Question

Four points A, B, C, and D are given on circle. Line segment AB and CD are parallel. Find the area of the figure formed by ABCD.

Answer 1

Construction: Draw perpendiculars OE and OF onto AE and CD respectively from the centre O.
In $\Delta$OEB,
${OE}^2$ + ${BE}^2$ = ${OB}^2$
[Using Pythagoras theorem]
${OE}^2$ = ${OB}^2$ - ${BE}^2$
${OE}^2$ = 25 - 9
(Perpendicular drawn from centre to a chord bisects the chord, so BE = 3 cm)
Therefore, OE = 4cm

Similarly In $\Delta$OFD,
${OF}^2$ + ${FD}^2$ = ${OD}^2$
${OF}^2$ = ${OD}^2$ - ${FD}^2$
${OF}^2=25-16=9$
(Perpendicular drawn from centre to a chord bisects the chord, so FD = 4 cm)
$\Rightarrow$ OF = 3 cm

Here, the figure formed by the points A, B, C, and D is a quadrilateral with one pair of sides parallel to each other.
Hence, ABCD is a trapezium.
We have,
Area of trapezium = $\frac{1}{2}$(perpendicular distance between two parallel sides)(sum of the lengths of the parallel sides)
=$\frac{1}{2}$ x (OE + OF) x (AB + CD)

= $\frac{1}{2}$(4 + 3)(8 + 6)

= 49 ${cm}^2$