In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. [2 MARKS]

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Solution

Concept: 1 Mark
Application: 1 Mark

Area of quadrant
$= \frac{90}{360} \times π \times 14 \times 14 = 153.94 c m^{2}$

Area of $Δ A B C = 0.5 \times 14^{2} = 98 c m^{2}$

The diameter $= 14 \times \sqrt{2} = 19.8 c m$ .

Area of semi-circle
$= 0.5 π \times {9.9}^{2} = 153.95 c m^{2}$

The area of segment
$= 153.94 - 98 = 55.94 c m^{2}$

The required area
$= 153.95 - 55.94 = 98.01 c m^{2}$ .

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In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. [2 MARKS]

Concept: 1 Mark Application: 1 Mark Area of quadrant =90360×π×14×14=153.94 cm2 Area of ΔABC=0.5×142=98 cm2 The diameter =14×√2=19.8 cm. Area of semi-circle =0.5 π×9.92=153.95 cm2 The area of segment =153.94−98=55.94 cm2 The required area =153.95−55.94=98.01 cm2.