How much paper of yellow shade is needed in $c m^{2}$ to make a kite given in Fig, in which ABCD is a square with diagonal 44 cm.

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Solution

$T h e k i t e i s d i v i d e d i n t w o p a r t s - u p p e r s q u a r e a n d l o w e r t r i a n g l e . T h e s q u a r e A B C D i s h a v i n g d i a g o n a l s o f l e n g t h 44. T h e a r e a o f t h e s q u a r e A B C D = \frac{1}{2} \times ({d i a g o n a l}^{2}) = \frac{1}{2} \times (44^{2}) = 22 \times 44 = 968 S i n c e d i a g o n a l s o f a s q u a r e a l w a y s d i v i d e t h e s q u a r e i n t o 4 e q u a l t r i a n g l e s . ∴ A r e a o f △ O A B = a r e a o f △ O B C = a r e a o f △ O C D = a r e a o f △ O A D = \frac{1}{4} \times 968 = 242 N o w i n t h e l o w e r t r i a n g l e o f k i t e, u s i n g H e r o n^{'} s f o r m u l a t h e a r e a o f t r i a n g l e w i t h s i d e s a, b & c i s g i v e n a s A = \sqrt[2]{[s (s - a) (s - b) (s - c)]} w h e r e 2 s = (a + b + c) . H e r e a = c = 20, b = 14, s = \frac{(20 + 20 + 14)}{2} = \frac{54}{2} = 27 A r e a o f l o w e r △ = \sqrt[2]{[27 \times (27 - 14) (27 - 20) (27 - 20)]} = \sqrt[2]{(27 \times 13 \times 7 \times 7)} = \sqrt[2]{(3 \times 3 \times 3 \times 13 \times 7 \times 7)} = 3 \times 7 \times \sqrt[2]{(3 \times 13)} = 21 \times \sqrt[2]{39} {U s e d i v i s i o n m e t h o d t o f i n d s q u a r e r o o t} = 21 \times 6.24 = 131.04 T o t a l y e l l o w p a p e r r e q u i r e d = a r e a o f △ O A B + a r e a o f △ O C D = 2 \times 242 = 484 T o t a l r e d p a p e r r e q u i r e d = a r e a o f △ O B C = 242 T o t a l g r e e n p a p e r r e q u i r e d = a r e a o f △ A O D + a r e a o f l o w e r t r i a n g l e = 242 + 131.04 = 373.04$

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How much paper of yellow shade is needed in cm2 to make a kite given in Fig, in which ABCD is a square with diagonal 44 cm.

How much paper of yellow shade is needed in $c m^{2}$ to make a kite given in Fig, in which ABCD is a square with diagonal 44 cm.