Midpoints cut each corner triangle’s area to 1/4 of its diagonal triangle. Summing opposite corner triangles removes 1/4 + 1/4 = 1/2 of the total 4-gon area, leaving exactly half for the inner Varignon parallelogram.
If the mid-points of the sides of a 4-gon (also known as a quadrilateral, but we prefer to call it a ‘4-gon’) are joined in order, prove that the area of the parallelogram thus formed will be half of the area of the given 4-gon. (You may wonder whether the 4-gon thus formed is always a parallelogram and if so, why? These questions will be tackled and answered in the chapter on quadrilaterals.)
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Let ABCD be the 4-gon and E, F, G, H the midpoints of AB, BC, CD, DA respectively.
Draw diagonal AC, dividing ABCD into triangles ABC and ADC.
In triangle ABC, E and F are midpoints, so triangle EBF is similar to ABC with ratio 1/2, meaning Area(EBF) = (1/4) x Area(ABC).
Similarly, Area(HDG) = (1/4) x Area(ADC).
Adding these:
Area(EBF) + Area(HDG) = (1/4) x [Area(ABC) + Area(ADC)] = (1/4) x Area(ABCD).
Similarly, using diagonal BD:
Area(HAE) + Area(GCF) = (1/4) x Area(ABCD).
Sum of four corner triangles = (1/4 + 1/4) x Area(ABCD) = (1/2) x Area(ABCD).
Subtracting corners gives:
Area(EFGH) = Area(ABCD) – (1/2) x Area(ABCD) = (1/2) x Area(ABCD).
Hence proved.
For more NCERT Solutions of Class 9 Ganita Manjari Chapter 6 Measuring Space: Perimeter and Area Question Answer (2026-27)
https://www.tiwariacademy.com/ncert-solutions/class-9/maths/ganita-manjari-chapter-6/