Question

Question 1
How much paper of each shade is needed to make a kite given in figure, in which ABCD is a square with diagonal 44cm.

Answer 1

We know that, all the sides of a square are always equal
i.e., AB = BC = CD = DA
$In\Delta ACD,AC=44cm,\angle D={90}^{\circ }$
Using Pythagoras theorem in $\Delta ACD$,
$A{C}^2=A{D}^2+D{C}^2$
$\Rightarrow {44}^2=A{D}^2+A{D}^2[\because DC=AD]$
$\Rightarrow 2A{D}^2=44\times 44$
$\Rightarrow A{D}^2=22\times 44\Rightarrow AD=\sqrt{22\times 44}$
[taking positive square root because length is always positive]
$\Rightarrow AD=\sqrt{2\times 11\times 4\times 11}$
$\Rightarrow AD=22\sqrt{2}cm$
So, AB = BC = CD = DA = $22\sqrt{2}cm$
$\therefore$ Area of square ABCD $=Side\times Side=22\sqrt{2}\times 22\sqrt{2}=968c{m}^2$
$\therefore$ Area of the red portion $=\frac{968}{4}=242c{m}^2$
[since, area of square is divided into four parts]
Now, area of the green portion $=\frac{968}{4}=242c{m}^2$
Area of the yellow portion $=\frac{968}{2}=484c{m}^2$
In $\Delta PCQ$, side PC = a = 20cm, CQ = b = 20cm and PQ = c = 14cm
$s=\frac{a+b+c}{2}=\frac{20+20+14}{2}=\frac{54}{2}=27cm$
$\therefore Areaof\Delta PCQ=\sqrt{s(s-a)(s-b)(s-c)}$ [by Heron’s formula]
$=\sqrt{27(27-20)(27-20)(27-14)}$
$=\sqrt{27\times 7\times 7\times 13}=\sqrt{3\times 3\times 3\times 7\times 7\times 13}$
$=21\sqrt{39}=21\times 6.24=131.04c{m}^2$
$\therefore$ Total area of the green portion = 242 + 131.04 = 373.04 $c{m}^2$
Hence, the paper required for each shade to make a kite is red paper 242 $c{m}^2$,
yellow paper 484 $c{m}^2$ and green paper 373.04 $c{m}^2$.