Answer :

Area of whole figure = ar || ABCD + ar || FGHI + ar DCIF + ar DEF + area semicircle


CD = 8 cm, BP = HQ = 4 cm, DE = EF = 5 cm, CI = 8 cm


ar ABCD = ar FGHI = base × height


ar ABCD = ar FGHI = BP×DC


ar ABCD = ar FGHI = 4×8


ar ABCD = ar FGHI = 32 cm2 eqn1 and eqn2


ar DCIF = area of square = side×side


ar DCIF = DC×CI


ar DCIF = 8×8


ar DCIF = 64 cm2 eqn3


Consider ∆DEF, EFDF and ∆DEF is isosceles


So, FL = LD




FL = LD = 4 cm


In ∆DEL, DLE = 90°



52 = EL2 + 42 (putting the values)


25 = EL2 + 16


25 – 16 = EL2


EL2 = 9



EL = 3 cm





Area of ∆DEF = 4×3


Area of ∆DEF = 12 cm2 eqn4



R = 4 cm




Area of semicircle = 3.14×8


Area of semicircle = 25.12 cm2 eqn5


Area of whole figure = eqn1 + eqn2 + eqn3 + eqn4 + eqn5


Area of whole figure = 32 + 32 + 64 + 12 + 25.12


Area of whole figure = 165.12 cm2


Area of the whole figure is 165.12 cm 2.


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