In the figure given, area of the shaded region is

1

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\frac{3}{2}

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2

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\frac{5}{2}

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Solution

The correct option is D $\frac{5}{2}$
$L e t t h e b i g g e r Δ b e A O B w i t h A = (0, 2), B = (3, 0) & O = (0, 0) . C l e a r l y ∠ O = 90^{o} a n d O A & O B l i e o n t h e a x e s . ∴ O A = T h e d i f f e r e n c e b e t w e e n t h e o r d i n a t e s = 2 - 0 = 2 u n i t s a n d O B = T h e d i f f e r e n c e b e t w e e n t h e a b s c i s s a e . ∴ O B = 3 - 0 = 3 u n i t s . ∴ a r Δ A O B = \frac{1}{2} O A \times O B = \frac{1}{2} 2 \times 3 s q u n i t s = 3 s q u n i t s . A g a i n w e t a k e t h e s m a l l e r Δ t o b e P O Q w i t h P = (0, 1), Q = (1, 0) a n d O = (0, 0) . B y t h e s i m i l a r a r g u m e n t s a s a b o v e w e h a v e O P = T h e d i f f e r e n c e b e t w e e n t h e o r d i n a t e s = 1 - 0 = 1 u n i t a n d O Q = T h e d i f f e r e n c e b e t w e e n t h e a b s c i s s a e = 1 - 0 = 1 u n i t . ∴ a r Δ P O Q = \frac{1}{2} O P \times O Q = \frac{1}{2} \times 1 \times 1 s q u n i t = \frac{1}{2} s q u n i t . ∴ T h e a r e a o f t h e s h a d e d p o r t i o n = (a r b i g g e r Δ) - (a r s m a l l e r Δ) = 3 - \frac{1}{2} s q u n i t = \frac{5}{2} s q u n i t s A n s - O p t i o n D$

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