Question

pqrs is a parallelogram. \\( \overline { p r } \\) and \\( \overline { q s } \\) are diagonals. match each element to its value. value of a length of \\( \overline { p r } \\) value of b length of \\( \overline { q s } \\)

Explanation:

Step1: Find value of \( a \)

In a parallelogram, the diagonals bisect each other. So, \( PT = TR \). Given \( PT = a + 4 \) and \( TR = 2a \).
Set \( a + 4 = 2a \)
Subtract \( a \) from both sides: \( 4 = 2a - a \)
So, \( a = 4 \)

Step2: Find length of \( \overline{PR} \)

\( PR = PT + TR \). We know \( PT = a + 4 \) and \( TR = 2a \), and \( a = 4 \).
\( PT = 4 + 4 = 8 \), \( TR = 2\times4 = 8 \)
\( PR = 8 + 8 = 16 \)

Step3: Find value of \( b \)

In a parallelogram, the diagonals bisect each other, so \( QT = TS \). Given \( QT = b \) and \( TS = 2b - 3 \).
Set \( b = 2b - 3 \)
Subtract \( b \) from both sides: \( 0 = b - 3 \)
So, \( b = 3 \)

Step4: Find length of \( \overline{QS} \)

\( QS = QT + TS \). We know \( QT = b = 3 \) and \( TS = 2b - 3 = 2\times3 - 3 = 3 \)
\( QS = 3 + 3 = 6 \)

Answer:

• value of \( a \): \( 4 \)
• length of \( \overline{PR} \): \( 16 \)
• value of \( b \): \( 3 \)
• length of \( \overline{QS} \): \( 6 \)

So the matches are:
16 → length of \( \overline{PR} \)
4 → value of \( a \)
3 → value of \( b \)
6 → length of \( \overline{QS} \)