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

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Answers (1)

$Red = 242 \; cm^{2}$

$Yellow = 484\; cm^{2}$

$Green = 373.14 \; cm^{2}$

$Given,\; ABCD\; is\; a\; square.$

We know that all sides of a square are equal

AB = BC = CD = DA and

$\angle A = \angle B = \angle C = \angle D = 90^{\circ} [\because \; All\; angles\; of\; a\; square\; are\; 90^{\circ}]$

$In \Delta ABC,\; using \; Pythagoras \; theorem$

$(AC)^{2} = (AB)^{2}+ (BC)^{2}$

$(44)^{2} = (AB)^{2}+ (BC)^{2}\; \; \; \; \; \; \; \; [\Theta \; AB = BC\; equal\; sides]$

$44\times 44=2\left ( AB \right )^{2}$

$\frac{44\times 44}{2}=\left ( AB \right )^{2}$

$(AB)^{2} = 22 \times 44$

Taking square root on both sides

$\sqrt{\left ( AB \right )^{2}}=\sqrt{22\times 44}$

$AB=\sqrt{22\times 2\times 22}$

$AB=22\sqrt{2}$

$AB = BC = CD = DA =22\sqrt{2}\; cm$

$Now,\; Area\; of \; square\; ABCD = (side)^{2}$

$=\left ( 22\sqrt{2}\right )^{2}=22\times 22\times \sqrt{2}\times \sqrt{2}$

$= 484\times \sqrt{2\times 2}=484\times 2$

$Area\; of\; square\; ABCD = 968\; cm^{2}$

But square ABCD is divided into four coloured squares.

$So,\; area\; of\; Yellow \; I =\frac{968}{4}= 242 cm^{2}$

$Area\; of\; Yellow \; II =\frac{968}{4}= 242 \; cm^{2}$

$Area\; of\; Green \; III =\frac{968}{4}= 242 \; cm^{2}$

$Area\; of\; Red \; IV =\frac{968}{4}= 242 \; cm^{2}$

$Total\; yellow\; area\; = 242\; cm^{2} + 242\; cm^{2} = 484\; cm^{2}$

We have to find the lower triangle of green colour as well.

$Let\; a = 20\; cm, b = 20\; cm, c = 14\; cm$

$Semi\; perimeter(s) = \frac{a+b+c}{2}$

$= \frac{20+20+14}{2}$

$= \frac{54}{2}=27$

Area of Triangular field:

$By\; heron's\; formula =\sqrt{S\left ( S-a \right )\left ( S-b \right )\left ( S-c \right )}$

$=\sqrt{27\left ( 27-20\right )\left ( 27-20 \right )\left ( 27-14 \right )}$

$= \sqrt{3\times 3\times 3\times 7\times 7\times 13}$

$= 21\sqrt{3\times 13}$

$= 131.14\; cm^{2}$

$So\; total\; green\; area\; = 242 + 131.14 = 373.14\; cm^{2}$

$Hence,\; paper\; required$

$Red = 242\; cm^{2}$

$Yellow = 484\; cm^{2}$

$green = 373.14\; cm^{2}$

Posted by

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