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Rectangle Problem | Geometry | PRMO-2017 | Question 13

Try this beautiful Rectangle Problem from Geometry from PRMO 2017, Question 13. You may use sequential hints to solve the problem.

Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.

Rectangle Problem – Geometry – PRMO 2017, Question 13


In a rectangle $A B C D, E$ is the midpoint of $A B ; F$ is a point on $A C$ such that $B F$ is perpendicular to $A C$; and FE perpendicular to BD. Suppose $\mathrm{BC}=8 \sqrt{3}$. Find AB.

  • $9$
  • $24$
  • $11$

Key Concepts


Geometry

Triangle

Trigonometry

Check the Answer


Answer:$24$

PRMO-2017, Problem 13

Pre College Mathematics

Try with Hints


Rectangle Problem

We have to find out the value of \(AB\). Join \(BD\). \(BF\) is perpendicular on \(AC\).

Let $\angle \mathrm{BAC}=\theta$
since $\mathrm{E}$ is mid point of hypotenous $\mathrm{AB}$ of right $\Delta \mathrm{AFB}$, therefore $A E=F E=B E$

Can you now finish the problem ……….

Rectangle Problem

Therefore

Therefore$ \angle E F A=\angle F A E=\theta$
and $\angle \mathrm{FEB}=\angle \mathrm{EAF}+\angle \mathrm{EFA}=2 \theta$
$\Rightarrow \angle E B D=90^{\circ}-\angle B E F=90^{\circ}-2 \theta$
But $\angle \mathrm{FAE}=\angle \mathrm{CAB}=\angle \mathrm{DBA}$
Therefore $\theta=90^{\circ}-2 \theta \Rightarrow \theta=30^{\circ}$
Therefore in $\Delta \mathrm{ABC}, \tan \theta=\frac{\mathrm{BC}}{\mathrm{AB}}$

Can you finish the problem……..

Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$

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