Diagonal of a rhombus are 20cm and21cm respectively,then find the side of rhombus and its perimeter​

1 thought on “Diagonal of a rhombus are 20cm and21cm respectively,then find the side of rhombus and its perimeter​”

1. S O L U T I O N :

   i. Let ABCD be the rhombus.

   AC = 20 cm, BD = 21 cm

   [tex]{\sf{A/Q\:=\; \dfrac{1}{2}\:AC\; ~~~ \bigg[Diagonals\; of\; rhombus\; bisect\:each\:other\bigg]}}[/tex]

   [tex]~~[/tex][tex]{\sf{\dfrac{1}{2}\:×\:20\:=\:10\:cm ~~~~~~~~~ (i)}}[/tex]

   [tex]\\[/tex]

   Also, BO

   [tex]{\sf{\dfrac{1}{2}\:BD\; ~~~ \bigg[Diagonals\; of\; rhombus\; bisect\:each\:other\bigg]}}[/tex]

   [tex]~~[/tex][tex]{\sf{\dfrac{1}{2}\:×\:20\:=\; \dfrac{21}{2}\:cm ~~~~~~~~~ (ii)}}[/tex]

   [tex]\\[/tex]

   ii. in ∆AOB, [tex]\angle[/tex]AOB = 90° [tex]{\sf{\bigg[Diagonals\: of\;a\: rhombus\:are\; perpendicular\;to\;each\; other\bigg]}}[/tex]

   [tex]\\[/tex]

   [tex]\therefore[/tex]AB = AO + BO [tex]{\sf{\bigg[Pythagoras\:theorem\bigg]}}[/tex]

   [tex]\\[/tex]

   [tex]{\sf{(10)^2\:+\; \bigg(\dfrac{21}{2}\bigg)^2 ~~~~~~ \bigg[From\:(i)\;and\:(ii)\bigg]}}[/tex]

   [tex]\\[/tex]

   [tex]{\sf{100\:+\; \dfrac{441}{4}}}[/tex]

   [tex]\\[/tex]

   [tex]{\sf{\dfrac{440\:+\:441}{4}}}[/tex]

   [tex]\\[/tex]

   [tex]\therefore[/tex][tex]{\sf{AB^2\:=\; \dfrac{841}{4}}}[/tex]

   [tex]\\[/tex]

   [tex]\therefore[/tex][tex]{\sf{AB\:=\; \sqrt\dfrac{841}{4}\; \bigg[Taking\; square\:root\:of\; both\;sides\bigg]}}[/tex]

   [tex]\\[/tex]

   [tex]~~~[/tex][tex]{\sf{\dfrac{29}{2}\:=\:14.5\:cm}}[/tex]

   [tex]\\[/tex]

   iii. Perimeter of ABCD

   [tex]{\sf{4\:×\:AB\:=\:4\:×\:14.5\:=\:58\:cm}}[/tex]

   [tex]\\[/tex]

   Hence,

   [tex]\therefore{\underline{\sf{The\:side\:and\: perimeter\: of\; the\; rhombus\:are\; \bf{14.5\:cm}\; \sf{and}\; \bf{58\:cm}\;\sf{respectively}.}}}[/tex]

   [tex]\\[/tex]

   [tex]~~~~[/tex][tex]\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}[/tex]

   [tex]~~~~~~~~~~~~~~~[/tex] ____________________

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