SSC CGL Tier-1 12-08-2017 Shift 1 Solved Paper

Δ‌ABC∼Δ‌ADE
and since,

DE

BC

=

1

3

The areas of Δ‌ADE and Δ‌ABC will be in the ratio

12

32

=

1

9

.
That means when the area of ΔADE is 20, the area of ΔABC will be 20×9=180
Now, area of trapezium BCED is 180−20=160
The line segment DC will divide the trapezium BCED into two triangles viz. ΔCED and ΔCBD in which ΔCED=

1

2

×DE× distance between the parallel lines DE and BC.
Area of ΔCBD=

1

2

×BC× distance between the parallel lines DE and BC. But DE=

1

2

BC, hence DC will divide the trapezium BCED is divided in the ratio 1: 3 and Area of ΔCED=

1

1+3

× areof the trapezium BCED =

1

4

×160=40cm2