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  • ABCD is a trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = 7/9 ar (XYBA)

ABCD is a trapezium in which AB\parallel DC, DC = 30 cm and AB = 50 cm. If X and Y are, respectively the mid-points of AD and BC, prove that ar (DCYX) = \frac{7}{9 } ar (XYBA)

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Solution.

Given: Trapezium ABCD with AB\parallel DC,
DC = 30 cm and AB = 50 cm.
X and Y are the mid-points of AD and BC, respectively.
To prove: ar(DCYX)=\frac{7}{9}ar(XYBA)


Construction, Join DY and extend it to meet AB produced at P.
Proof:
In \triangle DCY and \triangle PBY
CY = BY [Given, Y is the mid-point of BC]
\angle DCY = \angle PBY [alternate interior angles]
\angle 2 = \angle 3[Vertically opposite angles]
\angle DCY \cong \angle PBY [ASA congruence rule]
then, DC = BP [CPCT]
DC = 30 cm (given)
Q DC = BP = 30 cm
Now,
AP = AB + BP = 50 + 30 = 80 cm
In \triangle ADP; XY \parallel AP, X is the mid-point of AD and Y is the mid-point of DP.
XY=\frac{1}{2}AP=\frac{1}{2}\times 80=40cm       [mid-point theorem]
Let distance between AB, XY and XY, DC be h cm
Area of trapezium is given as \frac{1}{2} [Sum of parallel side × distance between them]
Now, area of trapezium DCYX = \frac{1}{2}h(30+40)

= \frac{1}{2}h(70)=35hcm^{2}

Area of trapezium XYBA = \frac{1}{2}h(40+50)=\frac{1}{2}h\times 90=45hcm^{2}
\frac{ar(DCYX)}{ar(XYBA)}=\frac{35h}{45h}=\frac{7}{9}
ar(DCYX)=\frac{7}{9}ar(XYBA)
Hence proved.

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