hint:Albert said:A square $ABCD$ , each side with length 13,
if points $E,F$ on $\overline{BC} ,\overline{AB} $ respectively,$\angle EDF=45^o,$and $\overline{EF}=11$
find length of $\overline{CE}=?$
The length of $\overline{CE}$ can be found using the Pythagorean Theorem, since the square has a right angle at point $C$. If we let $x$ represent the length of $\overline{CE}$, then we can set up the equation $x^2 + x^2 = 1^2$, since the side length of the square is 1 unit. Solving for $x$, we get $x = \frac{\sqrt{2}}{2} \approx 0.7071$ units.
The length of $\overline{CE}$ can be found by using the Pythagorean Theorem, since the square has a right angle at point $C$. If we let $x$ represent the length of $\overline{CE}$, then we can set up the equation $x^2 + x^2 = 1^2$, since the side length of the square is 1 unit. Solving for $x$, we get $x = \frac{\sqrt{2}}{2} \approx 0.7071$ units.
The formula for finding the length of $\overline{CE}$ in Square $ABCD$ with $45^o$ Angle is $x = \frac{\sqrt{2}}{2}$, where $x$ represents the length of $\overline{CE}$. This formula is derived from the Pythagorean Theorem, where $x$ is one of the legs of the right triangle formed by $\overline{CE}$ and the diagonal connecting points $A$ and $D$.
Yes, the length of $\overline{CE}$ in Square $ABCD$ with $45^o$ Angle is always the same, regardless of the size of the square. This is because the length of $\overline{CE}$ is determined by the angle and the side length of the square, which are constants. Therefore, the length of $\overline{CE}$ will always be $\frac{\sqrt{2}}{2}$ units.
Yes, the length of $\overline{CE}$ can still be found using the Pythagorean Theorem even if the side length of the square is not 1 unit. If we let $s$ represent the side length of the square, then the formula for finding the length of $\overline{CE}$ becomes $x = \frac{s\sqrt{2}}{2}$, where $x$ represents the length of $\overline{CE}$. This formula can be derived by setting up the equation $x^2 + x^2 = s^2$, since the diagonal of the square is equal to the side length times $\sqrt{2}$. Solving for $x$, we get $x = \frac{s\sqrt{2}}{2}$ units.