Adjoining fig, ABCD is a trapezium with $A B ∥ D C$ and $∠ B C D = 30^{0} .$ Fig. BGEC is a sector of a circle with centre C and AB = BC = 7 cm, DE = 4 cm and BF = 3.5 cm, then find the area of the shaded region. $(U s e π = \frac{22}{7})$

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Solution

$A B = 7 c m, D E = 4 c m, B F = 3.5 c m$
$B C = D E + E C = (4 + 7) c m = 11 c m$
Area pf trapezium $A B C D = \frac{1}{2} \times$ (Sum of parallelogram) $\times$ (Distance between them)
$= \frac{1}{2} \times (11 + 7) \times 3.5$
$= \frac{1}{2} \times 18 \times 3.5$
$= 31.5 {c m}^{2}$
Area of the sector $B G E C = \frac{30}{360} \times \frac{22}{9} \times 7 \times 7$
$= \frac{1}{12} \times 22 \times 7$
$= 12.83 {c m}^{2}$
Area of shaded region = Area of trapezium - Arc of sector
$(31.5 - 12.88) {c m}^{2}$
$= 18.67 {c m}^{2}$

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