Question

(i) ABCD is a trapezium in which AB || DC. Diagonals AC and BD intersect each other at O. Find the triangle which is equal to the area of $△$ BOC.

(ii) If AB = 6 cm, DC = 10 cm and perpendicular distance between AB and DC is 8 cm, then find the area of $\square$ ABCD.

Answer 1

(i)


It can be observed that $△$ DAC and $△$DBC lie on the same base DC and between the same parallels AB and CD.

$\therefore$ Area ($△$DAC) = Area ($△$DBC)

Subtracting Area ($△$DOC) on both the sides

Area ($△$DAC) − Area ($△$DOC) = Area ($△$DBC) − Area ($△$DOC)

Area ($△$AOD) = Area ($△$BOC)

(ii) Area of $\square$ ABCD $=\frac{(AB+CD)}{2}\times$ Perpendicular Distance between parallel sides.

Area of $\square$ ABCD $=\frac{(6+10)}{2}\times 8=8\times 8=64$ sq.cm