Similarity

In the figure, $\;$ $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\triangle TSR\right)} = \dfrac{3}{1}$, $\;$ $TS \parallel QR$.

Find

1. $PS : SR$
2. $TS : QR$
3. $\text{area } \left(\triangle PTS\right) : \text{area } \left(\triangle PQR\right)$
4. $\text{area } \left(\triangle PTS\right) : \text{area} \left(\text{quadrilateral } TQRS\right)$

1. Let height of $\triangle PTS = $ height of $\triangle TSR = h$
   
   From the figure,
   
   $\text{area} \left(\triangle PTS\right) = \dfrac{1}{2} \times PS \times h$
   
   $\text{area} \left(\triangle TSR\right) = \dfrac{1}{2} \times SR \times h$
   
   Given: $\;$ $\text{area } \left(\triangle PTS\right) : \text{area } \left(\triangle TSR\right) = 3 : 1$
   
   i.e. $\;$ $\dfrac{1}{2} \times PS \times h: \dfrac{1}{2} \times SR \times h = 3 : 1$
   
   i.e. $\;$ $PS : SR = 3 : 1$



2. Given: $\;$ $TS \parallel QR$
   
   $PR$ is a transversal
   
   $\therefore \;$ $\angle PST = \angle PRQ$ $\;\;\;$ [corresponding angles] $\;\;\; \cdots \; (1)$
   
   In $\triangle PTS$ and $\triangle PQR$,
   
   $\angle P = \angle P$ $\;\;\;$ [common angle]
   
   $\angle PST = \angle PRQ$ $\;\;\;$ [by (1)]
   
   $\therefore \;$ $\triangle PTS \sim \triangle PQR$ $\;\;\;$ [by AA axiom of similarity]
   
   $\implies$ $\dfrac{TS}{QR}= \dfrac{PS}{PR} = \dfrac{PT}{PQ}$ $\;\;\; \cdots (2)$
   
   $\;\;\;$ [corresponding sides of similar triangles are in proportion]
   
   $\because \;$ $\dfrac{PS}{SR} = \dfrac{3}{1}$ $\;\;\;$ [part (a)]
   
   $\therefore \;$ $\dfrac{PS}{PR} = \dfrac{PS}{PS + SR} = \dfrac{3}{3 + 1} = \dfrac{3}{4}$ $\;\;\; \cdots \; (3)$
   
   In view of equation $(3)$, equation $(2)$ becomes
   
   $TS : QR = 3 : 4$



3. $\because \;$ $\triangle PTS \sim \triangle PQR$
   
   $\implies$ $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \triangle PQR} = \dfrac{TS^2}{QR^2} = \dfrac{PS^2}{PR^2} = \dfrac{PT^2}{PQ^2}$
   
   [areas of two similar triangles are proportional to the squares of their corresponding sides]
   
   $\therefore \;$ $\text{area } \left(\triangle PTS\right) : \text{area } \left(\triangle PQR\right) = 3^2 : 4^2 = 9 : 16$



4. $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\text{quadrilateral } TQRS\right)} = \dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\triangle PQR\right) - \text{area } \left(\triangle PTS\right)}$
   
   i.e. $\;$ $\dfrac{\text{area } \left(\triangle PTS\right)}{\text{area } \left(\text{quadrilateral } TQRS\right)} = \dfrac{9}{16 - 9}$
   
   $\therefore \;$ $\text{area } \left(\triangle PTS\right) : \text{area} \left(\text{quadrilateral } TQRS\right) = 9 : 7$