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.

Answer 1

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