Question

1. given $delta mtwcongdelta bgk$, find the values of $x$ and $y$.
2. given $delta prscongdelta cfh$, find the values of $x$, $y$, and $z$.
3. if $delta abccongdelta def$, $ab = 8$, $bc = 19$, $ac = 14$, $ef = 4x - 1$, and $de = y - 6$, find the values of $x$ and $y$.
4. if $delta zmkcongdelta apy$, $mangle m=112^{circ}$, $mangle y = 41^{circ}$, $mangle k=(13x - 37)^{circ}$, and $mangle a=(2y + 7)^{circ}$, find the values of $x$ and $y$.
5. if $delta btscongdelta ghd$, $bs = 25$, $ts = 14$, $bt = 31$, $gd = 4x - 11$, $mangle s = 56^{circ}$, $mangle b = 21^{circ}$, and $mangle h=(7y + 5)^{circ}$, find the values of $x$ and $y$.

Explanation:

Step1: Recall congruent - triangle property

Corresponding angles and sides of congruent triangles are equal.

Step2: Solve for \(x\) in problem 8

In \(\triangle MTW\cong\triangle BGK\), \(\angle T\) and \(\angle G\) are corresponding angles. So \(4x - 3=45\).
Add 3 to both sides: \(4x=45 + 3=48\).
Divide both sides by 4: \(x = \frac{48}{4}=12\).

Step3: Solve for \(y\) in problem 8

\(\angle W\) and \(\angle K\) are corresponding angles. In \(\triangle MTW\), \(\angle M = 180-(4x - 3)-41\), and in \(\triangle BGK\), \(\angle B=180 - 45-(11y + 6)\). Since the triangles are congruent, and we already found \(x = 12\), \(\angle W = 41^{\circ}\), \(\angle K\) is also \(41^{\circ}\). So \(11y+6 = 41\).
Subtract 6 from both sides: \(11y=41 - 6 = 35\), \(y=\frac{35}{11}\).

Step4: Solve for \(y\) in problem 9

In \(\triangle PRS\cong\triangle CFH\), \(\angle R\) and \(\angle C\) are corresponding angles. So \(13y-1 = 90\) (since \(\angle C\) is a right - angle).
Add 1 to both sides: \(13y=90 + 1=91\).
Divide both sides by 13: \(y = 7\).

Step5: Solve for \(x\) in problem 9

\(PS\) and \(FH\) are corresponding sides. So \(2x-7 = 39\).
Add 7 to both sides: \(2x=39 + 7=46\).
Divide both sides by 2: \(x = 23\).

Step6: Solve for \(z\) in problem 9

\(\angle S\) and \(\angle H\) are corresponding angles. So \(6z-4=28\).
Add 4 to both sides: \(6z=28 + 4=32\).
Divide both sides by 6: \(z=\frac{32}{6}=\frac{16}{3}\).

Step7: Solve for \(x\) in problem 10

Since \(\triangle ABC\cong\triangle DEF\), \(BC = EF\). So \(19=4x - 1\).
Add 1 to both sides: \(4x=19 + 1=20\).
Divide both sides by 4: \(x = 5\).

Step8: Solve for \(y\) in problem 10

\(AB = DE\). So \(8=y - 6\).
Add 6 to both sides: \(y=8 + 6=14\).

Step9: Solve for \(x\) in problem 11

Since \(\triangle ZMK\cong\triangle APY\), \(\angle M\) and \(\angle P\) are corresponding angles. \(\angle M = 112^{\circ}\), and \(\angle P\) is also \(112^{\circ}\). \(\angle K\) and \(\angle Y\) are corresponding angles. So \(13x-37 = 41\).
Add 37 to both sides: \(13x=41 + 37=78\).
Divide both sides by 13: \(x = 6\).

Step10: Solve for \(y\) in problem 11

\(\angle A\) and \(\angle Z\) are corresponding angles. In \(\triangle ZMK\), \(\angle Z=180 - 112-(13x - 37)\) (with \(x = 6\)), \(\angle A = 2y+7\). First, in \(\triangle ZMK\) with \(x = 6\), \(\angle K = 41^{\circ}\), \(\angle M = 112^{\circ}\), so \(\angle Z=180-112 - 41=27^{\circ}\). Then \(2y+7 = 27\).
Subtract 7 from both sides: \(2y=27 - 7=20\).
Divide both sides by 2: \(y = 10\).

Step11: Solve for \(x\) in problem 12

Since \(\triangle BTS\cong\triangle GHD\), \(BT = GD\). So \(31=4x-11\).
Add 11 to both sides: \(4x=31 + 11=42\).
Divide both sides by 4: \(x=\frac{42}{4}=\frac{21}{2}\).

Step12: Solve for \(y\) in problem 12

\(\angle S\) and \(\angle D\) are corresponding angles. \(\angle H\) and \(\angle T\) are corresponding angles. In \(\triangle BTS\), we know \(\angle S = 56^{\circ}\), \(\angle B = 21^{\circ}\), so \(\angle T=180-(56 + 21)=103^{\circ}\). Then \(7y+5 = 103\).
Subtract 5 from both sides: \(7y=103 - 5=98\).
Divide both sides by 7: \(y = 14\).

Answer:

Problem 8: \(x = 12\), \(y=\frac{35}{11}\)
Problem 9: \(x = 23\), \(y = 7\), \(z=\frac{16}{3}\)
Problem 10: \(x = 5\), \(y = 14\)
Problem 11: \(x = 6\), \(y = 10\)
Problem 12: \(x=\frac{21}{2}\), \(y = 14\)