Question

1. if ( \triangle abc cong \triangle def ), ( ab = 8 ), ( bc = 19 ), ( ac = 14 ), ( ef = 4x - 1 ), and ( de = y - 6 ), find the values of ( x ) and ( y ).
2. if ( \triangle zmk cong \triangle apy ), ( mangle m = 112^{circ} ), ( mangle y = 41^{circ} ), ( mangle k = (13x - 37)^{circ} ), and ( mangle a = (2y + 7)^{circ} ), find the values of ( x ) and ( y ).
3. if ( \triangle bts cong \triangle ghd ), ( bs = 25 ), ( ts = 14 ), ( bt = 31 ), ( gd = 4x - 11 ), ( mangle s = 56^{circ} ), ( mangle b = 21^{circ} ), and ( mangle h = (7y + 5)^{circ} ), find the values of ( x ) and ( y ).

Explanation:

Step1: Use the property of congruent triangles (corresponding sides are equal) for problem 10

Since \(\triangle ABC\cong\triangle DEF\), then \(BC = EF\) and \(AB=DE\).
For \(BC = EF\), we have the equation \(19=4x - 1\).
Add \(1\) to both sides: \(19 + 1=4x-1 + 1\), so \(20 = 4x\).
Divide both sides by \(4\): \(x=\frac{20}{4}=5\).
For \(AB = DE\), we have the equation \(8=y - 6\).
Add \(6\) to both sides: \(y=8 + 6=14\).

Step2: Use the property of congruent triangles (corresponding angles are equal) and triangle - angle - sum theorem (\(m\angle A+m\angle P+m\angle Y = 180^{\circ}\)) for problem 11

Since \(\triangle ZMK\cong\triangle APY\), then \(m\angle K=m\angle Y\) and \(m\angle Z=m\angle A\).
Given \(m\angle Y = 41^{\circ}\), for \(m\angle K=m\angle Y\), we have the equation \(13x-37 = 41\).
Add \(37\) to both sides: \(13x-37 + 37=41 + 37\), so \(13x=78\).
Divide both sides by \(13\): \(x = 6\).
In \(\triangle ZMK\), \(m\angle Z=180^{\circ}-m\angle M - m\angle K=180^{\circ}-112^{\circ}-41^{\circ}=27^{\circ}\).
Since \(m\angle Z=m\angle A\), for \(m\angle A=(2y + 7)^{\circ}\), we have the equation \(2y+7 = 27\).
Subtract \(7\) from both sides: \(2y+7 - 7=27 - 7\), so \(2y=20\).
Divide both sides by \(2\): \(y = 10\).

Step3: Use the property of congruent triangles (corresponding sides and angles are equal) and triangle - angle - sum theorem (\(m\angle B+m\angle T+m\angle S = 180^{\circ}\)) for problem 12

Since \(\triangle BTS\cong\triangle GHD\), then \(TS = GD\) and \(m\angle T=m\angle H\).
Given \(TS = 14\), for \(TS = GD\), we have the equation \(4x-11 = 14\).
Add \(11\) to both sides: \(4x-11 + 11=14 + 11\), so \(4x=25\).
Divide both sides by \(4\): \(x=\frac{25}{4}=6.25\).
In \(\triangle BTS\), \(m\angle T=180^{\circ}-m\angle B - m\angle S=180^{\circ}-21^{\circ}-56^{\circ}=103^{\circ}\).
Since \(m\angle T=m\angle H\), for \(m\angle H=(7y + 5)^{\circ}\), we have the equation \(7y+5 = 103\).
Subtract \(5\) from both sides: \(7y+5 - 5=103 - 5\), so \(7y=98\).
Divide both sides by \(7\): \(y = 14\).

Answer:

1. \(x = 5\), \(y = 14\)
2. \(x = 6\), \(y = 10\)
3. \(x = 6.25\), \(y = 14\)