Question

$ABCD$ is a rhombus such that one of its diagonals is equal to its sides. Then the angles of rhombus $ABCD$ are-

• A. ${45}^{\circ },{135}^{\circ },{45}^{\circ },{135}^{\circ }$
• B. ${100}^{\circ },{80}^{\circ },{100}^{\circ },{80}^{\circ }$
• C. ${120}^{\circ },{60}^{\circ },{120}^{\circ },{60}^{\circ }$
• D. ${60}^{\circ },{60}^{\circ },{60}^{\circ },{60}^{\circ }$

Answer 1

Answer: (C)

${120}^{\circ },{60}^{\circ },{120}^{\circ },{60}^{\circ }$

When one side is equal to the diagonal $AC$ , then all sides are equal to $AC$ because all sides of rhombus are equal.

So, in $△ABC,$ all sides are equal.

Let measure of angles in $△ABC$ be $x$.

$\therefore$ $x+x+x={180}^o$ [Since, sum of angles is ${180}^o$ in triangle ]

$\Rightarrow$ $3x={180}^o$

$\therefore$ $x={60}^o$

$\therefore$ $\angle B={60}^o$

In $△ACD$,

Let measure of angles in $△ACD$ be $x$.

$\therefore$ $x+x+x={180}^o$

$\therefore$ $3x={180}^o$

$\therefore$ $x={60}^o$

$\therefore$ $\angle D={60}^o$

$\Rightarrow$ $\angle A={60}^o+{60}^o={120}^o$

$\Rightarrow$ $\angle C={60}^o+{60}^o={120}^o$

$\therefore$ The angles of rhombus $ABCD$ are ${120}^o,{60}^o,{120}^o,{60}^o$.