Diagonals of the rhombus

How long are the diagonals e, f in the diamond, if its side is 5 cm long and its area is 20 cm2?


e =  4.472 cm
f =  8.944 cm


a=5 cm S=20 cm2  S=ah h=S/a=20/5=4 cm  h=a sin(A) A=180πarcsin(h/a)=180πarcsin(4/5)53.1301   e=a 22 cosA=a 22 cos53.1301023542 =5 22 cos53.1301023542 =a 22 0.6=4.47214=4.472 cma=5 \ \text{cm} \ \\ S=20 \ \text{cm}^2 \ \\ \ \\ S=a h \ \\ h=S / a=20 / 5=4 \ \text{cm} \ \\ \ \\ h=a \cdot \ \sin(A) \ \\ A=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(h/a)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(4/5) \doteq 53.1301 \ ^\circ \ \\ \ \\ e=a \cdot \ \sqrt{ 2 - 2 \cdot \ \cos A ^\circ }=a \cdot \ \sqrt{ 2 - 2 \cdot \ \cos 53.1301023542^\circ \ }=5 \cdot \ \sqrt{ 2 - 2 \cdot \ \cos 53.1301023542^\circ \ }=a \cdot \ \sqrt{ 2 - 2 \cdot \ 0.6 }=4.47214=4.472 \ \text{cm}
f=a 2+2 cosA=a 2+2 cos53.1301023542 =5 2+2 cos53.1301023542 =a 2+2 0.6=8.94427=8.944 cmf=a \cdot \ \sqrt{ 2 + 2 \cdot \ \cos A ^\circ }=a \cdot \ \sqrt{ 2 + 2 \cdot \ \cos 53.1301023542^\circ \ }=5 \cdot \ \sqrt{ 2 + 2 \cdot \ \cos 53.1301023542^\circ \ }=a \cdot \ \sqrt{ 2 + 2 \cdot \ 0.6 }=8.94427=8.944 \ \text{cm}

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