22. Square ABCD, the area of triangle DEF is 6 square centimeters larger than that of triangle ABF. CD is 6 cm long, what is the length of de

S triangle BEC = s triangle EDF + s (DCBF) (1) s (ABCD) = s triangle AFB + s (DCBF) (2) by subtracting the two formulas, s triangle bec-s (ABCD) = s triangle edf-s triangle AFB = 6, so s triangle BEC = 6 + 16 = 22, because s triangle BEC = (BC times CE) divided by 2 = 22, (BC times CE) = 44, CE = 11, because CE = CD + de, so de = 7
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