Question

In the figure, the diagonals of a trapezium split it into four triangles.

What is the area of the whole trapezium?

Answer 1

Since quadrilateral ABCD is a trapezium, lines AB and CD are parallel.

So area of ∆ABD = 20 + 10 = 30 sq.cm

Area of ∆ABC = 30 sq.cm (Same base and the third vertex on a line parallel to the base)
Area of ∆ABC = 30 sq.cm
Area of ∆ABE = 20 sq.cm
∴ Area of ∆BEC = 30 – 20 = 10 sq.cm

From ∆ABC,
AE : EC = area of ∆ABE : Area of ∆BEC
AE : EC = 20 : 10 = 2 : 1

From ∆ACD,
AE : EC = 10 : area of ∆DEC
2 : 1 = 10: area of ∆DEC
Area of ∆ DEC = $\frac{10}{2}$ = 5 sq.cm

Total area of the trapezium = 20 + 10 + 10 + 5 = 45 sq.cm

So area of ∆ ABD = 20 + 10 = 30 sq.cm Area of ∆ ABC = 30 sq.cm (Same base and the third vertex on a line parallel to the base) Area of ∆ ABC = 30 sq.cm Area of ∆ ABE = 20 sq.cm Area of ∆ BEC = 30 – 20 = 10 sq.cm From ∆ ABC, AE : EC = area of ∆ ABE : Area of ∆ BEC AE : EC = 20 : 10 = 2 : 1 From ∆ ACD, AE : EC = 10 : area of ∆ DEC 2 : 1 = 10: area of ∆ DECRead more on Sarthaks.com - https://www.sarthaks.com/1330079/figure-diagonals-trapezium-split-four-triangles-area-yellow-triangle-square-centimetre