❍ Let’s Consider [tex]\bf{D_{1} \:and\:D_{2}} [/tex] be two diagonals of Rhombus. ⠀⠀⠀⠀⠀ The Formula for Side of Rhombus is given by : [tex]\dag\frak{\underline { As,\:We\:know\:that\::}}\\[/tex] [tex]\star\boxed{\pink{\sf{ \: Side_{(Rhombus)} = \sqrt { \bigg(\dfrac{D_{1}\:}{2}\bigg)^{2} + \bigg(\dfrac{D_{2}\:}{2}\bigg)^{2} }}}}\\[/tex] Where, [tex]D_{1} \:and\:D_{2} [/tex] are two diagonals of Rhombus ⠀⠀⠀⠀⠀⠀[tex]\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\[/tex] [tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(\dfrac{24\:}{2}\bigg)^{2} + \bigg(\dfrac{10\:}{2}\bigg)^{2} } }\\[/tex] [tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(\cancel{\dfrac{24\:}{2}}\bigg)^{2} + \bigg(\cancel {\dfrac{10\:}{2}}\bigg)^{2} }}\\[/tex] [tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(12\bigg)^{2} + \bigg( 5\bigg)^{2} } }\\[/tex] [tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { 144 + 25 } }\\[/tex] [tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { 169 }}\\[/tex] ⠀⠀⠀⠀⠀[tex]\underline {\boxed{\purple{ \mathrm { Side_{(Rhombus)}= 13\: cm}}}}\:\bf{\bigstar}\\[/tex] Therefore, ⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm { Side \:of\:Rhombus \:is\:\bf{13\: cm}}}}\\[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ [tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex] ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀[tex]\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}[/tex] ⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀ Reply
❍ Let’s Consider [tex]\bf{D_{1} \:and\:D_{2}} [/tex] be two diagonals of Rhombus.
⠀⠀⠀⠀⠀ The Formula for Side of Rhombus is given by :
[tex]\dag\frak{\underline { As,\:We\:know\:that\::}}\\[/tex]
[tex]\star\boxed{\pink{\sf{ \: Side_{(Rhombus)} = \sqrt { \bigg(\dfrac{D_{1}\:}{2}\bigg)^{2} + \bigg(\dfrac{D_{2}\:}{2}\bigg)^{2} }}}}\\[/tex]
Where,
⠀⠀⠀⠀⠀⠀[tex]\underline {\bf{\star\:Now \: By \: Substituting \: the \: Given \: Values \::}}\\[/tex]
[tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(\dfrac{24\:}{2}\bigg)^{2} + \bigg(\dfrac{10\:}{2}\bigg)^{2} } }\\[/tex]
[tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(\cancel{\dfrac{24\:}{2}}\bigg)^{2} + \bigg(\cancel {\dfrac{10\:}{2}}\bigg)^{2} }}\\[/tex]
[tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { \bigg(12\bigg)^{2} + \bigg( 5\bigg)^{2} } }\\[/tex]
[tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { 144 + 25 } }\\[/tex]
[tex]\qquad:\implies \tt {Side_{(Rhombus)} = \sqrt { 169 }}\\[/tex]
⠀⠀⠀⠀⠀[tex]\underline {\boxed{\purple{ \mathrm { Side_{(Rhombus)}= 13\: cm}}}}\:\bf{\bigstar}\\[/tex]
Therefore,
⠀⠀⠀⠀⠀[tex]\therefore {\underline{ \mathrm { Side \:of\:Rhombus \:is\:\bf{13\: cm}}}}\\[/tex]
⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀
[tex]\large {\boxed{\sf{\mid{\overline {\underline {\star More\:To\:know\::}}}\mid}}}\\\\[/tex]
⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀[tex]\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}[/tex]
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