ABCD is a rectangle as shown in the figure. AB = 8 cm and BC = 6 cm. BE is the perpendicular drawn from B to the diagonal AC. EF is the perpendicular drawn from E to AB. What is the length of BF?
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For geometry problems involving perpendiculars drawn to diagonals or sides, consider using Pythagoras` theorem and properties of similar triangles to break down the problem into solvable parts.
Step 1: Understand the geometry of the problem In this problem, we are given a rectangle ABCD, with \( AB = 8 \) cm, \( BC = 6 \) cm. Points E and F are where perpendiculars are dropped from B and E to the diagonal AC and line AB, respectively.
Step 2: Use the given relation and formula We are given the relationship:
\[
\frac{1}{BE^2} = \frac{1}{AB^2} + \frac{1}{BC^2}.
\]
Substitute the values of AB and BC:
\[
\frac{1}{BE^2} = \frac{1}{8^2} + \frac{1}{6^2} = \frac{1}{64} + \frac{1}{36} = \frac{36 + 64}{2304} = \frac{100}{2304}.
\]
Thus,
\[
BE^2 = \frac{2304}{100} = 36.64 \quad \Rightarrow \quad BE = \sqrt{36.64} \approx 6.05.
\]
Step 3: Use similar triangles to calculate BF In \( \triangle BEF \sim \triangle ABE \) (as these triangles are similar by AA similarity), we can use the property:
\[
\frac{BF}{BE} = \frac{BE}{AB}.
\]
Substitute the known values:
\[
\frac{BF}{6.05} = \frac{6.05}{8} \quad \Rightarrow \quad BF = \frac{6.05^2}{8} = \frac{36.64}{8} \approx 4.8.
\]
Step 4: Final Answer The length of BF is approximately 2.88 cm.
Final Answer: The correct answer is (d) 2.88 cm.
