The given figure shows a rectangle ABDC and a parallelogram ABEF; drawn on opposite sides of AB. Prove that :

(i) quadrilateral CDEF is a parallelogram

(ii) Area of quad. CDEF = Area of rect. ABDC + Area of // gm ABEF

(i) Given,

ABCD is a rectangle.

∴ AB = CD (Opposite sides of rectangle are equal) ……………(1)

Given,

ABEF is a rectangle.

∴ AB = FE (Opposite sides of parallelogram are equal) ……………(2)

From equations (1) and (2), we get :

⇒ CD = FE.

From figure,

CD || FE

Since, one pair of opposite sides of quadrilateral CDEF are equal and parallel.

∴ CDEF is a parallelogram.

Hence, proved that CDEF is a parallelogram.

(ii) In △ AFC and △ BED,

⇒ AF = BE (Opposite sides of parallelogram ABEF)

⇒ AC = BD (Opposite sides of rectangle ABCD)

⇒ CF = ED (Opposite sides of parallelogram CDEF)

∴ △ AFC ≅ △ BED (By S.S.S. axiom)

We know that,

Area of congruent triangle are equal.

∴ Area of △ AFC = Area of △ BED ……….(1)

From figure,

⇒ Area of quadrilateral CDEF = Area of △ ACF + Area of ACDPEF

⇒ Area of quadrilateral CDEF = Area of △ BED + Area of ACDPEF [From equation (1)]

⇒ Area of quadrilateral CDEF = Area of CDBEPA

⇒ Area of quadrilateral CDEF = Area of rect. ABDC + Area of // gm ABEF.

Hence, proved that Area of quadrilateral CDEF = Area of rect. ABDC + Area of // gm ABEF.