Question

the diagonal of a rhombus are 12 cm and 16 cm. find the length of each side of the rhombus​

Answer 1

$\large\underline{ \underline{ \sf \maltese{ \: Question:- }}}$

• The diagonal of a rhombus are 12 cm and 16 cm. Find the length of each side of the rhombus.

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$\large\underline{ \underline{ \sf \maltese{ \: Solution:- }}}$

• Let ABCD be a Rhombus in which $\sf\green{\: AC \: = \: 16cm\:}$ and $\sf\green{\: BD \: = \: 12cm \:}$

Since, the Diagonal of Rhombus bisect each other at Right Angles ,

Hence,

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$\</strong><strong>b</strong><strong>l</strong><strong>u</strong><strong>e</strong><strong>{\boxed{ \sf \</strong><strong>o</strong><strong>r</strong><strong>a</strong><strong>n</strong><strong>g</strong><strong>e</strong><strong>{</strong><strong>AO \: = \: \frac{1}{2}</strong><strong>\</strong><strong>:</strong><strong> </strong><strong>×</strong><strong> </strong><strong>\</strong><strong>:</strong><strong> AC</strong><strong> </strong><strong>}}}$

$\quad {:} \longrightarrow \sf{\sf{ AO \: = \: \frac{1}{2} \: \times \: 16}}$

$\quad {:} \longrightarrow \sf{\sf{ AO \: = \: \frac{1}{\cancel{2}} \: \times \: \cancel{16} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{AO \: = \: 8cm }}}$

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$\blue{\boxed{ \sf \orange{BO \: = \: \frac{1}{2}\: × \: BD }}}$

$\quad {:} \longrightarrow \sf{\sf{ BO \: = \: \frac{1}{2} \: \times \: 12}}$

$\quad {:} \longrightarrow \sf{\sf{ AO \: = \: \frac{1}{\cancel{2}} \: \times \: \cancel{12} }}$

$\qquad\quad {:} \longrightarrow \underline \red{\boxed{\sf{BO \: = \: 6cm }}}$

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$\blue{\boxed{ \sf \orange{For \: Finding \: the \: Side :- }}}$

Use Pythagoras Theorem in ∆ AOB

$\begin{gathered}\begin{gathered}\begin{gathered}: \implies \underline\blue{ \boxed{\displaystyle \sf \bold\green{\: {(AB)}^{2} \: = \: {(AO)}^{2} \: + \: {(OB)}^{2} }} }\\ \end{gathered}\end{gathered}\end{gathered}$

$\qquad \quad {:} \longrightarrow \sf{\sf{ {(AB)}^{2} \: = \: {(8)}^{2} \: + \: {(6)}^{2} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{{(AB)}^{2} \: = \: 64 \: + \: 36 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{{(AB)}^{2} \: = \: 100 }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{(AB) \: = \: \sqrt{100} }}$

$\qquad \quad {:} \longrightarrow \sf{\sf{(AB) \: = \: {10}^{2} }}$

$\bf \therefore \; AB \;= 10cm$

${:} \longrightarrow \underline \red{\boxed{\sf{So, \: The \: length \: of \: each \: side \: = \: 10 cm }}}$

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Note:-

• Diagram in attachment :)

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Answer 2

☯ Let ABCD be the rhombus & AC and BD bisect at point O.

⠀⠀⠀

$\frak{Where} \begin{cases} & \sf{AC = \bf{16\:cm}} \\ & \sf{BD = \bf{12\:cm}} \end{cases}$

$\setlength{\unitlength}{0.7cm}\begin{picture}(0,0)\linethickness{.4mm}\put(-3.5,1){\line(1,0){9}}\put(1,1){\line(0,1){3.5}}\qbezier(1,4.5)(1,4.5)(5.5,1)\put(1.02,1.02){\framebox(0.3,0.3)}\put(.7,4.8){\large\bf B}\put(-0,.3){\large\bf O}\put(5.8,.3){\large\bf C}\qbezier(1,4.5)(1,4.5)(-3.5,1)\qbezier(1,-3)(1,-3)(-3.5,1)\qbezier(1,-3)(1,-3)(5.5,1)\put(1,-3){\line(0,1){6}}\put(.7,-3.8){\large\bf D}\put(-4.5,0.5){\large\bf A}\end{picture}$

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$\dag\;{\underline{\frak{As\;we\;know\;that,}}}$

• Diagonals of rhombus bisects each other at right angles.

⠀⠀⠀

Therefore,

$:\implies\sf OB = \dfrac{BD}{2} = \cancel{\dfrac{12}{2}} = \bf{6\:cm}$

$:\implies\sf OC = \dfrac{AC}{2} = \cancel{\dfrac{16}{2}} = \bf{8\:cm}$

$\dag\;{\underline{\frak{Now,\:In\: \triangle\:BOC,}}}$

$\underline{\bigstar\:\boldsymbol{Using\: Pythagoras\: Theorem\::}}$

$\star\;{\boxed{\sf{\pink{(Hypotenuse)^2 = (Base)^2 + (Perpendicular)^2}}}}$

$:\implies\sf BC^2 = OC^2 + OB^2\\ \\ \\ :\implies\sf BC^2 = (8)^2 + (6)^2\\ \\ \\ :\implies\sf BC^2 = 64 + 36\\ \\ \\ :\implies\sf BC^2 = 100\\ \\ \\ :\implies\sf \sqrt{BC^2} = \sqrt{100}\\ \\ \\ :\implies\sf BC = \sqrt{100}\\ \\ \\ :\implies{\underline{\boxed{\frak{\purple{BC= 10\:cm}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Length\:of\:each\:side\:of\:a\: rhombus\:is\: {\textsf{\textbf{10\:cm}}}.}}}$

$\because\:{\underline{\sf{Each\:sides\:of\: rhombus\:are\:equal\:in\;length.}}}$