Question

4.
x =
pr =
ts =
if sq=11x-26 & pr=5x+28, find pr and ts.
show your work
show your work for credit!

Explanation:

Step1: Identify property of rectangle diagonals

In a rectangle, diagonals are equal and bisect each other. So \( SQ = PR \) (diagonals of rectangle are equal) and \( TS=\frac{1}{2}SQ \) (diagonals bisect each other). Given \( SQ = 11x - 26 \) and \( PR=5x + 28 \), set \( SQ=PR \):
\( 11x-26=5x + 28 \)

Step2: Solve for \( x \)

Subtract \( 5x \) from both sides:
\( 11x-5x-26=5x-5x + 28 \)
\( 6x-26 = 28 \)
Add 26 to both sides:
\( 6x-26 + 26=28 + 26 \)
\( 6x=54 \)
Divide by 6:
\( x=\frac{54}{6}=9 \)

Step3: Find \( PR \)

Substitute \( x = 9 \) into \( PR = 5x+28 \):
\( PR=5(9)+28=45 + 28=73 \)

Step4: Find \( SQ \) (to find \( TS \))

Substitute \( x = 9 \) into \( SQ=11x - 26 \):
\( SQ=11(9)-26=99 - 26=73 \)
Since \( TS=\frac{1}{2}SQ \) (diagonals bisect each other), \( TS=\frac{73}{2}=36.5 \)

Answer:

\( x = 9 \)
\( PR = 73 \)
\( TS = 36.5 \) (or \( \frac{73}{2} \))