Bunuel KarishmaB I solved the question correctly by applying a concept I know about inscribed triangles within rectangles (reasoned that parallelograms are also regular figures) but I'd appreciate your validation.
Concept: If a triangle inscribed in a rectangle shares a side with the rectangle, the area of the triangle is equal to half the area of the rectangle. In other words, the area of the two smaller triangles = area of main triangle.
Applying this rule to the parallelogram, I connected points E and C to create an inscribed triangle.
First small triangle / Main triangle + second small triangle = 1 / 3
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