jh05x21
幼苗
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设AB = a 梯形的高为h,△AOB边AB上的高为h1,△DOC边CD上的高为h2,
h1 = 2S(AOB)÷ AB = 50/a
h = 2S(ABC)÷ AB = 120/a
h2 = h - h2 = 120/a-50/a = 70/a
S(DOC) :S(AOB) = h2² :h1²
S(DOC) = 25 × (h2/h1)² = 25×49/25 = 49
而S(AOD)=S(BOC)=35
∴S(ABCD)=S(AOB)+S(BOC)+S(DOC)+S(AOD)=25+35+49+35 = 144
1年前
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