Trapezium

The lengths of a 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.

Result

x =  -23.667

Solution:

S=590 a=2x+3 c=x+8 h=x+4 S=a+c2⋅ h  590=((2x+3)+(x+8))/2⋅ (x+4) −1.5x2−11.5x+568=0 1.5x2+11.5x−568=0  a=1.5;b=11.5;c=−568 D=b2−4ac=11.52−4⋅1.5⋅(−568)=3540.25 D>0  x1,2=−b±D2a=−11.5±3540.253 x1,2=−3.83333333±19.833333333333 x1=16 x2=−23.666666666667   Factored form of the equation:  1.5(x−16)(x+23.666666666667)=0 x>0 x=x2=(−23.6667)≐−713≐−23.6667≐−23.667 a=2⋅ x+3=2⋅ (−23.6667)+3≐−1333≐−44.3333 c=x+8=(−23.6667)+8≐−473≐−15.6667 h=x+4=(−23.6667)+4≐−593≐−19.6667 S2=a+c2⋅ h=(−44.3333)+(−15.6667)2⋅ (−19.6667)=590S=590 \ \\ a=2x+3 \ \\ c=x+8 \ \\ h=x+4 \ \\ S=\dfrac{ a+c }{ 2 } \cdot \ h \ \\ \ \\ 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}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -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) \doteq - \dfrac{ 71 }{ 3 } \doteq -23.6667 \doteq -23.667 \ \\ a=2 \cdot \ x+3=2 \cdot \ (-23.6667)+3 \doteq - \dfrac{ 133 }{ 3 } \doteq -44.3333 \ \\ c=x+8=(-23.6667)+8 \doteq - \dfrac{ 47 }{ 3 } \doteq -15.6667 \ \\ h=x+4=(-23.6667)+4 \doteq - \dfrac{ 59 }{ 3 } \doteq -19.6667 \ \\ S_{2}=\dfrac{ a+c }{ 2 } \cdot \ h=\dfrac{ (-44.3333)+(-15.6667) }{ 2 } \cdot \ (-19.6667)=590

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