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2014-08-22T12:37:04+03:00
ΔABC ,mas<A=90
BC=3√13

AB/AC= 0,(6)= 6/9=2/3

AB/2=AC/3=k

AB=2k
AC=3k

BC²=AB²+AC²
(3√13)²= (2k)² + (3k)²

117= 4k² +9k²

117=13k²

k²= 9   =>k=3

AB=2*3=6
AC=3*3=9

A ΔABC= AB*AC/2=  6*9/2= 3*9=27