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.

Correct result:

u1 =  9.04
u2 =  15.04
a =  8.77

Solution:

S=u1u2/2 u2=u1+6  u12+6u1268=0 x2+6x136=0  a=1;b=6;c=136 D=b24ac=6241(136)=580 D>0  x1,2=b±D2a=6±5802=6±21452 x1,2=3±12.041594578792 x1=9.0415945787923 x2=15.041594578792   Factored form of the equation:  (x9.0415945787923)(x+15.041594578792)=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 12.041594578792 \ \\ x_{1} = 9.0415945787923 \ \\ x_{2} = -15.041594578792 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ (x -9.0415945787923) (x +15.041594578792) = 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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