本题不接受通过精确作图得出的解答。 图中显示了一个等腰三角形 PQR，其中 $PQ = QR$。三角形的顶点为 P(k, 4)、Q(5, 3) 和 R(9, 10)。

1. Calculate the square of the length of QR

$QR^2 = (9 - 5)^2 + (10 - 3)^2$
$QR^2 = 4^2 + 7^2 = 16 + 49 = 65$

2. Express the square of the length of PQ in terms of k and equate to $QR^2$

Since $PQ = QR$, $PQ^2 = QR^2$.
$PQ^2 = (k - 5)^2 + (4 - 3)^2$
$(k - 5)^2 + 1^2 = 65$

3. Solve the quadratic equation for k

$(k - 5)^2 = 64$
$k - 5 = \pm 8$
$k = 5 + 8 = 13 \text{ or } k = 5 - 8 = -3$

4. Determine the correct value of k based on the diagram

From the diagram, point P is to the left of the y-axis, so its x-coordinate must be negative ($k < 0$).
Therefore, $k = -3$.

1. Find the coordinates of P from part (i) using the given $PQ = QR$.

Using the distance formula for $PQ^2 = QR^2$:
$(5 - k)^2 + (3 - 4)^2 = (9 - 5)^2 + (10 - 3)^2$
$(5 - k)^2 + 1 = 16 + 49$
$(5 - k)^2 = 64$
$5 - k = \pm 8$
$k = -3 \text{ or } k = 13$
From the diagram, P is to the left of the y-axis, so $k < 0$. Thus, $k = -3$ and $P$ is $(-3, 4)$.

2. Set up an equation for the coordinates of S using $PS = SR$.

Since S lies on the y-axis, let its coordinates be $(0, y)$.
Given $PS = SR$, we have $PS^2 = SR^2$:
$(0 - (-3))^2 + (y - 4)^2 = (0 - 9)^2 + (y - 10)^2$
$9 + y^2 - 8y + 16 = 81 + y^2 - 20y + 100$

3. Solve for y to find the coordinates of S.

$25 - 8y = 181 - 20y$
$12y = 156$
$y = 13$
The coordinates of S are $(0, 13)$.

4. Find the equation of the line QS.

Using $Q(5, 3)$ and $S(0, 13)$:
Gradient of QS, $m = \frac{13 - 3}{0 - 5} = \frac{10}{-5} = -2$
Since S is the y-intercept, the equation of QS is:
$y = -2x + 13$

1. Find the coordinates of P and S from earlier parts

From part (i), $P$ is $(-3, 4)$.
From part (ii), $S$ is $(0, 13)$.
The coordinates of the vertices are $P(-3, 4)$, $Q(5, 3)$, $R(9, 10)$ and $S(0, 13)$.

2. Calculate the area of triangle PQR

Using the shoelace formula for $\triangle PQR$ with vertices $P(-3, 4)$, $Q(5, 3)$, $R(9, 10)$:
$\text{Area} = \frac{1}{2} |(-3)(3) + (5)(10) + (9)(4) - (4)(5) - (3)(9) - (10)(-3)|$
$\text{Area} = \frac{1}{2} |-9 + 50 + 36 - 20 - 27 + 30|$
$\text{Area} = \frac{1}{2} |60| = 30 \text{ units}^2$

3. Calculate the area of quadrilateral PQRS

Using the shoelace formula for quadrilateral $PQRS$ with vertices $P(-3, 4)$, $Q(5, 3)$, $R(9, 10)$, $S(0, 13)$:
$\text{Area} = \frac{1}{2} |(-3)(3) + (5)(10) + (9)(13) + (0)(4) - (4)(5) - (3)(9) - (10)(0) - (13)(-3)|$
$\text{Area} = \frac{1}{2} |-9 + 50 + 117 + 0 - 20 - 27 - 0 + 39|$
$\text{Area} = \frac{1}{2} |150| = 75 \text{ units}^2$

4. Find the ratio of the areas

$\text{Ratio} = \text{Area of } \triangle PQR : \text{Area of } PQRS$
$\text{Ratio} = 30 : 75$
Dividing by 15:
$\text{Ratio} = 2 : 5$