How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44 cm.

Explanation

We know that, all the sides of a square are always equal.
`i.e.," "AB=BC=CD=DA`
`" in " triangleACD, " "AC=44cm, angleD=90^(@)`
`"Using Pythagoras theorem in "triangleACD,`
`AC^(2)=AD^(2)+DC^(2)`
`rArr " "44^(2)=AD^(2)+AD^(2)" "[because DC=AD]`
`rArr " "2AD^(2)=44xx44`
`rArrr " "AD^(2)=22xx44rArrAD=sqrt(22xx44)`
`["taking positive square root because length is always positive"]`
`rArr " "AD=sqrt(2xx11xx4xx11)`
`rArr " "AD=22sqrt2cm`
`"So" " "AB=BC=CD=DA=22sqrt2cm`
`therefore" ""Area of square "ABCD="Side"xx"Side"=22sqrt2xx22sqrt2=968cm^(2)`
`therefore" ""Area of the red protion"=(968)/(4)=242cm^(2)`
`["since, area of square is divided into four parts"]`
`"Now, area of the green portion"=(968)/(4)=242cm^(2)`
`"In "trianglePCQ," ""side "PC=a=20cm, CQ=b=20cm and PQ=C=14cm`
`s=(a+b+c)/(2)=(20+20+14)/(2)=(54)/(2)=27cm`
`therefore" ""Area of "trianglePCQ=sqrt(s(s-a)(s-b)(s-c))" "["by Heron's formula"]`
`=sqrt(27(27-20)(27-20)(27-14))`
`=sqrt(27xx7xx7xx13)=sqrt(3xx3xx3xx7xx7xx13)`
`=21sqrt39=21xx6.24=131.04cm^(2)`
`therefore" ""Total area of the green portion"=242+131.04)=373.04cm^(2)`
`"Hence, the paper required for each shade to make a kite is red paper 242" cm^(2), "yellow paper 484" cm^(2) "and green paper 373.04" cm^(2)`.