Hence, of all the rectangles inscribed in the given circle, the square has the maximum area. Equations of the lines are y = 2x + 1, y = 3x + 1 and x + 4 Let $y_1$ = 2x + 1, $y_2$ = 3x + 1 Now area of the triangle bounded by the given lines, = $\,{}^{{\underset{0}{4}}}(y_2-y_1)$ dx = $\,{}^{\underset{0}{4}}[(3x + 1) - (2x + 1)]$ dx = $\,{}^{{\underset{0}{4}}}x$ dx = $\frac{1}{2}[x^2]^4_0$ = $\frac{1}{2}(4^2-0^2)$ = $\frac{1}{2}$ × 16 = 8 sq. units Thus, the area of the required triangular region is 8 square units.