A child draws the figure of an aeroplane as shown. Here, the wings ABCD and FGHI are parallelograms, the tail DEF is an isosceles triangle, the cockpit CKI is a semicircle and CDFI is a square. In the given figure, BP ⊥ CD, HQ ⊥ FI and EL ⊥ DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [Take π = 3.14.]

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 cm^{2}→ eqn1 and eqn2

ar DCIF = area of square = side×side

ar DCIF = DC×CI

ar DCIF = 8×8

ar DCIF = 64 cm^{2}→ eqn3

Consider ∆DEF, EF⊥DF and ∆DEF is isosceles

So, FL = LD

FL = LD = 4 cm

In ∆DEL, ∠DLE = 90°

5^{2} = EL^{2} + 4^{2} (putting the values)

25 = EL^{2} + 16

25 – 16 = EL^{2}

⇒ EL^{2} = 9

∴ EL = 3 cm

⇒ Area of ∆DEF = 4×3

∴ Area of ∆DEF = 12 cm^{2}→ eqn4

R = 4 cm

⇒ Area of semicircle = 3.14×8

∴ Area of semicircle = 25.12 cm^{2}→ 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 cm^{2}

Area of the whole figure is 165.12 cm ^{2}.

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