GATE 2025 IN GA Q.9: solution and answer

Official answer: DSource: Official answer key, GATE 2025 IN (IIT Roorkee)

1. Join $S$ to $L$ and look at $△PLS$. Both the rectangle and the square can be tied to the area of this one triangle.

2. Using the rectangle: take $PL$ as the base. $S$ lies on $MN$. In a rectangle $MN\parallel PL$, and the distance between these two lines is $LM$, because $LM\perp PL$. So the height from $S$ to $PL$ is $LM$, and $$\text{area}(△PLS)=\frac{1}{2}PL\cdot LM=\frac{1}{2}\text{area}(PLMN).$$ (A triangle standing on one side of a parallelogram, with its apex on the opposite side, has half the parallelogram's area.)

3. Using the square: take $PS=2\text{cm}$ as the base. $L$ lies on $QR$, and $QR\parallel PS$ at a distance of one side, $2\text{cm}$. So the height from $L$ to $PS$ is $2\text{cm}$, and $$\text{area}(△PLS)=\frac{1}{2}\times 2\times 2=2{\text{cm}}^2.$$

4. Setting the two expressions equal gives $\frac{1}{2}\text{area}(PLMN)=2$, so $$\text{area}(PLMN)=4{\text{cm}}^2,$$ wherever $L$ sits on $QR$.

5. Check with coordinates: put $Q=(0,0)$, $R=(2,0)$, $S=(2,2)$, $P=(0,2)$ and $L=(t,0)$. Then $PL=\sqrt{t^2+4}$. The line $PL$ is $2x+t(y-2)=0$ (both $P$ and $L$ satisfy it), and the distance from $S=(2,2)$ to it is $\frac{|4|}{\sqrt{t^2+4}}=LM$. So the area is $PL\cdot LM=\sqrt{t^2+4}\cdot \frac{4}{\sqrt{t^2+4}}=4{\text{cm}}^2$ for every $t$. The answer is option (D).