Question

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

Answer 1

Consider the figure below.



Given:
AB = 112 m
CD = 126 m
Radius of the circle = OB = OC = 65 m
We know that, the perpendicular drawn from the centre to the chord bisects the chord.
$\therefore \mathrm{EB}=\frac{\mathrm{AB}}{2}=\frac{112}{2}\mathrm{m}=56\mathrm{m}$

and FC = $\frac{\mathrm{DC}}{2}=\frac{126}{2}=63\mathrm{m}$

In right $△$OEB, we have:
$$\begin{gathered} {\mathrm{OB}}^2={\mathrm{OE}}^2+{\mathrm{EB}}^2(\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {65}^2={\mathrm{OE}}^2+{56}^2 \\ \Rightarrow {\mathrm{OE}}^2=4225-3136 \\ \Rightarrow {\mathrm{OE}}^2=1089 \\ \Rightarrow \mathrm{OE}=33\mathrm{m} \end{gathered}$$

In right $△$OFC, we have:
$$\begin{gathered} {\mathrm{OC}}^2={\mathrm{OF}}^2+{\mathrm{FC}}^2(\mathrm{Pythagoras}\mathrm{theorem}) \\ \Rightarrow {65}^2={\mathrm{OF}}^2+{63}^2 \\ \Rightarrow {\mathrm{OF}}^2=256 \\ \Rightarrow \mathrm{OF}=16\mathrm{m} \end{gathered}$$

Now, EF = OE – OF = (33 – 16) m = 17 m
Since the opposite sides of quadrilateral ABCD are parallel, it is a trapezium.

$$\begin{gathered} \therefore \mathrm{Area}\mathrm{of}\mathrm{trapezium}\mathrm{ABCD}=\frac{1}{2}\times \left(\mathrm{AB}\hspace{0.17em}+\mathrm{CD}\right)\times \mathrm{EF} \\ =\frac{1}{2}\left(112+126\right)\times 17 \\ =2023{\mathrm{m}}^2 \end{gathered}$$