Area enclosed within curve |x| + |y| = 1 is

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9.Straight Line

medium

The area enclosed within the curve $|x| + |y| = 1$ is

A

$\sqrt 2 $

B

$1$

C

$\sqrt 3 $

D

$2$

(IIT-1981)

Solution

(d) The given lines are $ \pm x \pm y = 1$

i.e. $x + y = 1,x – y = 1,x + y = – 1$and $x – y = – 1$

These lines form a quadrilateral whose vertices are $A( – 1,0),B(0, – 1),C(1,0)$and $D(0,1)$

Obviously $ABCD$ is a square.

Length of each side of this square is $\sqrt {{1^2} + {1^2}} = \sqrt 2 $

Hence area of square is $\sqrt 2 \times \sqrt 2 = 2sq.$units

Trick: Requird area = $\frac{{2{c^2}}}{{|ab|}} = \frac{{2 \times {1^2}}}{{|1 \times 1|}} = 2$.

Standard 11

Mathematics

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