Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: AK+KD=AD
=>\(KD=AD-AK=\frac23AD\)
=>\(S_{BDK}=\frac23\times S_{ABD}\)
=>\(S_{ABD}=15,5:\frac23=15,5\times\frac32=23,25\left(\operatorname{cm}^2\right)\)
Ta có: \(BD=\frac13\times BC\)
=>\(BC=3\times BD\)
=>\(S_{ABC}=3\times S_{ABD}=3\times23,25=69,75\left(\operatorname{cm}^2\right)\)
Ta có: AK+KD=AD
=>\(KD=AD-AK=\frac23AD\)
=>\(S_{BDK}=\frac23\times S_{ABD}\)
=>\(S_{ABD}=15,5:\frac23=15,5\times\frac32=23,25\left(\operatorname{cm}^2\right)\)
Ta có: \(BD=\frac13\times BC\)
=>\(BC=3\times BD\)
=>\(S_{ABC}=3\times S_{ABD}=3\times23,25=69,75\left(\operatorname{cm}^2\right)\)
Ta có: K là trung điểm của BC
=>\(BK=CK=\frac12\times BC\)
=>\(S_{AKB}=\frac12\times S_{ABC}=\frac12\times45=22,5\left(\operatorname{cm}^2\right)\)
Ta có: AP+PK=AK
=>\(PK=AK-AP=AK-\frac13\times AK=\frac23\times AK\)
=>\(S_{PKB}=\frac23\times S_{AKB}=\frac23\times22,5=15\left(\operatorname{cm}^2\right)\)
Kẻ \(EH\perp BG\), \(CF\perp BG\)
Ta có: \(S_{ABD}=S_{GBC}=\dfrac{1}{2}.AB.AD=\dfrac{1}{2}.S_{ABCD}\)
\(S_{BAG}=\dfrac{1}{2}.AB.AG=\dfrac{1}{2}.AB.\dfrac{1}{2}AD=\dfrac{1}{4}.AB.AD=\dfrac{1}{2}S_{ABD}\)
\(S_{GEB}=\dfrac{1}{2}.AG.EB=\dfrac{1}{2}.AG.\dfrac{1}{2}.AB=\dfrac{1}{4}.AG.AB=\dfrac{1}{2}S_{ABG}\)
\(\Rightarrow S_{GEB}=\dfrac{1}{2}.\dfrac{1}{2}.\dfrac{1}{2}S_{ABCD}=\dfrac{1}{8}S_{ABCD}=\dfrac{1}{4}S_{GBC}\)
\(\Leftrightarrow\dfrac{1}{2}.EH.BG=\dfrac{1}{4}.\dfrac{1}{2}CF.BG\)
\(\Leftrightarrow EH=\dfrac{1}{4}CF\)
Lại có: \(S_{OBE}=\dfrac{1}{2}OB.EH=\dfrac{1}{2}OB.\dfrac{1}{4}CF=\dfrac{1}{4}S_{OBC}\)
Ta có: \(S_{CBE}=S_{OBE}+S_{OBC}=S_{OBE}+4S_{OBE}=5S_{OBE}\)
\(S_{CBE}=5.10=50\left(cm^2\right)\)
Mà \(S_{CBE}=\dfrac{1}{2}S_{CBA}=\dfrac{1}{4}S_{ABCD}\Rightarrow S_{ABCD}=200\left(cm^2\right)\)