Question

ABCD is a square of side 12 cm and E is a point in the interior of ABCD. If ∠EDC = ∠ECD = 15°, then find the perimeter of ΔAEB. Please tell

Answer 1

Perimeter of ΔAEB is 36 cm

Step-by-step explanation:

As shown in the attached figure

In $\Delta EDC \text{ and } \Delta BCE$

$\because \angle EDC=\angle ECD =15^\circ$

$\therefore ED=EC$

$\angle ADE=90^\circ-15^\circ=75^\circ$

$\angle BCE=90^\circ-15^\circ=75^\circ$

Therefore,

$\angle ADE=\angle BCE$

In $\Delta ADE \text{ and }\Delta BCE$

$\because AD=BC=12$ cm

$ED=EC$

$\angle ADE=\angle BCE$

$\therefore \Delta ADE \cong \Delta BCE$     (SAS congruity rule)

$\therefore \angle AED=\angle BED=\theta$ (assume)

And $AE=BE$

$\implies \angle EAB=\angle EBA$

i.e. $\Delta AEB$ is an isosceles triangle

Around point E

$150^\circ +\theta+\theta+\angle AEB=360^\circ$

$\implies \angle AEB=210^\circ-2\theta$

Again in $\Delta AEB$

$\angle EAB+\angle EBA+\angle AEB=180^\circ$

$\implies 2\angle EAB+210^\circ-2\theta=180^\circ$

$\implies 2\angle EAB=-30^\circ+2\theta$

$\implies \angle EAB=\theta-15^\circ$

$\therefore \angle EBA=\theta-15^\circ$

$\angle DAE=90^\circ-\angle EAB$

$\implies \angle DAE=105^\circ-\theta$

In $\Delta EDC$

By sine rule

$\frac{\sin 150^\circ}{12}=\frac{\sin 15^\circ}{ED}$

$\implies \frac{\sin 30^\circ}{12}=\frac{\sin 15^\circ}{ED}$

$\implies ED=24\sin 15^\circ$

In $\Delta AED$

By sine rule

$\frac{\sin\theta}{12}=\frac{\sin (105^\circ-\theta)}{24\sin 15^\circ}$

$\implies \frac{24\sin 15^\circ}{12}=\frac{1}{\sin\theta}(\sin 105^\circ\cos \theta-\cos 105^\circ\sin \theta)$

$\implies 2\sin 15^\circ=\sin 75^\circ\cot\theta-\cos (90^\circ+15^\circ)$

$\implies 2\sin 15^\circ=\sin 75^\circ\cot\theta+\sin 15^\circ$

$\implies \sin 15^\circ=\sin (90^\circ-15^\circ)\cot\theta$

$\implies \sin 15^\circ=\cos 15^\circ\cot\theta$

$\implies \cot\theta=\frac{\sin 15^\circ}{\cos 15^\circ}$

$\implies \cot\theta=\tan 15^\circ$

$\implies \cot\theta=\tan (90^\circ-75^\circ)$

$\implies \cot\theta=\cot 75^\circ$

$\implies \theta=75^\circ$

$\therefore\angle EAB=\angle EBA=75^\circ-15^\circ=60^\circ$

Thus $\Delta AEB$ is an equilateral triangle

Therefore,

$EA=EB=AB=12$ cm

Thus, the perimeter of $\Delta AEB$

$P=12+12+12$

$\implies P=36$ cm

Hope this answer is helpful.

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