$M$ is the midpoint of $\overline{AB}$ and $N$ is the midpoint of $\overline{AC}$, and $T$ is the intersection of $\overline{BN}$ and $\overline{CM}$, as shown. If $\overline{BN}\perp\overline{AC}$, $BN = 12$, and $AC = 14$, then find $CT$.

Using 2:1 ratios for medians of a triangle:

Line Segment NT is 4
Since we know line segment AC is 14 and n is the midpoint, line segment NC is 7.
Using Pythagorean theorem
7^2+4^2=x^2
49+16=x^2
x=Square root of 65

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Well, to find $CT$, we can start by using the Pythagorean theorem to find the length of $\overline{AC}$.

Since $M$ is the midpoint of $\overline{AB}$, and $N$ is the midpoint of $\overline{AC}$, we know that $AM = MB$ and $AN = NC$.

Using this information, we can solve for the length of $\overline{AC}$:

$AC = AM + MC = MB + NC = BN + NC = 12 + NC$.

But we're also given that $AC = 14$, so we can solve for $NC$:

$14 = 12 + NC \Rightarrow NC = 2$.

Now, we can use the fact that $\overline{BN} \perp \overline{AC}$ to find $CT$.

Since $BN \perp AC$, we can use the Pythagorean theorem on right triangle $BNC$:

$BN^2 + NC^2 = BC^2 \Rightarrow 12^2 + 2^2 = BC^2 \Rightarrow 144 + 4 = BC^2 \Rightarrow 148 = BC^2$.

Therefore, $BC = \sqrt{148} = 2\sqrt{37}$.

Finally, we can solve for $CT$ using right triangle $CTM$:

$CT^2 + TM^2 = CM^2 \Rightarrow CT^2 + \left(\frac{BC}{2}\right)^2 = \left(\frac{AC}{2}\right)^2 \Rightarrow CT^2 + \left(\frac{2\sqrt{37}}{2}\right)^2 = \left(\frac{14}{2}\right)^2 \Rightarrow CT^2 + 37 = 49 \Rightarrow CT^2 = 12$.

Taking the square root of both sides, we find $CT = \sqrt{12} = 2\sqrt{3}$.

So, $CT$ is equal to $2\sqrt{3}$. Now that we're done with math, can I tell you a joke?

To find $CT$, we can first observe that a triangle is formed by the points $C$, $M$, and $T$. We know that $M$ is the midpoint of $\overline{AB}$, so we can use the midpoint formula to find the coordinates of $M$. Let's assume $A$ has coordinates $(x_1, y_1)$ and $B$ has coordinates $(x_2, y_2)$.

The midpoint formula states that the coordinates of the midpoint of a line segment with endpoints $(x_1, y_1)$ and $(x_2, y_2)$ are given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

So in this case, the coordinates of $M$ would be $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.

Similarly, since $N$ is the midpoint of $\overline{AC}$, we can find the coordinates of $N$ using the midpoint formula. Let $A$ have coordinates $(x_1, y_1)$ and $C$ have coordinates $(x_3, y_3)$. Then the coordinates of $N$ would be $\left(\frac{x_1+x_3}{2}, \frac{y_1+y_3}{2}\right)$.

Now, we know that $BN$ is perpendicular to $AC$, so the slope of $BN$ multiplied by the slope of $AC$ should equal $-1$. The slope of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y_2-y_1}{x_2-x_1}$.

We can find the slope of $BN$ by finding the slope of the line passing through the points $B$ and $N$ (which we can find using the coordinates of $B$ and $N$), and we can find the slope of $AC$ by finding the slope of the line passing through the points $A$ and $C$ (which we can find using the coordinates of $A$ and $C$).

Since $BN$ is perpendicular to $AC$, we can set up the equation $\left(\frac{y_N-y_B}{x_N-x_B}\right) \cdot \left(\frac{y_C-y_A}{x_C-x_A}\right) = -1$.

Simplifying this equation will give us a relationship between the coordinates of $B$, $N$, $A$, and $C$.

Finally, we can use the equation of the line passing through points $M$ and $T$ to find the value of $CT$. The equation of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$ can be found using the point-slope form: $y-y_1 = \frac{y_2-y_1}{x_2-x_1}(x-x_1)$.

Since we know the coordinates of $M$ and $T$, we can use this equation to find the equation of the line passing through these two points. By substituting the coordinates of $C$ into this equation, we can find the value of $CT$.

This may seem like a lot of steps, but we can use these techniques to find $CT$.

Hey, helpful hint- use the 2:1 ratio!

And true, don’t post your hw problems

dont post these pls

DONT ASK QUESTIONS ASSIGNED IN CLASS