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Chapter 8 Quadrilaterals Ex 8.1 Solutions

Question - 11 : -

11. In ╬ФABC and╬ФDEF, AB = DE, AB || DE, BC = EF and BC || EF. Vertices A, B and C are joinedto vertices D, E and F respectively (see Fig. 8.22).

Show that

(i) quadrilateralABED is a parallelogram

(ii) quadrilateralBEFC is a parallelogram

(iii) AD || CF andAD = CF

(iv) quadrilateralACFD is a parallelogram

(v) AC = DF

(vi) ╬ФABC тЙЕ ╬ФDEF.

Answer - 11 : -

(i) AB = DE and AB || DE (Given)

Two opposite sides of a quadrilateral are equal and parallel toeach other.

Thus, quadrilateral ABED is a parallelogram


(ii) Again BC = EF and BC || EF.

Thus, quadrilateral BEFC is a parallelogram.


(iii) Since ABED and BEFC are parallelograms.

тЗТ AD =BE and BE = CF (Opposite sides of a parallelogram are equal)

, AD = CF.

Also, AD || BE and BE || CF (Opposite sides of a parallelogramare parallel)

, AD || CF


(iv) AD and CF are opposite sides of quadrilateral ACFD whichare equal and parallel to each other. Thus, it is a parallelogram.


(v) Since ACFD is a parallelogram

AC || DF and AC = DF


(vi) In ╬ФABC and ╬ФDEF,

AB = DE (Given)

BC = EF (Given)

AC = DF (Opposite sides of a parallelogram)

, ╬ФABC тЙЕ ╬ФDEF[SSS congruency]

Question - 12 : -

12. ABCD is atrapezium in which AB || CD and AD = BC (see Fig. 8.23). Show that

(i) тИаA = тИаB

(ii) тИаC = тИаD

(iii) ╬ФABC тЙЕ ╬ФBAD

(iv) diagonal AC =diagonal BD

[Hint : Extend ABand draw a line through C parallel to DA intersecting AB produced at E.]

Answer - 12 : -

To Construct: Draw a line through C parallel to DA intersectingAB produced at E.

(i) CE = AD (Opposite sides of a parallelogram)

AD = BC (Given)

, BC = CE

тЗТтИаCBE = тИаCEB

also,

тИаA+тИаCBE = 180┬░ (Angles on the same side oftransversal and тИаCBE = тИаCEB)

тИаB +тИаCBE = 180┬░ ( As Linear pair)

тЗТтИаA = тИаB


(ii) тИаA+тИаD = тИаB+тИаC =180┬░ (Angles on the same side of transversal)

тЗТтИаA+тИаD = тИаA+тИаC (тИаA = тИаB)

тЗТтИаD = тИаC


(iii) In ╬ФABC and ╬ФBAD,

AB = AB (Common)

тИаDBA = тИаCBA

AD = BC (Given)

, ╬ФABC тЙЕ ╬ФBAD[SAS congruency]


(iv)Diagonal AC = diagonal BD by CPCT as ╬ФABC
тЙЕ╬ФBA.

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