Câu hỏi của Minh Bình

Question

Cho tứ giác ABCD có AC cắt BD tại O. Biết AC=4. BD=5, góc AOB=60 độ. Tính SABCD

Nguyễn Lê Phước Thịnh · Accepted Answer

$S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB=\dfrac{1}{2}\cdot OA\cdot OB\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OA\cdot OB$$S_{OBC}=\dfrac{1}{2}\cdot OB\cdot OC\cdot sinBOC$$=\dfrac{1}{2}\cdot OB\cdot OC\cdot sin120=\dfrac{\sqrt{3}}{4}\cdot OB\cdot OC$$S_{ODC}=\dfrac{1}{2}\cdot OD\cdot OC\cdot sinDOC$$=\dfrac{1}{2}\cdot OD\cdot OC\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OD\cdot OC$$S_{AOD}=\dfrac{1}{2}\cdot OA\cdot OD\cdot sinAOD$$=\dfrac{1}{2}\cdot OA\cdot OD\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OA\cdot OD$$S_{ABCD}=S_{AOB}+S_{AOD}+S_{COD}+S_{COB}$$=\dfrac{\sqrt{3}}{4}\left(OA\cdot OB+OB\cdot OC+OD\cdot OC+OD\cdot OA\right)$$=\dfrac{\sqrt{3}}{4}\cdot\left(OB\cdot AC+OD\cdot AC\right)$$=\dfrac{\sqrt{3}}{4}\left(AC\cdot BD\right)=\dfrac{\sqrt{3}}{4}\cdot4\cdot5=5\sqrt{3}$Câu trả lời: $S_{OAB}=\dfrac{1}{2}\cdot OA\cdot OB\cdot sinAOB=\dfrac{1}{2}\cdot OA\cdot OB\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OA\cdot OB$$S_{OBC}=\dfrac{1}{2}\cdot OB\cdot OC\cdot sinBOC$$=\dfrac{1}{2}\cdot OB\cdot OC\cdot sin120=\dfrac{\sqrt{3}}{4}\cdot OB\cdot OC$$S_{ODC}=\dfrac{1}{2}\cdot OD\cdot OC\cdot sinDOC$$=\dfrac{1}{2}\cdot OD\cdot OC\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OD\cdot OC$$S_{AOD}=\dfrac{1}{2}\cdot OA\cdot OD\cdot sinAOD$$=\dfrac{1}{2}\cdot OA\cdot OD\cdot sin60=\dfrac{\sqrt{3}}{4}\cdot OA\cdot OD$$S_{ABCD}=S_{AOB}+S_{AOD}+S_{COD}+S_{COB}$$=\dfrac{\sqrt{3}}{4}\left(OA\cdot OB+OB\cdot OC+OD\cdot OC+OD\cdot OA\right)$$=\dfrac{\sqrt{3}}{4}\cdot\left(OB\cdot AC+OD\cdot AC\right)$$=\dfrac{\sqrt{3}}{4}\left(AC\cdot BD\right)=\dfrac{\sqrt{3}}{4}\cdot4\cdot5=5\sqrt{3}$

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