Question

In the figure, ABCD is a trapezium in which AB || DC and DC = 40 cm and AB = 60 cm. If X and y are, respectively the midpoints of AD and BC, Prove that:
(i) XY = 50 cm
(ii) DCYX is a trapezium
(iii) ar (trap. $DCYX)=\frac{9}{11}$ ar (trap. XYBA)

Answer 1

Given : In the figure, ABCD is a trapezium in which AB || DC
DC = 40 cm, AB = 60 cm
X and Y are the mid points of AD and BC respectively


Proof: In trap . ABCD, AB || DC
(i) DC = 40 cm, AB = 60 cm
$\therefore XY=\frac{1}{2}(AB+DC)=\frac{1}{2}(60+40)cm=\frac{100}{2}=50cm$
(ii) AB || DC
$\because$ X and Y are the mid points of AD and BC
$\therefore$ XY || AB || DC
$\therefore$ DCYX is a trapezium
(ii) AB || DC
$\because$ X and Y are the mid points of AD and BC
\therefore XY || AB || DC
\therefore DCYX is a trapezium
(iii) \because X and Y are the mid points of AD and BC respectively
\therefore Trapezium DCYX and trap. ABYX have the same height
let their height be h cm, then
ar (trap. DCYX) $=\frac{1}{2}(DC+XY)\times h=\frac{1}{2}(40+50)h=\frac{1}{2}\times 90hc{m}^2=45h{m}^2$
and area (trap. ABYX) $=\frac{1}{2}(AB+XY)h=\frac{1}{2}(60+50)\times hc{m}^2=\frac{1}{2}\times 110hc{m}^2=55hc{m}^{2}Now,\frac{ar(trap.DCXY)}{ar(trap.ABYX)}.=\frac{45h}{55h}=\frac{9}{11}\Rightarrow ar(trap.DCYX)=\frac{9}{11}ar(trap.ABYX)$