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(D) Given the parabola $y^2 = 4x$. The ordinates of the vertices are $y_1 = 1, y_2 = 2, y_3 = 4$.Since the points lie on the parabola,their $x$-coordi

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$\frac{3}{4}$

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(D) Given the parabola $y^2 = 4x$. The ordinates of the vertices are $y_1 = 1, y_2 = 2, y_3 = 4$.Since the points lie on the parabola,their $x$-coordinates are given by $x = \frac{y^2}{4}$.For $y_1 = 1$,$x_1 = \frac{1^2}{4} = \frac{1}{4}$.For $y_2 = 2$,$x_2 = \frac{2^2}{4} = 1$.For $y_3 = 4$,$x_3 = \frac{4^2}{4} = 4$.The vertices of the triangle are $(\frac{1}{4}, 1), (1, 2),$ and $(4, 4)$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by $\frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$.Area $= \frac{1}{2} |\frac{1}{4}(2 - 4) + 1(4 - 1) + 4(1 - 2)|$.Area $= \frac{1}{2} |\frac{1}{4}(-2) + 3 + 4(-1)|$.Area $= \frac{1}{2} |-\frac{1}{2} + 3 - 4| = \frac{1}{2} |-\frac{1}{2} - 1| = \frac{1}{2} |-\frac{3}{2}| = \frac{3}{4}$.

The area of the triangle formed inside the parabola $y^2 = 4x$ whose vertices have ordinates $1, 2,$ and $4$ is:

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