# Trapezoid 15

Area of trapezoid is 266. What value is x if bases b1 is 2x-3, b2 is 2x+1 and height h is x+4

Result

x =  10

#### Solution:

$A=266 \ \\ A=(b_{1}+b_{2})h/2 \ \\ A=((2x-3)+(2x+1))(x+4)/2 \ \\ (2x-3+2x+1)(x+4)/2=266 \ \\ \ \\ 2x^2 +7x -270=0 \ \\ \ \\ a=2; b=7; c=-270 \ \\ D=b^2 - 4ac=7^2 - 4\cdot 2 \cdot (-270)=2209 \ \\ D>0 \ \\ \ \\ x_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -7 \pm \sqrt{ 2209 } }{ 4 } \ \\ x_{1,2}=\dfrac{ -7 \pm 47 }{ 4 } \ \\ x_{1,2}=-1.75 \pm 11.75 \ \\ x_{1}=10 \ \\ x_{2}=-13.5 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 2 (x -10) (x +13.5)=0 \ \\ x>0 \ \\ x=x_{1}=10$

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