Question

find the value of x in parallelogram qrst.
q
6x+29
t
r
2x+33
s
x =
submit

Explanation:

Step1: Recall parallelogram side property

In a parallelogram, opposite sides are equal. So \(QR = ST\) and \(QT = RS\). Here, \(QT = 6x + 29\) and \(RS = 2x + 33\), so set them equal: \(6x + 29 = 2x + 33\).

Step2: Solve for x

Subtract \(2x\) from both sides: \(6x - 2x + 29 = 33\) → \(4x + 29 = 33\).
Subtract 29 from both sides: \(4x = 33 - 29\) → \(4x = 4\).
Divide by 4: \(x = \frac{4}{4} = 1\).

Answer:

\(x = 1\)