Question

If the lengths of diagonals of a rhombus are a cm and b cm, then the perimeter of rhombus is equal to

• A. $\frac{1}{2}(a+b)$cm
  
• B. $\frac{1}{2}(a^2+b^2)$
  
• C. $\frac{1}{2}\sqrt{a^2+b^2}$
  
• D. $2\sqrt{a^2+b^2}$
  

Answer 1

The correct option is D $2\sqrt{a^2+b^2}$



Consider a rhombus ABCD such that its diagonals AC and BD intersect each other at O.
Let AC = a cm and BD = b cm.
Since the diagonals of a rhombus bisects each other at right angles.
$\therefore AO=OC=\frac{a}{2}$cm and $BO=OD=\frac{b}{2}$cm and $\angle DOC=90°$
Now, in $\Delta DOC$, by Pythagoras theorem,
$(DC{)}^2=(OD{)}^2+(OC{)}^2$
$\Rightarrow (DC{)}^2=(\frac{b}{2}{)}^2+(\frac{a}{2}{)}^2$
$\Rightarrow DC=\sqrt{\frac{b^2+a^2}{4}}$
$\Rightarrow DC=\sqrt{\frac{a^2+b^2}{2}}$
Now, $AB=BC=CD=DA=\frac{1}{2}\sqrt{a^2+b^2}$ ($\therefore$ ABCD is a rhombus)
$\therefore$ Perimeter of a rhombus ABCD = $4\times \frac{1}{2}\sqrt{a^2+b^2}$
$=2\sqrt{a^2+b^2}$
Hence, the correct answer is option (d).