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(A) Let the floor be the $x$-axis and the vertex of the angle be the origin $(0,0)$. The wall is a line making an angle of $135^{\circ}$ with the floo

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$\frac{\pi l^2}{4}$

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(A) Let the floor be the $x$-axis and the vertex of the angle be the origin $(0,0)$. The wall is a line making an angle of $135^{\circ}$ with the floor,so its equation is $y = -x$ (or $x+y=0$).Let the ladder have endpoints $P(x_1, 0)$ on the floor and $Q(x_2, y_2)$ on the wall. Since $Q$ is on $y = -x$,$Q$ is $(x_2, -x_2)$.The length of the ladder is $l$,so $(x_1 - x_2)^2 + (0 - (-x_2))^2 = l^2$,which simplifies to $(x_1 - x_2)^2 + x_2^2 = l^2$.The mid-point $(h, k)$ of the ladder is $\left(\frac{x_1+x_2}{2}, \frac{0-x_2}{2}\right) = \left(\frac{x_1+x_2}{2}, -\frac{x_2}{2}\right)$.From this,$x_2 = -2k$ and $x_1+x_2 = 2h$,so $x_1 = 2h - x_2 = 2h + 2k$.Substituting into the length equation: $(2h+2k - (-2k))^2 + (-2k)^2 = l^2$,which is $(2h+4k)^2 + 4k^2 = l^2$.Expanding gives $4h^2 + 16hk + 16k^2 + 4k^2 = l^2$,or $4h^2 + 16hk + 20k^2 = l^2$.The general equation of an ellipse is $Ax^2 + Bxy + Cy^2 = F$. The area of such an ellipse is given by $\frac{2\pi F}{\sqrt{4AC - B^2}}$.Here $A=4, B=16, C=20, F=l^2$. The denominator is $\sqrt{4(4)(20) - (16)^2} = \sqrt{320 - 256} = \sqrt{64} = 8$.Area $= \frac{2\pi l^2}{8} = \frac{\pi l^2}{4}$.

$A$ wall is inclined to the floor at an angle of $135^{\circ}$. $A$ ladder of length $l$ is resting on the wall. As the ladder slides down,its mid-point traces an arc of an ellipse. Then,the area of the ellipse is

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