Question

what is the side of a rhombus whose diagonal is 24 cm and 10 cm​

Answer 1

Answer:

question and answer both are there

Answer 2

❍ Let's Consider $D_{1} \:and\:D_{2}$ be two diagonals of Rhombus.

⠀⠀⠀⠀⠀ The Formula for Side of Rhombus is given by :

$\dag\frak{\underline { As,\:We\:know\:that\::}}$

$\star\boxed{\pink{\sf{ \: Side_{(Rhombus)} = \sqrt { \bigg(\dfrac{D_{1}\:}{2}\bigg)^{2} + \bigg(\dfrac{D_{2}\:}{2}\bigg)^{2} }}}}$

Where,

• $D_{1} \:and\:D_{2}$ are two diagonals of Rhombus

⠀⠀⠀⠀⠀⠀$\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}$

$\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(\dfrac{24\:}{2}\bigg)^{2} + \bigg(\dfrac{10\:}{2}\bigg)^{2} } }$

$\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(\cancel{\dfrac{24\:}{2}}\bigg)^{2} + \bigg(\cancel {\dfrac{10\:}{2}}\bigg)^{2} }}$

$\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(12\bigg)^{2} + \bigg( 5\bigg)^{2} } }$

$\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { 144 + 25 } }$

$\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { 169 }}$

⠀⠀⠀⠀⠀$\underline {\boxed{\purple{ \mathrm {  Side_{(Rhombus)}= 13\: cm}}}}\:\bf{\bigstar}$

Therefore,

⠀⠀⠀⠀⠀$\therefore {\underline{ \mathrm {  Side \:of\:Rhombus \:is\:\bf{13\: cm}}}}$

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$\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}$

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀$\begin{gathered}\boxed{\begin {array}{cc}\\ \dag\quad \Large\underline{\bf Formulas\:of\:Areas:-}\\ \\ \star\sf Square=(side)^2\\ \\ \star\sf Rectangle=Length\times Breadth \\\\ \star\sf Triangle=\dfrac{1}{2}\times Breadth\times Height \\\\ \star \sf Scalene\triangle=\sqrt {s (s-a)(s-b)(s-c)}\\ \\ \star \sf Rhombus =\dfrac {1}{2}\times d_1\times d_2 \\\\ \star\sf Rhombus =\:\dfrac {1}{2}p\sqrt {4a^2-p^2}\\ \\ \star\sf Parallelogram =Breadth\times Height\\\\ \star\sf Trapezium =\dfrac {1}{2}(a+b)\times Height \\ \\ \star\sf Equilateral\:Triangle=\dfrac {\sqrt{3}}{4}(side)^2\end {array}}\end{gathered}$

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