Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used.

Explanation

(i) For the triangle marked I
sides,a=5cm,b=5cm,c=1cm
Semi Perimeter
`s=(a+b+c)/2`
`=(5+5+1)/2`
`=11/2`
By using Heron,s formula
Area of triangle` I=sqrt(s(s-a)(s-b)(s-c)``=sqrt(5.5(5.5-5)(5.5-5)(5.5-1)`
`=sqrt(5.5xx0.5xx0.5xx4.5`
`=sqrt(6.1875`
`=2.5cm^2`
Area of triangle`=2.5cm^2`
(ii)For the rectangle marked II
Area of rectangle=length `xx` breath
`=6.5cmxx1cm`
`=6.5cm^2`
(iii) For trapezium marked III
Draw AE is perpendicular to BC and AF parallel to DC
`AD=FC=1CM`
`BF=BC-FC`
`=2cm-1cm`
`=1cm`
Here `triangle ABF` is
equilateral triangle,Eis the
mid-point of BF
So,BE=EF
`EF=(BF)/2`
`=1/2`
`=0.5`
In `triangleAEF`
`AF^2=AE^2+EF^2`
`L^2=AE^2+0.5^2`
`AE^2=sqrt(1-0.5^2)`
`AE^2=sqrt(0.75)`
`AE=0.9cm`(approx)
Area of trapezium=`1/2xx`sum of parallel `sides xx distance`
between them
`=1/2xx(BC+AD)xxAE`
`=1/2xx3xx0.9`
Area of trapezium` = 1.4cm^2`
(iv)For the triangle marked IV and V
Area of the triangle=`1/2xxbasexxheight`
`=1/2xx6xx1.5`
`=4.5 cm^2`
Area of the triangle `=4.5cm^2+4.5cm^2=9cm^2`
Total area of the paper used=`2.5cm^2+6.5cm^2+1.4cm^2+9cm^2`
Total area of the paper used=`19.4 cm^2`