、如右图所示,点E和点F分别是长方形ABCD的边AD和CD的中点,三角形BFE的面积是15dm,求长方形ABCD的面积
2个回答
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在AB也 取中点P, 连到F
与BE交点记作Q
因为中位线 PQ = 1/2 AE
PQ = 1/2 AE = 1/4 AD
FQ = FP - PQ = AD - 1/4 AD = 3/4 AD
三角形BPQ BFQ高相等,面积之比=底边之比
S-BPQ /S-BFQ = FQ/PQ = 1/4 / 3/4 = 1/3
三角形BFQ EFQ 等底等高面积相等
S-BFQ = 1/2 * 15= 15/2
S-BPQ = 15/2 * 1/3 = 5/2
S-BPQ + S-BFQ = 15/2 + 5/2 = 10
= 1/4 S-ABCD
所以S-ABCD = 4 * 10 = 40
与BE交点记作Q
因为中位线 PQ = 1/2 AE
PQ = 1/2 AE = 1/4 AD
FQ = FP - PQ = AD - 1/4 AD = 3/4 AD
三角形BPQ BFQ高相等,面积之比=底边之比
S-BPQ /S-BFQ = FQ/PQ = 1/4 / 3/4 = 1/3
三角形BFQ EFQ 等底等高面积相等
S-BFQ = 1/2 * 15= 15/2
S-BPQ = 15/2 * 1/3 = 5/2
S-BPQ + S-BFQ = 15/2 + 5/2 = 10
= 1/4 S-ABCD
所以S-ABCD = 4 * 10 = 40
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