Question

In the given figure, ABCE is a rhombus and $△ACD$ is a right triangle with one of the interior angle as ${30}^o.$
What type of triangle is $△CED?$

• A. Isosceles triangle
• B. Equilateral triangle
• C. Scalene triangle

Answer 1

The correct option is B Equilateral triangle

Opposite angles of a rhombus are equal in measure.

Hence, $\angle BAE=\angle BCE$
The diagonals of a parallelogram bisect the respective interior angles.
$\therefore \frac{\angle BAE}{2}=\frac{\angle BCE}{2}$
$\Rightarrow \angle CAE=\angle ACE={30}^O$

For $△CAD,\angle ACD={90}^o$
$\Rightarrow \angle ACE+\angle ECD={90}^o$
$\Rightarrow {30}^o+\angle ECD={90}^o$
$\Rightarrow \angle ECD={90}^o-{30}^o={60}^o$

Also, $\angle CAD+\angle ACD+\angle D={180}^o$
$\Rightarrow {30}^o+{90}^o+\angle D={180}^o$
$\Rightarrow \angle D={180}^o-{30}^o-{90}^o={60}^o$

For $△ECD,$
As two interior angles are ${60}^o$ hence the third angle $\angle CED$ will be also ${60}^o.$

Hence, it is an equilateral triangle.