Answer:
[tex]A = 50cm {}^{2} [/tex]
Step-by-step explanation:
[tex]let \: the \: two \: sides \: be : \\ p = 10 \: cm \\ q = x \: cm \: (yet \: unknown) \\ then \: the \: area,\: \boxed{A} \: is \: given \: by : \\ A = \frac{1}{2} pq = \frac{1}{2} 10 \times q. \\ so \: lets \: find \: q : \\ if \: we \: bisect \: the \: rhombus, \: then \: applying \: the \: laws \: of \: sum \: of \: angles : \\ 180 - (80 + 50) = 50. \\ applying \: the \: sin \: rule : \\ \frac{ \sin(50) }{10} = \frac{ \sin(50) }{q} \\ \\ q \sin(50 ) = 10 \sin(50) \\ q = \frac{10 \sin(50) }{ \sin(50) } \\ q = 10. \\ ......wait....what {?} \: all \: of \: this \: to \: just \: to \\ \: know \: that \: q = 10. \\ so \: \\ A = \frac{1}{2} pq = \frac{1}{2} \times 10 \times 10 \\ A = 50cm {}^{2} [/tex]
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