Question

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

Answer 1

In the given figure, ABCD is a square with diagonal 44 cm.
∴ AB = BC = CD = DA. ....(1)

In right angled ∆ABC,
AC² = AB² + BC² (Pythagoras Theorem)
⇒ 44² = 2AB²
⇒ 1936 = 2AB²
⇒ AB² = $\frac{1936}{2}$
⇒ AB² = 968
⇒ AB = $22\sqrt{2}$ cm ...(2)

∴ Sides of square = AB = BC = CD = DA = $22\sqrt{2}$ cm

Area of square ABCD = (side)²
= ($22\sqrt{2}$)²
= 968 cm² ...(3)

Area of red portion = $\frac{968}{4}=242{\mathrm{cm}}^2$
Area of yellow portion = $\frac{968}{2}=484{\mathrm{cm}}^2$
Area of green portion = $\frac{968}{4}=242{\mathrm{cm}}^2$

Now, in ∆AEF,
The sides of the triangle are of length 20 cm, 20 cm and 14 cm.
∴ Semi-perimeter of the triangle is
$s=\frac{20+20+14}{2}=\frac{54}{2}=27\mathrm{cm}$

∴ By Heron's formula,
$$\begin{gathered} \mathrm{Area}\mathrm{of}∆AEF=\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)} \\ =\sqrt{27\left(27-20\right)\left(27-20\right)\left(27-14\right)} \\ =\sqrt{27\left(7\right)\left(7\right)\left(13\right)} \\ =21\sqrt{39} \\ =131.04{\mathrm{cm}}^2...\left(4\right) \end{gathered}$$

Total area of the green portion = 242 + 131.04 = 373.04 cm²

Hence, the paper required of each shade to make a kite is red paper 242 cm², yellow paper 484 cm² and green paper 373.04 cm².