如图所示为两个全等的等腰三角形 PQS 和 RSQ。 PQ 和 RS 平行于 x 轴。

1. Use the horizontal sides and congruence to determine the coordinates of $Q$.

Since $PQ$ is parallel to the $x$-axis and $P=(-2,1)$, point $Q$ has $y$-coordinate $1$.

Since $S$ is on the $y$-axis and $RS$ is horizontal at $y=4$, $S=(0,4)$.

So
$RS=13-0=13.$

The triangles are congruent, so the corresponding horizontal sides are equal:
$PQ=RS=13.$

Starting from $P(-2,1)$ and moving $13$ units to the right,
$x_Q=-2+13=11.$

Hence $Q=(11,1)$.

1. Find the gradient of $PR$ using $P(-2,1)$ and $R(13,4)$.

$m=\frac{4-1}{13-(-2)}=\frac{3}{15}=\frac{1}{5}$

2. Substitute the gradient and one point into $y=mx+c$ to find the equation.

Using $y=\frac{1}{5}x+c$ and point $P(-2,1)$,
$1=\frac{1}{5}(-2)+c$
$c=1+\frac{2}{5}=\frac{7}{5}$
So the equation is
$y=\frac{1}{5}x+\frac{7}{5}.$

1. Identify the point on the $x$-axis vertically below $R$.

Since $RS$ is parallel to the $x$-axis, $RS$ is horizontal.

Therefore $RW$ is vertical. Point $R$ is $(13,4)$, so a vertical line through $R$ has $x=13$.

Since $W$ is on the $x$-axis, its $y$-coordinate is $0$.

Hence $W=(13,0)$.

1. Find the two horizontal side lengths $PQ$ and $RS$.

Using $P(-2,1)$, $Q(11,1)$, $R(13,4)$ and $S(0,4)$,

$PQ = 11-(-2)=13$

$RS = 13-0=13$

2. Find one slanted side length using Pythagoras' theorem.

For $PS$,

horizontal difference $=0-(-2)=2$

vertical difference $=4-1=3$

$PS=\sqrt{2^2+3^2}=\sqrt{13}$

Since the triangles are congruent, $QR=PS=\sqrt{13}$.

3. Add all four side lengths to find the perimeter of $PQRS$.

$\text{Perimeter}=PQ+QR+RS+SP$

$=13+\sqrt{13}+13+\sqrt{13}$

$=26+2\sqrt{13}=33.211\ldots$

$\approx 34$