Question

The area, in sq. units, enclosed by the lines \(x=2\), \(y=|x-2|+4\), the \(X\)-axis and the \(Y\)-axis is equal to

• A. 8
• B. 12
• C. 10
• D. 6

Answer 1

The Correct Option is C

To determine the area enclosed by the lines \(x=2\), \(y=|x-2|+4\), the \(X\)-axis, and the \(Y\)-axis, we will analyze the geometry formed by these lines. The equation \(y=|x-2|+4\) describes a V-shaped graph. It can be split into two linear equations:

• When \(x \geq 2\), \(y=x-2+4=x+2\). 
• When \(x < 2\), \(y=-(x-2)+4=-x+6\).

Thus, we have two lines: \(y=x+2\) and \(y=-x+6\). These intersect the line \(x=2\) as follows:

• For \(y=x+2\), when \(x=2\), \(y=2+2=4\).
• For \(y=-x+6\), when \(x=2\), \(y=-2+6=4\).

Both lines intersect the vertical line \(x=2\) at \(y=4\), confirming the V's vertex is on this line. The line \(y=x+2\) intersects the \(X\)-axis at \(x=-2\) (setting \(y=0\)). The line \(y=-x+6\) intersects the \(Y\)-axis at \(y=6\) (setting \(x=0\)).

Now, calculate the area of the triangle formed: vertices at \((0,0)\), \((0,6)\), and \((-2,0)\).

The base of the triangle is along the \(Y\)-axis from \((0,0)\) to \((0,6)\), so base = 6 units. The height is along the \(X\)-axis from \((0,0)\) to \((-2,0)\), so height = 2 units.

The area of a triangle is given by:

\[\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}=\frac{1}{2} \times 6 \times 2=6\text{ units}^2\]

However, we observe that the region enclosed by the conditions also encompasses another triangle with vertices: \((+2,4)\), \((0,0)\), and \((0,6)\) formed under the positive section of the V.

Vertex 1	Vertex 2	Vertex 3
\((0,0)\)	\((0,6)\)	\((2,4)\)

Next, calculate the area of this triangle:

Base = from \((0,6)\) to \((0,0)\) = 6 units, Height from \((2,4)\) to the \(Y\)-axis = 2 units.

Area = \(\frac{1}{2} \times\ 6 \times\ 2 = 6\text{ units}^2\).

Thus, combining areas: \(6 + 4\ = 10 \text{sq. units}\).

Therefore, the area enclosed by the given lines is \(\boxed{10}\text{ sq. units.}‍\)

Answer 2

 

The required area can be considered as a trapezium with vertices: A(0, 0), B(2, 0), C(2, 4) and D(0, 6). 

From the coordinates: 
AB = 2 units (horizontal base), 
BC = 4 units (vertical side), 
AD = 6 units (slant side, but we use height for area).

Here, the two parallel sides (heights from horizontal AB) are: 
From point B to C = 4 units, 
From point A to D = 6 units.

The distance between the parallel sides (i.e., horizontal length) is: AB = 2 units.

Area of a trapezium is given by: 
\( \text{Area} = \frac{1}{2} \times (\text{Sum of parallel sides}) \times \text{height} \)

So, 
\( \text{Area} = \frac{1}{2} \times (4 + 6) \times 2 \) 
\( = \frac{1}{2} \times 10 \times 2 = 10 \, \text{cm}^2 \)

Therefore, the correct option is (C): 10.