GATE 2024 EE GA Q.7: solution and answer

Official answer: marks to all candidatesSource: Official answer key, GATE 2024 EE (IISc Bengaluru)

1. Rhombus property: all four sides are equal, so $AD=CD=5\text{cm}$ (the rhombus is AFCD, so $CD$ and $DA$ are both sides). Its diagonals $AC$ and $DF$ bisect each other at right angles, which is why $\angle DBC=90^{\circ }$ and why $B$ is the midpoint of $DF$: $BF=DB$. The figure shows $D,B,F,E$ on one vertical line, with $B$ between $D$ and $E$.

2. Triangle $DAE$ is right-angled at $A$ (given $\angle DAE=90^{\circ }$). By Pythagoras: $$DE=\sqrt{A{D}^2+A{E}^2}=\sqrt{5^2+12^2}=\sqrt{169}=13\text{cm}$$

3. $B$ lies between $D$ and $E$ on the same line, so $$DB=DE-BE=13-10=3\text{cm}$$

4. The diagonals of a rhombus bisect each other, so $BF=DB=3\text{cm}$. This is option (A).

5. A caution about the data: they cannot all hold in one figure. The diagonals are perpendicular, so $AB$ is perpendicular to $DE$ and triangles $ABD$ and $ABE$ are both right-angled at $B$. Triangle $ABE$ gives $A{B}^2=A{E}^2-B{E}^2=144-100=44$. But $AB$ is a leg of triangle $ABD$, whose hypotenuse is $AD=5$, so $A{B}^2=25-D{B}^2$, which is less than 25. Both cannot be true. Put another way: with $DB=3$ and $AD=5$, right triangle $ABD$ gives $AB=4$, and then right triangle $ABE$ would make $AE=\sqrt{4^2+10^2}=\sqrt{116}\approx 10.8\text{cm}$, not 12. The official answer key awards marks to all candidates for this question. The route above uses the stated right angle $\angle DAE$, the rhombus side and $BE$, and gives 3, option (A).