Question

ABCD is a squared of side 4 cm . If E is a point in the interior of the square such that $\Delta CED$ is equilateral, then find area of $\Delta ACE$ $\left(inc{m}^2\right)$

Answer 1

$\Rightarrow$ Side of a square, $ABCD=4cm$

It is given that $△CED$ is an equilateral triangle.

$\therefore$ $EC=CD=DE=4cm$

$\Rightarrow$ $\angle ECD={60}^o.$ [ An angle of an equilateral triangle ]

$AC$ is a diagonal of a square $ABCD$

$\therefore$ $\angle ACD={45}^o.$

$\Rightarrow$ $\angle ECA=\angle ECD-\angle ACD$

$\Rightarrow$ $\angle ECA={60}^o-{45}^o$

$\Rightarrow$ $\angle ECA={15}^o$

In $△ACE,$ draw perpendicular $EM$ the base $AC.$

Now in $△EMC,$

$\Rightarrow$ $\sin {15}^o=\frac{P}{H}=\frac{EM}{EC}$

$\Rightarrow$ $\sin {15}^o=\frac{EM}{4}$

$\Rightarrow$ $\frac{(\sqrt{3}-1)}{2\sqrt{2}}=\frac{EM}{4}$

$\Rightarrow$ $(\sqrt{3}-1)\times \frac{\sqrt{2}}{(2\sqrt{2}\times \sqrt{2})}=\frac{EM}{4}$ [ By rationalizing the denominator ]

$\Rightarrow$ $\frac{\sqrt{2}(\sqrt{3}-1)}{4}=\frac{EM}{4}$

$\Rightarrow$ $EM=\sqrt{2}(\sqrt{3}-1)$

Diagonal of a square, $\left(AC\right)=\sqrt{2}a$

$AC=\sqrt{2}\times 4=4\sqrt{2}$

Diagonal of a square $=4\sqrt{2}cm$

Now, in $△AEC,$

$\Rightarrow$ Area of $△AEC=\frac{1}{2}\times AC\times EM$

$=\frac{1}{2}\times 4\sqrt{2}\times \sqrt{2}(\sqrt{3}-1)$

$=4(\sqrt{3}-1)c{m}^2$