Question

the side of a rhombus is 5cm if the length of one diagonal of the rhombus is 8cm, then find the length of the other diagonal​

Answer 1

Given :

• The side of rhombus = 5 cm
• One of the length of diagonal of rhombus = 8 cm

To Find :

• The length of the other diagonal

Solution :

The relation between the diagonals and sides of a rhombus is given by ,

$\\ \star \: {\boxed{\sf{\purple{ \bigg( \dfrac{d_1}{2} \bigg)^{2} + \bigg( \dfrac{d_2}{2} \bigg)^{2} = {s}^{2} }}}}$

$\\ \\ \sf{where}\begin{cases} & \sf{d_1 \& \: d_2 \: are \: diagonal \: of \: rhombus} \\ \\ & \sf{s \: is \: side \: of \: the \: rhombus}\end{cases}$

Substituting the values we have ,

$\\ : \implies \sf \bigg( \dfrac{8}{2} \bigg)^{2} + { \bigg( \dfrac{d_2}{2} \bigg) }^{2} = {(5)}^{2}$

$\\ : \implies \sf \: {(4)}^{2} + { \bigg( \dfrac{d_2}{2} \bigg)}^{2} = 25$

$\\ : \implies \sf 16 + { \bigg( \dfrac{d_2}{2} \bigg)}^{2} = 25$

$\\ : \implies \sf \: { \bigg( \dfrac{d_2}{2} \bigg)}^{2} = 25 - 16$

$\\ : \implies \sf \: { \bigg( \dfrac{d_2}{2} \bigg)}^{2} = 9$

$\\ : \implies \sf \: \bigg( \dfrac{d_2}{2} \bigg) = \sqrt{9}$

$\\ : \implies \sf \:\dfrac{d_2}{2} = 3$

$\\ : \implies \sf \:d_2 = 3 \times 2$

$\\ : \implies \sf{\underline{\boxed {\mathfrak{\pink{d_2 = 6 \: cm}}}}} \: \bigstar$

Hence ,

• The length of the other diagonal of the given rhombus is 6 cm

Answer 2

$\huge{ \underline{ \bf{Given \: : }}}$

• side of rhombus = 5cm
• one diagonal = 8cm

$\huge{ \underline{ \bf{To \: Find \: : }}}$

• the length of the other diagonal

$\huge{ \underline{ \underline{ \mathrm{Solution \: : - }}}}$

As we know,

The relation between the diagonals and sides of a rhombus is:

$\purple \leadsto\red{ \underline{ \underline{ \boxed{ \tt{ \green{( \frac{d _1 }{2} ) {}^{2} + ( \frac{d _2 }{2} ) {}^{2} = {s}^{2} }}}}}}$

$\tt \to( \cancel \frac{8}{2} ) {}^{2} + ( \frac{d _2 }{2} ) {}^{2} = {5}^{2}$

$\tt \to(4 {)}^{2} +( \frac{d _2 }{2} ) {}^{2} = 25$

$\tt \to16 + ( \frac{d _2 }{2} ) {}^{2} = 25$

$\tt \to( \frac{d _2}{2} ) {}^{2} = 25 - 16$

$\tt \to \: \frac{d _2 }{2} = \sqrt{9}$

$\tt \to \: d_2 = 3 \times 2$

$\tt \to \: d_2 = 6$

$\pink \star{ \green{ \boxed{ \red{ \tt \: d_2 = 6cm}}}}$

Therefore, the length of the other diagonal is 6cm.