UPSC CDS 2 2018 Math Paper

Let T be the parallelogram ABDC and S be the parallelogram EFGH.In parallelogram ABHF:
Area of Δ FEH =

1

2

× Area of parallelogram ABHF …..(1)
In parallelogram CDHF:
Area of ΔFGH =

1

2

× Area of parallelogram CDHF …..(2)
Adding (1) and (2), we get:
Area of Δ FEH + Area of Δ FGH =

1

2

× (Area of parallelogram ABHF + Area of parallelogram CDHF)
⇒ Area of parallelogram EFGH =

1

2

× Area of parallelogram ABCD
⇒ Area of parallelogram S =

1

2

× Area of parallelogram T.
Thus, statement (1) is correct
In ΔACD :
FG=

1

2

×AD
Similarly,EH=

1

2

×AD
∴ FG + EH = AD ...(3)
In ΔABC :
FE=

1

2

×BC
Similarly,GH=

1

2

×BC
∴ FE + GH = BC ...(4)
Adding (3) and (4), we get:
FG + EH + FE + GH = AD + BC
Thus, statement (2) is wrong.