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Rectangle Problem | Geometry | PRMO-2017 | Question 13

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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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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