Question

ex 3
if mh = 8, find the length of ah.
$8^{2}+8^{2}=c^{2}$

Explanation:

Step1: Identify square side length

Given $MH = 8$, and $MH$ is a side of square $MHAT$, so all sides are 8: $MH=HA=AT=TM=8$.

Step2: Use Pythagorean theorem

For right triangle $MHA$, $AH$ is a side, wait correction: $AH$ is a side, no—wait, $MH$ is side, $AH$ is a side? No, wait square $MHAT$: vertices $M, H, A, T$, so sides $MH, HA, AT, TM$. Wait no, diagonals are $MA$ and $HT$. Wait the problem says $MH=8$, find $AH$. Wait in square, all sides are equal. Wait no, wait the figure: $MH$ is a side, $AH$ is a side? No, wait no—wait, no, in square $MHAT$, $MH$ and $AH$ are adjacent sides, so they are equal? No, wait no, the diagonal is $HT$? Wait no, the green line is $AH$? Wait no, the problem says "find the length of $AH$", $MH=8$. Wait, no, $MHAT$ is a square, so $MH$ is a side length of 8, and $AH$ is also a side? That can't be. Wait no, I misread: $MH$ is a side, $AH$ is a diagonal? Wait no, the vertices: $M$ connected to $H$ and $A$, $H$ connected to $T$ and $A$, $A$ connected to $T$, $T$ connected to $M$. So $MH$ is a side, $AH$ is a side? No, that would make $MH$ and $AH$ adjacent sides of the square, so $AH=MH=8$. But the blue writing has $8^2+8^2=c^2$, which is for diagonal. Wait, maybe $MH$ is half the diagonal? No, the problem says $MH=8$. Wait, no, in a square, diagonals bisect each other, so if $S$ is the intersection, $MS=SH=AS=ST$. Wait, if $MH=8$ is a side, then diagonal $AH$? No, $AH$ is a side. Wait, no, let's reorient: square $MHAT$, so sides are $MH, HT, TA, AM$. Oh! That's the mistake. So $MH$ is a side, $HT$ is a side, $TA$ is a side, $AM$ is a side. Then $AH$ is a diagonal of the square.

Corrected Step1: Confirm square side

Square $MHAT$ has side length $MH = 8$.

Corrected Step2: Calculate diagonal length

For square diagonal $d = s\sqrt{2}$, where $s$ is side length. Here, $AH$ is the diagonal, so:
$$AH = 8\sqrt{2}$$
Wait no, wait if $MH$ is a side, then the diagonal from $A$ to $H$: in square $MHAT$, vertices $M, H, T, A$, so the sides are $MH$, $HT$, $TA$, $AM$. So $MH=8$, right angle at $H$, so triangle $MHA$ is right triangle with legs $MH=8$ and $HA=8$? No, no, $HA$ would be a side. I'm confused. Wait the problem says "If $MH=8$, find the length of $AH$." The figure is a square, so $MH$ and $AH$ are adjacent sides, so they are equal? But the blue writing has Pythagoras, so maybe $MH$ is half the diagonal? No, the problem says $MH=8$. Wait, no, maybe $MH$ is a diagonal? No, the figure shows $M$ connected to $H$, $A$, $T$; $H$ connected to $A$, $T$; $A$ connected to $T$. So $MHAT$ is a square, so $MH=HA=AT=TM=8$, so $AH=8$. But the blue writing is $8^2+8^2=c^2$, which is $c=8\sqrt{2}$. So maybe $MH$ is a side, and $AH$ is the diagonal. That must be it: the square has side $MH=8$, and $AH$ is the diagonal connecting $A$ to $H$, crossing the other diagonal $MT$ at $S$.

Final Step1: Define square side

Let square side $s = MH = 8$.

Final Step2: Compute diagonal $AH$

Use Pythagorean theorem on right $\triangle MHA$:
$$AH = \sqrt{MH^2 + MA^2}$$
Wait no, $MH$ and $MA$ are sides? No, $MH$ and $MH$? No, in square, adjacent sides are equal, so legs are both 8:
$$AH = \sqrt{8^2 + 8^2} = \sqrt{64+64} = \sqrt{128} = 8\sqrt{2}$$

Answer:

$8\sqrt{2}$