Question

1. if s is the mid - point of rt, rs = 5x + 17, and st = 8x - 31, find rs.
2. if line y bisects ac, ab = 4 - 5x, and bc = 2x + 25, find ac.
3. if m∠def = 117°, find the value of x.
4. if m∠pqs = 16°, m∠sqr=(9x + 17)°, and m∠pqr=(12x - 6)°, find m∠pqr.

Explanation:

Step1: Use mid - point property for question 11

Since \(S\) is the mid - point of \(\overline{RT}\), \(RS = ST\). So, \(5x+17 = 8x - 31\).

Step2: Solve for \(x\)

Subtract \(5x\) from both sides: \(17=8x - 5x-31\), which simplifies to \(17 = 3x-31\). Then add 31 to both sides: \(17 + 31=3x\), so \(48 = 3x\). Divide both sides by 3, \(x = 16\).

Step3: Find \(RS\)

Substitute \(x = 16\) into the expression for \(RS\): \(RS=5x + 17=5\times16+17=80 + 17=97\).

Step4: Use angle - bisector property for question 12

Since line \(y\) bisects \(\overline{AC}\), \(AB = BC\). So, \(4-5x=2x + 25\).

Step5: Solve for \(x\)

Add \(5x\) to both sides: \(4=2x+5x + 25\), which simplifies to \(4=7x + 25\). Subtract 25 from both sides: \(4-25=7x\), so \(- 21=7x\). Divide both sides by 7, \(x=-3\).

Step6: Find \(AC\)

\(AC=AB + BC\). Since \(AB = BC\), \(AC = 2AB\). Substitute \(x=-3\) into the expression for \(AB\): \(AB=4-5\times(-3)=4 + 15 = 19\). So, \(AC = 2\times19=38\).

Step7: Use angle - sum property for question 20

\(m\angle DEF=m\angle DEG+m\angle GEF\). So, \(117=(12x + 1)+(5x-3)\).

Step8: Simplify the equation

Combine like - terms: \(117=12x+5x+1 - 3\), which simplifies to \(117 = 17x-2\). Add 2 to both sides: \(117 + 2=17x\), so \(119 = 17x\). Divide both sides by 17, \(x = 7\).

Step9: Use angle - sum property for question 21

\(m\angle PQR=m\angle PQS+m\angle SQR\). So, \(12x-6=16+(9x + 17)\).

Step10: Simplify the equation

\(12x-6=16 + 9x+17\), which simplifies to \(12x-6=9x + 33\). Subtract \(9x\) from both sides: \(12x-9x-6=33\), so \(3x-6=33\). Add 6 to both sides: \(3x=33 + 6=39\). Divide both sides by 3, \(x = 13\).

Step11: Find \(m\angle PQR\)

Substitute \(x = 13\) into the expression for \(m\angle PQR\): \(m\angle PQR=12x-6=12\times13-6=156 - 6=150^{\circ}\)

Answer:

1. \(RS = 97\)
2. \(AC = 38\)
3. \(x = 7\)
4. \(m\angle PQR=150^{\circ}\)