MAT Exam 2022 Paper

Concept:Use the area of a right triangle to find the altitude, then apply similarity of triangles to get the required segment length.Explanation:In rectangle $\mathrm{ABCD}$, $\triangle \mathrm{ABC}$ is right-angled at $\mathrm{B}$. Using Pythagoras theorem, $\mathrm{AC}=\sqrt{8^2+6^2}=10$ cm. Area of $\triangle \mathrm{ABC} = \frac{1}{2}\times \mathrm{AB}\times \mathrm{BC} = \frac{1}{2}\times 8\times 6 = 24$ sq cm. This area can also be written as $\frac{1}{2}\times \mathrm{AC}\times \mathrm{BE}$. So, $\frac{1}{2}\times 10\times \mathrm{BE}=24$, which gives $\mathrm{BE}=4.8$ cm. Now, $\mathrm{EF}\perp \mathrm{AB}$ and $\mathrm{BE}\perp \mathrm{AC}$, so $\angle \mathrm{BFE}=90^\circ$ and $\angle \mathrm{BEA}=90^\circ$. Also, $\angle \mathrm{FBE}=\angle \mathrm{EBA}$ because both are the same angle at $\mathrm{B}$. Hence, $\triangle \mathrm{BFE}\sim \triangle \mathrm{BEA}$ by AA similarity. Corresponding sides give $\frac{\mathrm{BF}}{\mathrm{BE}}=\frac{\mathrm{BE}}{\mathrm{AB}}$. Thus, $\mathrm{BF}=\frac{(4.8)^2}{8}=\frac{23.04}{8}=2.88$ cm.Answer:$\mathrm{BF}=2.88$ cm, which is Option D.