Question

find the area of the rhombus whose perimeter is 80m and One of whose diagonal is 24m​

Answer 1

$\huge\tt{\bold{\purple{\underbrace{\red{\mathfrak{ɾҽզմíɾҽժ⠀ αղsաҽɾ♡༄}}}}}}$

$\huge{\fbox{\pink{Given:-}}}$

Perimeter of the Rhombus = 80m

Diagonal of the Rhombus = 24 m

⠀

$\huge{\fbox{\pink{To find:-}}}$

Area of the Rhombus.

⠀

$\huge{\fbox{\pink{Solution:-}}}$

Perimeter = 80 m,

So, length of a side = 80/4

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀=20m,

and Diagonal of a rhombus (d1) = 24m,

As, we know that,

The halves of diagonals and a side of a rhombus from a right angled triangle with side as hypotenuse.

Let the length of the other diagonal (d2) =2x.

${x}^{2} + ( \frac{22}{2} ) ^{2} = {20}^{2} \\ \implies {x}^{2} + 144 \: = 400 \\ \implies {x}^{2} = 400 - 144 \\ \implies {x}^{2} = 256 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \: x = \sqrt{256} \: \: \: \: \: \: \: \: \: \: \: \: \\ \implies \: x = 16 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

The measure of the another diagonal of the rhombus (d2) = 2 x 16

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀ = 32 m

$area \: of \: the \: rhombus = \frac{1}{2} \times d1 \times d2 \\ \implies area \: of \: the \: rhombus = \frac{1}{2} \times 32 \times 24 \\ \implies \:area \: of \: the \: rhombus = 384 {m}^{2} \: \: \: \: \: \: \: \: \: \: \:$

⠀

$\huge{\fbox{\pink{Formula Used:-}}}$

$area \: of \: the \: rhombus = \frac{1}{2} \times d1 \times d2$

Answer 2

$\huge{\underline{\underline{\mathfrak{\red{Given\::}}}}}$

• $\bf{Perimeter\:of\:rhombus\:=\:80\:m}$
• $\bf{One\:diagonal\:of\:rhombus\:=\:24\:m}$

${ }$

$\huge{\underline{\underline{\mathfrak{\red{Find\::}}}}}$

• $\bf{Area\:of\:rhombus}$

${ }$

$\huge{\underline{\underline{\mathfrak{\red{Solution\::}}}}}$

• $\bf{\bold{Perimeter\: of\: rhombus \:= \:4 \:\times\: side}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\::\:\Longrightarrow{\sf{4a\:=\:80}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\::\:\Longrightarrow{\sf{a\:=\:20\:m}}$

$\:\:\:\:\:\sf{Let\:AC\:=\:25\:m}$

$\:\:\:\:\:\sf{\therefore\:OA\:=\:{\frac{1}{2}}}$

$\:\:\:\:\:\sf{AC\:=\:{\frac{1}{2}}\:\times\:24\:=\:12\:m}$

$\:\:\:\:\:\bf{\bold{In\:\triangle\:AOB}}$

$\:\:\:\:\:\bf{OB}^{2}\:{=\:AB}^{2}\:{-\:OA}^{2}$

• $\bf{\bold{[By\:using\:Pythagoras\:therorem]}}$

$\:\:\:\:\:\::\:\Longrightarrow{\sf{OB\:=\:\sqrt{20}^{2}\:{-\:12}^{2}}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\::\:\Longrightarrow{\bf{\sqrt{400\:-\:144}}}$

$\:\:\:\:\:\:\:\:\:\:\:\:\::\:\Longrightarrow{\bf{\sqrt{256}}\:=\:16\:m}$

• $\:\:\:\:\:\bf{Also\:BO\:=\:OD}$

$\:\:\:\:\:\bf{[Diagonal\:of\:rhombus\:bisect\:each\:other\:at\:90}^{°}]$

$\:\:\:\:\:\:\::\:\Longrightarrow{\bf{\therefore\:BD\:=\:20\:B\:=\:2\:\times\:16\:=\:32\:m}}$

• $\sf{\therefore\:Area\:of\:rhombus\:=\:{\frac{1}{2}}\:\times\:BD\:\times\:AC}$

$\:\:\:\:\:\:\::\:\Longrightarrow\sf{\therefore\:Area\:of\:rhombus\:=\:{\frac{1}{2}}\:\times\:32\:\times\:24\:=\:384\:m}^{2}$

${ }$

$\:\:\:\:\therefore{\underline{\sf{Hence,\:the\:area\:of\:rhombus\:is\:{\bold{384\:m}^{2}}}}}$