Question

3.5 см
3.5 cm
3.5 cm
3.5 см
find the area of above figure​

Answer 1

Answer:

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}}722∗36090∗(3.5)2

\mathit{\frac{22}{7*4} * \frac{49}{4} }7∗422∗449

\mathit{\frac{22*49}{7*4*4}}7∗4∗422∗49

\mathit{\frac{11*7}{2*4}cm^2}2∗411∗7cm2

\mathit{\frac{77}{8}cm^2}877cm2

And, Area of ΔOBD

\bold{\mathit{\frac{1}{2}*OB*OD}}21∗OB∗OD

\mathit{\frac{1*2*3.5}{2}}21∗2∗3.5

\mathit{\frac{7}{2}cm^2}27cm2

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

\mathit{(\frac{77}{8} - \frac{7}{2})cm^2}(877−27)cm2

\mathit{\frac{77-28}{8}cm^2}877−28cm2

\mathit{\frac{49}{8}cm^2}849cm2

HENCE, Area of shaded region = 6.125cm²