Question

if $qr = 10z - 44$, $ps = -5z + 85$, and $tu = 11z - 34$, what is $tu$?
$tu = \square$

Explanation:

Step1: Identify the relationship

From the diagram, \( PS = QR \) (since the segments are marked as equal, and the figure suggests a midline or equal - length segment relationship in a trapezoid - like figure). So we set \( 10z - 44=-5z + 85 \).

Step2: Solve for z

Add \( 5z \) to both sides: \( 10z+5z - 44=-5z + 5z+85 \), which simplifies to \( 15z-44 = 85 \).
Add 44 to both sides: \( 15z-44 + 44=85 + 44 \), so \( 15z=129 \).
Divide both sides by 15: \( z=\frac{129}{15}=\frac{43}{5} = 8.6 \)? Wait, no, wait. Wait, \( 85 + 44=129 \)? Wait, \( 85+44 = 129 \), but \( 129\div15 = 8.6 \)? Wait, no, maybe I made a mistake. Wait, \( 10z-44=-5z + 85 \). Let's re - do:

\( 10z+5z=85 + 44 \)

\( 15z=129 \)? Wait, \( 85 + 44=129 \), but \( 129\div15 = 8.6 \). Wait, but then let's check the other relationship. Wait, also, \( TU=QR + PS \)? No, from the diagram, the midline (or the segment \( TU \)) should be related to \( QR \) and \( PS \). Wait, actually, in a trapezoid, the midline length is the average of the two bases. Wait, if \( PS \) and \( QR \) are the two bases, and \( TU \) is the midline, then \( TU=\frac{PS + QR}{2} \). But we also know that \( PS = QR \) (from the segment markings). Wait, the markings on \( UP = PQ \) and \( TS=SR \), so \( PS \) is the midline between \( QR \) and \( TU \)? Wait, no, maybe \( PS \) is parallel to \( QR \) and \( TU \), and \( P \) and \( S \) are mid - points. So \( PS=\frac{QR + TU}{2} \)? Wait, I think I messed up the first relationship.

Wait, let's look at the segment markings: \( UP = PQ \) (two marks) and \( TS = SR \) (two marks). So \( P \) is the mid - point of \( UQ \) and \( S \) is the mid - point of \( TR \). So by the midline theorem for trapezoids (or the midline of a trapezoid: the segment connecting the mid - points of the non - parallel sides is parallel to the two bases and its length is the average of the lengths of the two bases). So here, the two bases are \( QR \) and \( TU \), and the midline is \( PS \). So the formula is \( PS=\frac{QR + TU}{2} \).

We know that \( QR = 10z - 44 \), \( PS=-5z + 85 \), \( TU = 11z-34 \).

So \( -5z + 85=\frac{(10z - 44)+(11z-34)}{2} \)

Multiply both sides by 2: \( 2(-5z + 85)=(10z - 44)+(11z-34) \)

\( -10z + 170=10z+11z-44 - 34 \)

\( -10z + 170=21z-78 \)

Add \( 10z \) to both sides: \( 170=31z-78 \)

Add 78 to both sides: \( 170 + 78=31z \)

\( 248 = 31z \)

Divide both sides by 31: \( z = 8 \)

Ah, that's better. I made a mistake in the initial relationship. So \( z = 8 \).

Step3: Calculate TU

Now that we have \( z = 8 \), substitute into the formula for \( TU \): \( TU=11z-34 \)

\( TU=11\times8-34 \)

\( TU = 88-34 \)

\( TU = 54 \)

Answer:

\( 54 \)