Answer :

Since, the kite is in the shape of a square.

Each diagonal of square = 32 cm ...(Given)

We know that, the diagonals of a square bisect each other at right angle.



For Part I :
Area of Part I
= $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 32 × 16
= 16 × 16 = 256${cm}^2$

For Part II :
Area of Part II
= $\frac{1}{2}$ × Base × Height
= $\frac{1}{2}$ × 32 × 16
= 16 × 16 = 256${cm}^2$

For Part III :
It is a triangle with sides 6 cm, 6 cm and 8 cm.
Now, we know that,

$s=\frac{a+b+c}{2}$
$?$ $s=\frac{6+6+8}{2}=\frac{20}{2}=10cm$

Now, Area of triangle
= $\sqrt{10(10?6)(10?6)(10?8)}$
(Since, Heron's formula [area = $\sqrt{s(s?a)(s?b)(s?c)}$])
= $\sqrt{10\times 4\times 4\times 2}$

= $\sqrt{5\times 2\times 4\times 4\times 2}$
= $2\times 4\sqrt{5}{cm}^2$
= $8\sqrt{5}{cm}^2$
= 8 × 2.24 cm = 17.92 ${cm}^2$

Hence, the area of colour used for Paper I = 256 ${cm}^2$

The area of colour used for Paper II = 256 ${cm}^2$

The area of colour used for Paper III = 17.92 ${cm}^2$