Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used.

Explanation

The given picture has been divided into five regions marked I to V, as shown.
Area of Region I This region is a triangle in which a = 5 cm, b = 5 cm, c = 1 cm.
`therefore" "s=(1)/(2)(5+5+1)cm=(11)/(2)cm`.
`therefore" "(s-a)=((11)/(2)-5)cm=(1)/(2)cm`,
`(s-b)=((11)/(2)-5)cm=(1)/(2)cm`,
`(s-c)=((11)/(2)-1)cm =(9)/(2)cm`.
`therefore" ""area of Region I" = sqrt(s(s-a)(s-b)(s-c))`
`=sqrt((11)/(2)xx(1)/(2)xx(1)/(2)xx(9)/(2))cm^(2)=sqrt((99)/(16))cm^(2)`
`=(sqrt(99))/(4)cm^(2)=(9.94)/(4)cm^(2)=2.49 cm^(2)`.
Area of Region II This region is a rectangle of length 6.5 cm and breadth 1 cm.
`therefore" ""area of Region II"=(6.5 xx 1)cm^(2) = 6.5 cm^(2)`.
Area of Region III This region is an isosceles trapezium ABCD in which AB = 2 cm, BC = 1 cm, DC = 1 cm and AD = 1 cm.
Draw DE `bot` AB and CF `bot` AB.
Then, EF = DC = 1 cm.
And, AE = BF = 0.5 cm.
`DE^(2) = AD^(2) - AE^(2)={1^(2)-((1)/(2))^(2)}cm^(2)=(1-(1)/(4))cm^(2) = (3)/(4) cm^(2)`.
`therefore" "DE = (sqrt(3))/(2)`cm.

Area of Region III = `(1)/(2) xx (AB + DC) xx DE`
`={(1)/(2)xx(2+1)xx(sqrt(3))/(2)}cm^(2)=(3 sqrt(3))/(4)cm^(2)`
`=(3 xx 1.732)/(4)cm^(2)=(5.196)/(4)cm^(2)`
`=1.299 cm^(2) ~~ 1.3 cm^(2)`.
Area of Region IV This region is a right triangle with base 1.5 cm and height 6 cm.
`therefore" ""area of Region IV"=((1)/(2)xx "base" xx "height")`
`=((1)/(2)xx(3)/(2)xx6)cm^(2) = (9)/(2)cm^(2) = 4.5 cm^(2)`.
Area of Region V This is a right triangle with base 1.5 cm and height 6 cm.
`therefore" ""area of Region V"=((1)/(2)xx(3)/(2)xx6)cm^(2)=(9)/(2)cm^(2)=4.5 cm^(2)`.
`therefore" "` total area of the paper used
`=(2.49 + 6.5 + 1.3 + 4.5 + 4.5)cm^(2) = 19.29 cm^(2)`.