Question

Four points A, B, C, D are given on a circle. Line segment AB and CD are parallel. Find the area of the figure formed by joining these points (in ${cm}^2$).

• A. 98
• B. 49
• C. 7
• D. 28

Answer 1

The correct option is B

49

Find the perpendicular distance of lines AB and CD from centre.

Apply Pythagoras theorem to find the perpendicular distance of the two lines.

In $\Delta$OEB,

${OE}^2$ + ${BE}^2$ = ${OB}^2$

${OE}^2$ = 25 - 9 (perpendicular drawn from Centre to a chord bisects the chord therefore BE = 3 cm)

OE = 4 cm

Similarly In $\Delta$OFD,

${OF}^2$ + ${FD}^2$ = ${OD}^2$

${OF}^2$ = 25 - 16

$\Rightarrow$ OF = 3 cm

Area of trapezium,

= $\frac{1}{2}$(perpendicular distance between two parallel sides)(sum of the length of the parallel sides)

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

= $49{cm}^2$