A child draws the figure of an aeroplane as given. Here the wings EDCF and AGHB are parallelograms the tail ADK is an isosceles triangle, the cockpit BLC is a semi circle and the portion ABCD is a square. Let $F P ⊥ C D & H Q ⊥ A B$ AB = 6 cm, KD = 5 cm, FP = HQ = 2 cm. The area of the figure is

98.14 c m^{2}

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87.25 c m^{2}

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84.63 c m^{2}

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91.56 c m^{2}

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Solution

The correct option is A $98.14 c m^{2}$
Area of $E D C F = D C \times F P = 12 s q c m$
So, area of $A G H B = 12 s q c m$ .
Area of $A B C D = 6 \times 6 = 36 s q c m$
Area of triangle KDC = $\frac{6 \sqrt{5^{2} - \frac{36}{4}}}{2} = 3 \sqrt{16} = 12 s q c m$
Area of semicircle = $\frac{π {.3}^{2}}{2} = 14.14 s q c m$
So, the total area $= 86.14 s q c m .$

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Similar questions

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A child draws the figure of an aeroplane as given. Here the wings EDCF and AGHB are parallelograms the tail ADK is an isosceles triangle, the cockpit BLC is a semi circle and the portion ABCD is a square. Let FP⊥CD&HQ⊥AB AB = 6 cm, KD = 5 cm, FP = HQ = 2 cm. The area of the figure is

The correct option is A 98.14cm2Area of EDCF=DC×FP=12sqcmSo, area of AGHB=12sqcm.Area of ABCD=6×6=36sqcmArea of triangle KDC = 6√52−3642=3√16=12sqcm Area of semicircle = π.322=14.14sqcmSo, the total area =86.14sqcm.