A) Construction of PQR and SQR:

Here are the solutions for both parts of question 13:

a) Construction of $\triangle PQR$ and $\triangle SQR$:

Steps for Construction:

1. Draw a line segment $PQ = 7$ cm.
2. At point $Q$, use a protractor to construct an angle of $120^\circ$. Let this ray be $QX$.
3. Along the ray $QX$, measure and mark a point $R$ such that $QR = 5$ cm.
4. Join $P$ and $R$ to complete $\triangle PQR$.

Steps to construct $\triangle SQR$ with area equal to $\triangle PQR$:

1. Draw a line through point $P$ parallel to the base $QR$. To do this:
   • Draw an arc with center $Q$ and radius $QP$.
   • Draw an arc with center $R$ and radius $RP$.
   • Alternatively, draw a line through $P$ parallel to $QR$ by constructing an angle equal to $\angle PQR$ at $P$ on the opposite side of $PQ$, or by using a compass to transfer the distance between $P$ and $QR$.
2. Extend the line segment $RQ$ to a suitable length.
3. Choose any point $S$ on the line drawn parallel to $QR$ (from step 1) such that $S$ is not $P$ and $S$ is not on the line $QR$. For example, you can extend the parallel line and pick a point $S$ on it.
4. Join $S$ to $Q$ and $R$.
5. $\triangle SQR$ is the required triangle.

Reason: Triangles on the same base (or equal bases) and between the same parallel lines have equal areas. Here, $\triangle PQR$ and $\triangle SQR$ share the same base $QR$ and their vertices $P$ and $S$ lie on a line parallel to $QR$. Therefore, Area($\triangle PQR$) = Area($\triangle SQR$).

b) Relationship between $\triangle ABC$ and $\triangle BDE$:

Given:

• $BC = CD$
• $AE \parallel BD$

Relationship:
Area($\triangle ABC$) = $\frac{1}{2}$ Area($\triangle BDE$)

Reason:

1. Since $BC = CD$, point $C$ is the midpoint of the line segment $BD$.
2. In $\triangle ABD$, $AC$ is a median to the side $BD$. A median divides a triangle into two triangles of equal area. Therefore, Area($\triangle ABC$) = Area($\triangle ACD$). This implies Area($\triangle ABD$) = Area($\triangle ABC$) + Area($\triangle ACD$) = 2 $\times$ Area($\triangle ABC$).
3. Given that $AE \parallel BD$.
4. Consider $\triangle ABD$ and $\triangle EBD$. They share the same base $BD$.
5. Since their vertices $A$ and $E$ lie on a line parallel to their common base $BD$, their areas are equal. Therefore, Area($\triangle ABD$) = Area($\triangle EBD$).
6. From steps 2 and 5, we can conclude: 2 $\times$ Area($\triangle ABC$) = Area($\triangle EBD$) Area($\triangle ABC$) = $\frac{1}{2}$ Area($\triangle EBD$)

The final answer is $\boxed{\text{Area}(\triangle ABC) = \frac{1}{2} \text{Area}(\triangle BDE)}$.