In the quadrangle ABCD, the areas are known: S1 of the triangle ABO = 10, S2 of the triangle BOC = 20, S3 of the triangle COD = 60. Find the area ABCD (i.e., the intersection point of the diagonals).
Consider triangles ABO and BOC, the areas of which are 10 cm2 and 20 cm2, respectively.
The triangles ABO and BOC have a total height BH, then the ratio of the bases of the triangles is equal to the ratio of their bases. CO / AO = Svos / Savo = 20/10 = 2.
In the triangles SOD and AOD, the total height of the DC, then CO / AO = Scod / Saod.
2 = Sod / Saod.
Saod = Scod / 2 = 60/2 = 30 cm2.
Then Savsd = Saov + Svos + Sod + Saod = 10 + 20 + 60 + 30 = 120 cm2.
Answer: The area of the AVSD quadrangle is 120 cm2.
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