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Question
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.
Solution
CD = 8 cm
BP = HQ = 4 cm
DE = EF = 5 cm
Area of the parallelogram ABCD `="B"xx"H"`
= 8 × 4
= 32 sq. cm
Area of parallelogram ABCD = B × H
= 8 × 4
= 32 sq.cm
Area of parallelogram FGHI = B × H
= 8 × 4
= 32 sq. cm
Area of the square = Side2
= 82 = 64 sq.cm
In Δ ELF, we have:
EL2 = 52 - 42
EL2 = 9
EL = 3 cm
Area of ΔDEF `=1/2xx"B"xx"H"`
`= 1/2xx8xx3`
= 12 sq.cm
Area of the semicircle = `1/2pi"r"^2`
`=1/2xx22/7xx16`
= 25.12 sq.cm
∴ Total Area = Area of the parallelogram ABCD + Area of the parallelogram FGHI + Area of the triangle DEF + Area of the semicircle CKI + Area of the square
Total Area = 165.12 cm2
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