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

Other useful links

Related

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

Other useful links

Related

Leave a ReplyCancel reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Knowledge Partner

Cheenta is a knowledge partner of Aditya Birla Education Academy

Cheenta Academy

Aditya Birla Education Academy

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

HALL OF FAME BLOG OFFLINE CLASS

CHEENTA ON DEMAND BOSE OLYMPIAD

Cheenta Academy Private Limited |

support@cheenta.com

Menu

Math Olympiad Program