GATE 2022 EC GA Q.3: solution and answer

Official answer: BSource: Official answer key, GATE 2022 EC (IIT Kharagpur)

1. The shortest distance between the two parallel sides is the perpendicular height $h$ of the trapezium. Drop perpendiculars from S and R onto PQ, meeting it at M and N, so that SRNM is a rectangle and $MN=SR=6\text{cm}$.

2. The two end pieces of PQ then add up to the difference of the parallel sides: $PM+NQ=PQ-SR=11-6=5\text{cm}$.

3. Slide the rectangle SRNM out, i.e. move the leg QR a distance of 6 cm along PQ towards P so that it meets the leg SP at S. This leaves a triangle with sides $SP=3$, $QR=4$ and base $PM+NQ=5$ (all in cm). Its height on the base of 5 is the same $h$.

4. Converse of Pythagoras' theorem: $3^2+4^2=9+16=25=5^2$, so this triangle has a right angle between the sides 3 and 4.

5. Work out the area of this right triangle in two ways: $\frac{1}{2}\times 3\times 4=\frac{1}{2}\times 5\times h$, so $h=\frac{12}{5}=2.4\text{cm}$.

6. Check with lengths: let $PM=a$ and $NQ=b$, with $a+b=5$. From the two right triangles SMP and RNQ, $h^2=9-a^2=16-b^2$, so $b^2-a^2=7$. Since $b^2-a^2=(b-a)(b+a)$, this gives $b-a=\frac{7}{5}=1.4$, hence $a=1.8$ and $b=3.2$. Both are positive, so the feet M and N do lie on PQ as assumed. Then $h^2=9-3.24=5.76$ (and also $16-10.24=5.76$), so $h=2.4\text{cm}$.

7. The shortest distance between PQ and SR is $2.40\text{cm}$, which is option (B).