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