The Application and Improvement of Mathematics on Page P43 of Primary School Summer Learning and Life by Jiangsu Education Publishing House
Using equal area and equal height to solve the problem, let d be the high DF of BC, where d =50 cm2 = be× df ÷ 2 = 20× df ÷ 2-df = 5cm. Second, let C be the high CG of AD, and the S triangle DEC=S triangle CED, so CE × DF ÷ 2 = DE× CG ÷ 2-60× 5 ÷ 2 = 40× CG ÷ 2-So CG=7.5cm 3, and S C = AE× CG ÷ 2.