8 \\ h=x+4 \\ S=\frac{a+c}{2}\cdot h \\ 590=((2x+3)+(x+8))/2\ast (x+4) \\ 590=((2x+3)+(x+8))/2\cdot (x+4) \\ -1.5x^2-11.5x+568=0 \\ 1.5x^2+11.5x-568=0 \\ a=1.5;b=11.5;c=-568 \\ D=b^2-4ac=11.5^2-4\cdot 1.5\cdot (-568)=3540.25 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-11.5\pm \sqrt{3540.25}}{3} \\ {x}_{1,2}=-3.83333333\pm 19.833333333333 \\ {x}_1=16 \\ {x}_2=-23.666666666667 \\ \text{ Factored form of the equation: } \\ 1.5(x-16)(x+23.666666666667)=0 \\ x>0 \\ x=x_2=(-23.6667)=-\frac{71}{3}=-23\frac{2}{3}\doteq -23.6667 \\ a=2\cdot x+3=2\cdot \frac{-71}{3}+3=-\frac{133}{3}=-44\frac{1}{3}\doteq -44.3333 \\ c=x+8=\frac{-71}{3}+8=\frac{-71}{3}+\frac{8\cdot 3}{3}=\frac{-71}{3}+\frac{24}{3}=\frac{-71+24}{3}=\frac{-47}{3}=-\frac{47}{3}=-15\frac{2}{3}\doteq -15.6667 \\ h=x+4=\frac{-71}{3}+4=\frac{-71}{3}+\frac{4\cdot 3}{3}=\frac{-71}{3}+\frac{12}{3}=\frac{-71+12}{3}=\frac{-59}{3}=-\frac{59}{3}=-19\frac{2}{3}\doteq -19.6667 \\ {S}_2=\frac{a+c}{2}\cdot h=\frac{\frac{-133}{3}+\frac{-47}{3}}{2}\cdot \frac{-59}{3}=\frac{(-44.3333)+(-15.6667)}{2}\cdot (-19.6667)=590 \end{gathered}$$

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