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

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

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Allison 3 months 2021-07-24T13:10:57+00:00 1 Answers 0 views 0

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    2021-07-24T13:12:52+00:00

    S O L U T I O N :

    i. Let ABCD be the rhombus.

    AC = 20 cm, BD = 21 cm

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

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

    \\

    Also, BO

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

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

    \\

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

    \\

    \thereforeAB = AO + BO {\sf{\bigg[Pythagoras\:theorem\bigg]}}

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

    \\

    {\sf{100\:+\; \dfrac{441}{4}}}

    \\

    {\sf{\dfrac{440\:+\:441}{4}}}

    \\

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

    \\

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

    \\

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

    \\

    iii. Perimeter of ABCD

    {\sf{4\:×\:AB\:=\:4\:×\:14.5\:=\:58\:cm}}

    \\

    Hence,

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

    \\

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

    ~~~~~~~~~~~~~~~ ____________________

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