Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.
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.
Geometry
Triangle
Trigonometry
But try the problem first...
Answer:$24$
PRMO-2017, Problem 13
Pre College Mathematics
First hint
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 ..........
Second Hint
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........
Final Step
Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
Try this beautiful Rectangle Problem from Geometry, from PRMO 2017.
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.
Geometry
Triangle
Trigonometry
But try the problem first...
Answer:$24$
PRMO-2017, Problem 13
Pre College Mathematics
First hint
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 ..........
Second Hint
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........
Final Step
Therefore $A B=B C \cot \theta=8 \sqrt{3} \cot 30^{\circ}=24$
AMC - AIME Boot Camp... for brilliant students.
Use our exclusive one-on-one plus group class system to prepare for Math Olympiad
