Diagonals of a rhombus 2

One diagonal of a rhombus is greater than other by 4 cm . If the area of the rhombus is 96 cm2, find the side of the rhombus.

Result

a =  10 cm

Solution:

u=4+v=4+12=16 S=96 cm2  S=uv2=(4+v)v2   96=(4+v)⋅ v/2 −0.5v2−2v+96=0 0.5v2+2v−96=0  a=0.5;b=2;c=−96 D=b2−4ac=22−4⋅0.5⋅(−96)=196 D>0  v1,2=−b±D2a=−2±1961=−2±141 v1,2=−2±14 v1=12 v2=−16   Factored form of the equation:  0.5(v−12)(v+16)=0  v=v1=12 cm u=4+v=4+12=16 cm  a=(u/2)2+(v/2)2=(16/2)2+(12/2)2=10 cmu=4 + v=4 + 12=16 \ \\ S=96 \ \text{cm}^2 \ \\ \ \\ S=\dfrac{ uv }{ 2 }=\dfrac{ (4+v)v }{ 2 } \ \\ \ \\ \ \\ 96=(4+v) \cdot \ v/2 \ \\ -0.5v^2 -2v +96=0 \ \\ 0.5v^2 +2v -96=0 \ \\ \ \\ a=0.5; b=2; c=-96 \ \\ D=b^2 - 4ac=2^2 - 4\cdot 0.5 \cdot (-96)=196 \ \\ D>0 \ \\ \ \\ v_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -2 \pm \sqrt{ 196 } }{ 1 }=-2 \pm 14 \sqrt{ 1 } \ \\ v_{1,2}=-2 \pm 14 \ \\ v_{1}=12 \ \\ v_{2}=-16 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 0.5 (v -12) (v +16)=0 \ \\ \ \\ v=v_{1}=12 \ \text{cm} \ \\ u=4+v=4+12=16 \ \text{cm} \ \\ \ \\ a=\sqrt{ (u/2)^2+(v/2)^2 }=\sqrt{ (16/2)^2+(12/2)^2 }=10 \ \text{cm}

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