MathematicsA kite is in the shape of a square with a diagonal of  32cm   and an isosceles triangle of base  8cm   and sides  6cm   each is to be made of three different shades as shown in the figure. ____ paper of each shade has been used in it?

A kite is in the shape of a square with a diagonal of  32cm   and an isosceles triangle of base  8cm   and sides  6cm   each is to be made of three different shades as shown in the figure. ____ paper of each shade has been used in it?


IMG_256

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    Solution:

    Let us find the area of part 1 first
    (i) Area of part 1
    Let us assume that the triangle ΔABF  in different orientation as follows
    IMG_256
    We know that the all sides of square are equal so, let us assume that
    BA=AF=a  
    We are given that the length of diagonal as  32cm   so, we have
    BF=32cm  
    IMG_256
    The Pythagoras theorem is given as b 2 = a 2 + c 2  .
    By using the above theorem to triangle  ΔABF   we get,
    B A 2 +A F 2 =B F 2  
    By substituting the required values we get
    a 2 + a 2 = 32 2 2 a 2 =1024 a= 512 =16 2 cm  
    Let us assume that the area of part 1 as  A 1  .
    We know that the formula of area of triangle as
    A= 1 2 (base)(height)  
    By using the above formula to ΔABF  we get,
    A 1 = 1 2 (a)(a)= a 2 2  
    By substituting the required values in above equation we get
    A 1 = (162 2 ) 2 2 A 1 = 512 2 =256c m 2  
    Therefore the area of part 1 is  256c m 2  Now, let us solve the second part.
    (i) Area of part 2
    Let us assume that the area of part 2 as  A 2  
    We know that the diagonal of a square divides the square into two triangles of same area that is
    A 2 = A 1 =256c m 2  
    Therefore the area of part 2 is 256c m 2  .
    (iii) Area of part 3
    Let us take the triangle  ΔCDE   and draw the height as shown below.
    IMG_256
    We are given that the dimensions of each triangle are 8cm,6cm,6cm  .
    So, from the triangle  ΔCDE   we have,
    CE=CD=6cm DE=8cm  
    We know that the altitude of isosceles triangle will be the median
    So, the point ‘H’ is mid – point of ‘DE’
    So, we can the length of ‘HE’ as
    HE= DE 2 HE= 8 2 cm=4cm  
    IMG_256The Pythagoras theorem is given as, b 2 = a 2 + c 2  .
    By using the above theorem to triangle  ΔCHE   we get,
    C H 2 +H E 2 =C E 2  
    By substituting the required values we get
    C H 2 + 4 2 = 8 2 C H 2 =6416 CH= 48 =4 3 cm  
    We know that the formula of area of triangle as
    A= 1 2 (base)(height)  
    By using the above formula to ΔCDE  as A 3  ,
    A 3 = 1 2 (DE)(CH)  
    By substituting the required values in above equation we get
    A 3 = 1 2 8 4 3 A 3 =16 3 c m 2  
    Therefore the area of third part is  16 3 c m 2  .
     
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