Question

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

Answer 1

S O L U T I O N :

i. Let ABCD be the rhombus.

AC = 20 cm, BD = 21 cm

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

$~~$${\sf{\dfrac{1}{2}\:×\:20\:=\:10\:cm ~~~~~~~~~ (i)}}$

Also, BO

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

$~~$${\sf{\dfrac{1}{2}\:×\:20\:=\; \dfrac{21}{2}\:cm ~~~~~~~~~ (ii)}}$

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

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

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

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

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

$\therefore$${\sf{AB^2\:=\; \dfrac{841}{4}}}$

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

$~~~$${\sf{\dfrac{29}{2}\:=\:14.5\:cm}}$

iii. Perimeter of ABCD

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

Hence,

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

$~~~~$$\qquad\quad\therefore{\underline{\textsf{\textbf{Hence, Proved!}}}}$

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