Q22 of 77 Page 15

In Fig. 15.85, ABCD is a trapezium in which AB||DC and DC=40 cm and AB=60 cm. If X and Y are, respectively, the mid points of AD and BC, Prove that


(i) XY =50 cm (ii) DCYX is a trapezium


(iii) ar(trap. DCYX) = ar(trap. XYBA)

(i) Join DY and extend it to meet AB produced at P

∠BYP = ∠CYD (Vertically opposite angles)


∠DCY = ∠PBY (Since DC || AP)


BY = CY (Since Y is the mid-point of BC)


Hence, by A.S.A. congruence rule


ΔBYP ≅ ΔCYD


DY = YP


And,


DC = BP


Also,


X is the mid-point of AD


Therefore,


XY || AP


And,


XY = AP


XY = (AB + BP)


XY = (AB + DC)


XY = (60 + 40)


= × 100


= 50 cm


(ii) We have,


XY || AP


XY || AB and AB || DC


XY || DC


DCYX is a trapezium.


(iii) Since X and Y are the mid-points of AD and BC respectively


Therefore,


Trapezium DCYX and ABYX are of same height and assuming it as 'h' cm


Area (Trapezium DCYX) = (DC + XY) * h


= (40 + 50) h


= 45h cm2


Area (Trapezium ABYX) = (AB + XY) * h


= (60 + 50) * h


= 55h cm2


So,


=


=


Area of trapezium​ DCYX = Area of trapezium ABXY


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In Fig. 15.84, PSDA is a parallelogram in which PQ=QR=RS and AP||BQ||CR. Prove that


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 23

In Fig. 15.86, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. AE intersects BC in F. Prove that


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(ii) ar(Δ BDE) = ar(Δ BAE)


(iii) ar(Δ BFE) = ar(Δ AFD)


(iv) ar(Δ ABC) = ar(Δ BEC)


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D is the mid-point of side BC of Δ ABC and E is the mid-point of BD. If O is the mid-point of AE, prove that ar(ΔBOE) = ar(Δ ABC)