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 $\perp$CD, HQ $\perp$FI and EL $\perp$DF. If CD = 8 cm, BP = HQ = 4 cm and DE = EF = 5 cm, find the area of the whole figure. [ Take $\pi$ = 3.14]

Answer 1

Total area = Area of parelleogram ABCD + Area of parallelogram FGHI + Area of square CDFI + Area of isosceles triangle DEF + Area of semicircle CKI

Area of semicircle CKI = $\frac{1}{2}\pi r^2=\frac{1}{2}\times 3.14\times 4^2=25.12c{m}^2$

In triangle EFL, $\angle ELF={90}^o,EL=\sqrt{5^2-4^2}=\sqrt{25-16}=\sqrt{9}=3cm=height$

Area of isosceles triangle DEF = $\frac{1}{2}\times base\times height=\frac{1}{2}\times 4\times 3=6c{m}^2$

Area of square CDFI =$a^2=8^2=64c{m}^2$

Area of parallelogram ABCD = Area of parallelogram FGHI = $base\times height=8\times 4=32c{m}^2$

Total area $=25.12+6+64+32=127.12c{m}^2$