Question

if $\triangle qrscong\triangle mnp$, $qr = 11$, $qs = 15$, $mp = 2x + 1$, $mangle s=84^{circ}$, $mangle r = 32^{circ}$, and $mangle m=(17y - 4)^{circ}$, find the values of $x$ and $y$.

Explanation:

Step1: Use corresponding - parts of congruent triangles are equal (CPCTC) for sides

Since $\triangle QRS\cong\triangle MNP$, corresponding sides are equal. So $MP = QS$.
Given $QS = 15$ and $MP=2x + 1$, we set up the equation $2x+1 = 15$.
Subtract 1 from both sides: $2x=15 - 1=14$.
Divide both sides by 2: $x=\frac{14}{2}=7$.

Step2: Use corresponding - parts of congruent triangles are equal (CPCTC) for angles

Since $\triangle QRS\cong\triangle MNP$, corresponding angles are equal. So $\angle M=\angle Q$.
We know that in a triangle, the sum of interior angles of a triangle is $180^{\circ}$, but we can also use the fact that $\angle M=\angle Q$. And since $\triangle QRS\cong\triangle MNP$, $\angle M=\angle Q$, $\angle N=\angle R$, $\angle P=\angle S$.
We are given $\angle S = 84^{\circ}$, $\angle R = 32^{\circ}$, and $\angle M=(17y - 4)^{\circ}$.
First, find $\angle Q$ in $\triangle QRS$: $\angle Q=180^{\circ}-\angle R-\angle S=180^{\circ}-32^{\circ}-84^{\circ}=64^{\circ}$.
Then, since $\angle M=\angle Q$, we set up the equation $17y - 4=64$.
Add 4 to both sides: $17y=64 + 4=68$.
Divide both sides by 17: $y=\frac{68}{17}=4$.

Answer:

$x = 7$, $y = 4$