Question

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

• A. $\sqrt{2}$
• B. $1$
• C. $\sqrt{3}$
• D. $2$

Answer 1

The correct option is D

$2$

Explanation for correct option:

Step 1: Analysing the cases:

Given, curve is $\left|x\right|+\left|y\right|=1$

There are two cases in this question:

Case $1$: When $x>0$

Here the equation of the curve will be $x+\left|y\right|=1$. Now if $y>0$ then, $x+y=1$ and if $y<0$ then, $x-y=1$

Case $2$: When $x<0$

Here the equation of the curve will be $-x+\left|y\right|=1$. Now if $y>0$ then, $-x+y=1$ and if $y<0$ then, $-x-y=1$.

Hence the graph of the given curve will be as shown:

Step 2: Finding the area:

Here we have two triangles with base $b=2$ and height $h=1$.

Therefore area of the triangle $ABC$ is,

$$\begin{gathered} \text{Area of}△\left(\mathrm{ABC}\right)=\frac{1}{2}\times \text{Base}\times \text{Height} \\ =\frac{1}{2}\times 2\times 1 \\ =1\text{square unit} \end{gathered}$$

Therefore the total area is equal to,

$$\begin{gathered} \text{Total area}=2\times \text{Area(△ABC)} \\ =2\times 1 \\ =2\text{square units} \end{gathered}$$

Therefore the area enclosed within the curve $\left|x\right|+\left|y\right|=1$ is equal to $2\mathrm{square}\mathrm{units}$.

Hence, option (D) is the correct answer.