Question

Plz solve dis one ....

Answer 1

Here, it's given that, ABCD is a trapezium with AB || DC

Now, Join DY and produce it to meet AB produced at P.

In ∆BYP and ∆CYD

∠BYP  = ∠CYD    (vertically opposite angles)

 ∠DCY  = ∠PBY  (∵ Alternate opposite angles as DC || AP and BC is the transversal)

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

Thus ∆BYP ≅ ∆CYD

⇒ DY = YP and DC = BP

⇒ Y is the mid point of AD

∴ XY || AP and XY =1/2 AP  (mid point theorem)

⇒ XY =1/2 AP
=1/2(AB + BP)
=1/2(AB + DC)
=1/2(50 + 30)
=1/2× 80 cm
= 40 cm

Since X and Y are the mid points of AD and BC respectively.

∴ trapezium DCYX and ABYX are of same height, say h cm

Now

ar(DCXY) 1/2(DC+XY)×h (30+40) 70
_______ =___________= _______= __
ar(XYBA) 1/2(AB+XY)×h (50+40) 90

⇒ ar(DCXY) = 7/9 ar(XYBA)

hope it helps :)

Answer 2

Formula to be used:
(i) Mid segment of a trapezium = $\frac{\text{ Sum of parallel sides } }{2}$
(ii) Area of a parallelogram =$\frac{\text{ Sum of parallel sides } }{2} *h$, where h is the perpendicular distance between the two parallel sides.
(iii)The line joining the midpoints of the two non-parallel sides is the mid segment of the trapezium.
(iv) Mid segment is always parallel  to the other two parallel sides.
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XY is the mid segment of the trapezium ABCD.
The two parallel sides are AB=50cm and CD =30cm.

Length of the mid segment XY =$\frac{\text{ Sum of parallel sides } }{2}= \frac{AB+CD }{2}=\frac{50+30}{2}=40$

 
Mid segment divides the height ( perpendicular distance between the two parallel sides) in two halves.

The mid segment divides the trapezium ABCD into two trapezium DCYX and trapezium XYBA with equal heights.

Let the height be h.

Find the area of the trapezium DCYX:
The two parallel sides are  DC =30 and YX = 40 and the height is h.
Ar(DCYX) =$\frac{\text{ Sum of parallel sides } }{2} *h= \frac{DC+YX}{2} *h= \frac{30+40}{2} *h=\frac{70}{2} *h=35h$

Find the area of the trapezium XYBA:
The two parallel sides are  XY =40 and AB = 50 and the height is h.
Ar(XYBA) =$\frac{\text{ Sum of parallel sides } }{2} *h= \frac{XY+AB}{2} *h= \frac{40+50}{2} *h=\frac{90}{2} *h=45h$


Therefore,
Ar(DCYX) :Ar(XYBA) =35h : 45h
$\frac{Ar(DCYX)}{Ar(XYBA)}= \frac{35h}{45h}$
Reduce the fraction $\frac{35h}{45h}$ :
$\frac{Ar(DCYX)}{Ar(XYBA)}= \frac{7}{9}$
Multiply both sides by Ar(XYBA):
$Ar(DCYX)= \frac{7}{9}}Ar(XYBA)$ (Proved)