Rhombus

The rhombus with area 68 has one diagonal is longer by 6 than second one.
Calculate the length of the diagonals and rhombus sides.

Result

u1 =  9.04
u2 =  15.04
a =  8.77

Solution:

S=u1u2/2 u2=u1+6  u12+6u1−2⋅68=0 x2+6x−136=0  a=1;b=6;c=−136 D=b2−4ac=62−4⋅1⋅(−136)=580 D>0  x1,2=−b±D2a=−6±5802=−6±21452 x1,2=−3±12.0415945788 x1=9.04159457879 x2=−15.0415945788   Factored form of the equation:  (x−9.04159457879)(x+15.0415945788)=0  u1>0 u1=9.04 S = u_1 u_2 /2 \ \\ u_2 = u_1 + 6 \ \\ \ \\ u_1^2+6 u_1 - 2 \cdot 68 = 0 \ \\ x^2 +6x -136 =0 \ \\ \ \\ a=1; b=6; c=-136 \ \\ D = b^2 - 4ac = 6^2 - 4\cdot 1 \cdot (-136) = 580 \ \\ D>0 \ \\ \ \\ x_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ -6 \pm \sqrt{ 580 } }{ 2 } = \dfrac{ -6 \pm 2 \sqrt{ 145 } }{ 2 } \ \\ x_{1,2} = -3 \pm 12.0415945788 \ \\ x_{1} = 9.04159457879 \ \\ x_{2} = -15.0415945788 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -9.04159457879) (x +15.0415945788) = 0 \ \\ \ \\ u_1 > 0 \ \\ u_1 = 9.04 \ \\
u2=u1+6=15.04 u_2 = u_1 + 6 = 15.04 \ \\
a2=(u1/2)2+(u2/2)2 a=(u1/2)2+(u2/2)2=8.77a^2 = (u_1/2)^2+(u_2/2)^2 \ \\ a = \sqrt{ (u_1/2)^2+(u_2/2)^2 } = 8.77

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