Question

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) = 9/11 ar(trap. XYBA)

Answer 1

Answer:

Step-by-step explanation:

i) join DY and extend it to P joining AB.

So in triangle DCY and Triangle BPY

angle CYD = Angle BYP (Vertically opposite angles)

Angle DCY = Angle YBP (DC||AP and DP is a transversal)

And, CY=BY (Y is the midpoint of BC)

So, triangle DCY≅ triangle BYP

By CPCT DC=BP=40cm

So AP=AB+BP=60+40 cm=100cm

BY midpoint theorem

XY=1/2 AP

= 1/2×100cm

=50cm

ii.) We know that Line joining mid points of opposite sides is parallel to the third side.

Hence XY ||AB

AB||DC

SO, XY||DC

Therefore DCXY IS a trapezium.

(iii) Since x and y are the midpoint of DA and CB respectively Trapezium DCXY and ABYX are of the same height say h.

Area of trapezium DCXY= 1/2×(DC+XY)×h

= 1/2×(40+50)×h

= 1/2×90×h

= 45h

Area of trapezium ABXY=1/2×(AB+XY)×h

=1/2×(60+50)×h

=1/2×110×h

=55h

Therefore area of DCXY/area of ABXY= 45h/55h

= 9/11

Hence proved.

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