Question

The area of the region satisfying the inequalities \(|x|-y≤1,y≥0\) and \(y≤1\) is [This Question was asked as TITA]

• A. 3 Square units
• B. 2 Square units
• C. 5 Square units
• D. 4 Square units

Answer 1

The Correct Option is A

The graph of \(|x|-y≤1,y≥0\) and \(y≤1\) is as follows:

 

To find the area of quadrilateral ABCD, we subtract the areas of triangles EAD and BFC from the area of rectangle EFCD:

Area of ABCD = Area of EFCD − Area of △EAD − Area of △BFC 

Using the formula:
\( \text{Area of ABCD} = EF \times FC - \frac{1}{2} \times EA \times ED - \frac{1}{2} \times BF \times FC \)

Substituting the values:
\( = 4 \times 1 - \frac{1}{2} \times 1 \times 1 - \frac{1}{2} \times 1 \times 1 \)

\( = 4 - 0.5 - 0.5 = 3 \) square units

Final Answer: \( \boxed{3} \) square units

Answer 2

Area of the region contained by the lines | x | -y ≤ 1, y ≥ 0 and y ≤ 1 is the two triangle and the one rectangle in white region.
So, we have calculate these area to get the total area.
Total Area = Area of rectangle + 2 × Area of triangle
= \(2+(\frac{1}{2}\times2\times1)=3\)
Therefore, the correct option is (A) : 3 Square units.