Đáp án: $420cm^2$
Giải thích các bước giải:
Kẻ $AE\perp CD, OF\perp CD, BG\perp CD$
$\to AE//OF//BG$
Do khoảng cách từ $A, B, O$ đến $CD$ theo thứ tự bằng $20cm, 15cm, 12cm$
$\to AE=20cm, OF=12cm, BG=15cm$
$\to S_{OCD}=\dfrac12OF\cdot CD=\dfrac12\cdot 12\cdot 24=144$
Ta có $AE//OF$
$\to \dfrac{CO}{CA}=\dfrac{OF}{AE}=\dfrac{3}{5}$
$\to \dfrac{CO}{CA-CO}=\dfrac{3}{5-3}$
$\to \dfrac{CO}{AO}=\dfrac32$
$\to \dfrac{S_{ADO}}{S_{OCD}}=\dfrac32$
$\to S_{ADO}=\dfrac32S_{OCD}=\dfrac32\cdot 144=216$
Lại có $BG//OF$
$\to \dfrac{DO}{DB}=\dfrac{OF}{OG}=\dfrac{4}{5}$
$\to \dfrac{DO}{DB-OB}=\dfrac4{5-4}$
$\to \dfrac{DO}{OB}=4$
$\to \dfrac{S_{OCD}}{S_{OBC}}=4$
$\to S_{OBC}=\dfrac14S_{OCD}=36$
Lại có $\dfrac{S_{BOA}}{S_{BOC}}=\dfrac{OA}{OC}=\dfrac23$
$\to S_{BOA}=\dfrac23S_{BOC}=24$
$\to S_{ABCD}=S_{OAB}+S_{OBC}+S_{OCD}+S_{ODA}$
$\to S_{ABCD}=24+36+144+216=420$
