Question

Two parallel chords $AB$ and $CD$ are drawn on the same side of centre $O$ of a circle. Radius of circle is $65m$, length of chords are $112m$ and $126m$. Find the area of $\square ABCD$.

• A. $2023m^2$
• B. $2123m^2$
• C. $2223m^2$
• D. $2323m^2$

Answer 1

The correct option is A $2023m^2$

$OM$ and $ON$ are $\perp$ drawn on $AB$ and $CD$ respectively.

Also, $OM$ and $ON$ bisects the chords $AB$ and $CD$ respectively.
Now, $AB=112$
$⟹AM=\frac{AB}{2}$
$=\frac{112}{2}=56$
$\therefore AM=56cm=MB$
Also, $CD=126$
$⟹CN=\frac{CD}{2}$
$=\frac{126}{2}=63$
$\therefore CN=63cm=ND$
And $OC=65=OA$ ........ (Radii of same circle)
In $△OAM,\angle OMA={90}^o$
$\therefore O{A}^2=A{M}^2+O{M}^2$ ........ [By Pythagoras theorem]
$⟹O{M}^2=O{A}^2-A{M}^2$
$={65}^2-{56}^2$
$=4225-3136=1089$
$\therefore OM=\sqrt{1089}=33$
$\therefore OM=33m$
In $△OCN,\angle ONC={90}^o$
$\therefore O{C}^2=C{N}^2+O{N}^2$ ........... [By Pythagoras theorem]
$⟹O{N}^2=O{C}^2-C{N}^2$
$={65}^2-{63}^2$
$=4225-3969=256$
$\therefore ON=\sqrt{256}$
$\therefore ON=16m$
$\therefore$ Altitude of trapezium $△BCD$ is $MN=OM-ON$
$=33-16$
$=17m$
$\therefore$ Area of trapezium $ABCD=\left(\frac{AB+CD}{2}\right)\times MN$
$=\left(\frac{112+126}{2}\right)\times 17$
$=2023m^2$
Thus, Area of trapezium $ABCD$ is $2023m^2$.

Hence, option A is correct.