Question

In the following figure, CD = 5 cm, BE = 10 cm, AE = 12 cm, \(\angle DAB = \angle DCB\), and \(\angle DAE = \angle DBC = 90^\circ\). Points AFCD create a rhombus. The length of BF (in cm) is:
\includegraphics[width=0.3\linewidth]{7.png}

• A. 3
• B. 2
• C. 4
• D. 1

Hint:

For geometric problems, use symmetry and properties like diagonals of rhombus intersecting at \(90^\circ\) to simplify calculations.

Answer 1

The Correct Option is A

Step 1: From the properties of the rhombus, diagonals intersect at \(90^\circ\): \(CD = AD = AF = FC = 5\).
From provided question:
(i) \ CD = 5 \ cm
(ii) \ AE = 12 \ cm
(iii) \ BE = 10 \ cm
(iv) \ \angle DAB = \angle DCB \ \text{and} \ \angle DAE = \angle DBC = 90^{\circ}
\text{then} \ \angle DBC = \angle DBA = \angle ABF = \angle FBC = 90^{\circ}
\text{In} \ \triangle DAE:
(DE)^{2} = (DA)^{2} + (AE)^{2}
DE = \sqrt{(5)^{2} + (12)^{2}} = 13 \ cm
DE = DB + BE \Rightarrow DB = DE - BE
DB = 13 - 10 = 3 \ cm
\ DB = 3 \ cm = BF