Question

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

Answer 1

We know that, all the sides of a square are always equal.

i.e., AB = BC = CD = DA

In ΔACD, AC = 44 cm, ∠D = 90°

Using Pythagoras theorem in ΔACD,

AC² = AD² + DC²

⇒ 44² = AD² + AD²   ...[∵ DC = AD]

⇒ 2AD² = 44 × 44

⇒ AD² = 22 × 44

⇒ AD = `sqrt(22 xx 44)`  ...[Taking positive square root because length is always positive]

⇒ AD = `sqrt(2 xx 11 xx 4 xx 11)`

⇒ AD = `22sqrt(2)` cm

So, AB = BC = CD = DA = `22sqrt(2)` cm

∴ Area of square ABCD = Side × Side

= `22sqrt(2) xx 22sqrt(2)`

 = 968 cm²

∴ Area of the red portion = `968/4` = 242 cm²  ...[Since, area of square is divided into four parts]

Now, area of the green portion = `968/4` = 242 cm²

Area of the yellow portion = `968/2` = 484 cm²

In ΔPCQ, side PC = a = 20 cm, CQ = b = 20 cm and PQ = c = 14 cm

`s = (a + b + c)/2`

= `(20 + 20 + 14)/2`

= `54/2`

= 27 cm

∴ Area of ΔPCQ = `sqrt(s(s - a)(s - b)(s - c))`  ...[By Heron’s formula]

= `sqrt(27(27 - 20)(27 - 20)(27 - 14))`

= `sqrt(27 xx 7 xx 7 xx 13)`

= `sqrt(3 xx 3 xx 3 xx 7 xx 7 xx 13)`

= `21sqrt(39)`

= 21 × 6.24

= 131.04 cm²

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

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