UPSC 2017 CDS I Math Paper

AC || ED. So, the perpendicular distance between them, say h cm, is constant.ABDE is a parallelogram
herefore, area of ABDE =AB×h=5hcm2
Area of ΔBDE =

1

2

×DE×h=

5h

2

cm2
Area of ΔBCD=

1

2

×BC×h=

7h

2

cm2
Therefore, area of ABDE:Area of ΔBDE : Area of ΔBCD=5h:

5h

2

:

7h

2

=10:5:7