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:

S=590 a=2x+3 c=x+8 h=x+4 S=a+c2 h 590=((2x+3)+(x+8))/2(x+4)  590=((2x+3)+(x+8))/2 (x+4) 1.5x211.5x+568=0 1.5x2+11.5x568=0  a=1.5;b=11.5;c=568 D=b24ac=11.5241.5(568)=3540.25 D>0  x1,2=b±D2a=11.5±3540.253 x1,2=3.83333333±19.8333333333 x1=16 x2=23.6666666667   Factored form of the equation:  1.5(x16)(x+23.6666666667)=0 x>0 x=x2=(23.6667)=71323.6667=23.6667 a=2 x+3=2 (23.6667)+3=133344.3333 c=x+8=(23.6667)+8=47315.6667 h=x+4=(23.6667)+4=59319.6667 S2=a+c2 h=(44.3333)+(15.6667)2 (19.6667)=590

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