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

The line \(y=|x-2|+4\) intersects the y -axis at \((0,6)\) and intersects \(x=2\) at \((2,4)\)
The other vertices are \((0,0)\) and \((2,0)\)
The figure formed is a trapezium of parallel sides \(6\) and \(4\) and the distance between the parallel sides is \(2\).
Required answer =\(\frac{1}{2}×2×(6+4)=10\)
So, the correct option is (C): \(10\)

Answer 2

The required area can be considered as Trapezium with vertices as A(0,0), B(2,0), C(2,4) and D(0,6)
where AB=2, BC=4 and AD = 6
Now, Area of Trapezium = \(\frac{1}{2}\times\text{sum of opposite sides}\times\text{height}\)
= \(\frac{1}{2}\times(4+6)\times2\)

= \(\frac{1}{2}\times20\)
= 10 cm²
So, the correct option is (C) : 10.