Question

Diagonals of a parallelogram divide it into four triangles of equal area

• A. True
• B. False

Answer 1

The correct option is A

True

Draw the diagonals in the parallelogram as shown in the figure.

We know that the diagonals of a parallelogram bisect each other.

Triangle ADC and parallelogram ABCD are between same parallel lines and are on the same base, So Area(ΔADC)= $\frac{1}{2}$ of Area(ABCD)------------(1)

Similarly , Area($△$BCD)= $\frac{1}{2}$ of Area(ABCD)

Consider $△$ADC, DO is the median of $△$ADC.

So, Area ($△$AOD) = Area($△$DOC)-----------------(2)

From (1) and (2), Area($△$AOD) = Area($△$DOC)= $\frac{1}{4}$ Area(ABCD)

Similarly in $△$BCD

Area($△$BOC) = Area($△$DOC) = $\frac{1}{4}$ Area(ABCD) -----(3)

From (2) and (3),

Area($△$AOB) = $\frac{1}{4}$ Area(ABCD)

Hence showed that diagonals of a parallelogram divide it into four triangles of equal area.