The area, in sq. units, enclosed by the lines $$x=2,y=\mid x-2\mid+4$$, the X-axis and the Y-axis is equal to
The required figure is a trapezium with vertices A(0,0), B(2,0), C(2,4) and D(0,6)
AB = 2 BC = 4 and AD = 6
Area of trapezium = $$\frac{1}{2}\left(sum\ of\ the\ opposite\ sides\right)\cdot height$$ = $$\frac{1}{2}\left(4+6\right)\cdot2\ =\ 10$$
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