Question

Find the area of rombus whose perimeter is 164and a diagonal is of length 80cm

Answer 1

$\large\boxed{\bold{\red{Answer:-360cm{}^{2}}}}$

$\huge\mathbb{\underline{\underline{SOLUTION:-}}}$

$\bold{Given}\begin{cases}\sf{perimeter\:of\: rhombus}\\ \sf{one \:of\:its\: diagonal=80cm}\end{cases}$

$\sf\underline{\underline{\red{According\:to\: Question:-}}}$

$\sf perimeter\:of\: rhombus=4\times side\\ \\ \sf\implies 164=4\times side\\ \\ \sf\implies \cancel {\frac{164}{4}}=side\\ \\ \sf\implies side=41cm$

$\boxed{\mathfrak{\purple{side\:of\: rhombus=41cm}}}$

$\sf\underline{\tt{\red{Area \:of\: Rhombus=\frac{1}{2}\times (product\:of\: diagonal)}}}$

• we have to find the value of 2nd diagonal

So ,by the Pythagoras formula

$\sf AC{}^{2}=BC{}^{2}+AB{}^{2}$

$\bold{Given}\begin{cases}\sf{side=41cm}\\ \sf{one\: diagonal=80cm}\end{cases}$

• let the 2nd diagonal be x

$\sf \leadsto (\frac{80}{2}){}^{2}+(\frac{x}{2}){}^{2}=41{}^{2}\\\\ \sf\leadsto (\frac{6400}{4})+(\frac{x{}^{2}}{4})=1681\\ \\ \sf\leadsto 6400+x{}^{2}=1681\times 4\\ \\ \sf\leadsto x{}^{2}=6724-6400\\ \\ \sf\leadsto x{}^{2}=324\\ \\ \sf\implies x=\sqrt{324}\\ \\ \sf \implies x=18cm$

• $\sf The \:diagonal_2\:of\: rhombus=18cm$

$\rm\underline{\red{Area\:of\: rhombus=\frac{1}{2}\times d_1\times d_2}}$

$\bold{we\:have}\begin{cases}\sf{side=41cm}\\ \sf{diagonal_1=40cm}\\ \sf{diagonal_2=18cm}\end{cases}$

$\sf Area\:of\: Rhombus=\frac{1}{2}\times \cancel{18}\times 40\\ \\ \sf\implies Area\:of\: Rhombus=9\times 40\\ \\ \sf\implies Area\:of\: rhombus=360cm{}^{2}$

$\boxed{\purple{\mathfrak{Area=360cm{}^{2}}}}$

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