Question

Construct a rhombus $ABCD$ whose length of one side $AB$ is $5cm$ and $\angle DAB=\frac{\pi }{2}$. Find the difference between the length of its diagonals.

Answer 1

Given : $ABCD$ is a rhombus.

$AB=5cm$

$⟹AB=BC=CD=DA=5cm$ ..........(1) (In rhombus all sides are equal)

$\angle DAB=\frac{\pi }{2}={90}^o$ ..... (2)

$⟹\angle DCB={90}^o$ ...... (Opposite angles are equal)

$⟹\angle DAB+\angle ABC={180}^o$ ........ (Adjacent angles are supplementary)

$⟹\angle ABC={90}^o$

Similarly, $\angle BDA={90}^o$

$AC$ and $BD$ are diagonals of $ABCD$

In $△DAB$, $\angle DAB={90}^o$

$\therefore$ By Pythagoras theorem,

$B{D}^2=A{B}^2+A{D}^2=5^2+5^2=50$

$\therefore BD=\sqrt{50}$ ............. (3)

In $△ADC$, $\angle ADC={90}^o$

By Pythagoras theorem we have,

$A{C}^2=A{D}^2+D{C}^2=5^2+5^2=50$

$\therefore AC=\sqrt{50}$ ............... (4)

By using (3) and (4),

$AC-BD=0$

Hence, difference between two diagonals is $0$.