Question

In figure, OACB is a quadrant of a circle with centre O and radius 3.5 cm. If OD = 2 cm., find the area of the shaded region. (use π = 22/7)

Answer 1

ACB is a quadrant, subtend at 90° angle at O.

So, ø = 90° , r = 3.5cm

now,

Area of quadrant OACB

$\sf{\bold{\frac{22}{7}*\frac{90}{360} * (3.5)^2}}$

$\mathit{\frac{22}{7*4} * \frac{49}{4} }$

$\mathit{\frac{22*49}{7*4*4}}$

$\mathit{\frac{11*7}{2*4}cm^2}$

$\mathit{\frac{77}{8}cm^2}$

And, Area of ΔOBD

$\bold{\mathit{\frac{1}{2}*OB*OD}}$

$\mathit{\frac{1*2*3.5}{2}}$

$\mathit{\frac{7}{2}cm^2}$

Area of the shaded region = Area of quadrant OACB − Area of ΔOBD

$\mathit{(\frac{77}{8} - \frac{7}{2})cm^2}$

$\mathit{\frac{77-28}{8}cm^2}$

$\mathit{\frac{49}{8}cm^2}$

HENCE, Area of shaded region = 6.125cm²

Answer 2

Solution :

i ) Dimensions of the sector OACB :

Radius ( r ) = OB = OA = 3.5 cm

sector angle ( x ) = 90°

Area of the sector = ( x/360 ) × πr²

= ( 90/360 ) × ( 22/7 ) × 3.5²

= ( 1/4 ) × ( 22/7 ) × 3.5 × 3.5

= 9.625 cm² ----( 1 )

ii ) Dimensions of the right ∆BOD,

base ( b ) = 3.5 cm

height ( h ) = OD

h = 2 cm ( given )

Area ∆BOD = ( bh )/2

= ( 3.5 × 2 )/2

= 3.5 cm² ---------( 2 )

iii ) Area of shaded region = ( 1 ) - ( 2 )

= 9.625 - 3.5

= 6.125 cm²

******