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user100 [1]
3 years ago
5

A rhombus ha a perimeter of 80 meters and the length of one diagonal is 24 meters.find the area of the rhombus

Mathematics
2 answers:
damaskus [11]3 years ago
5 0
<h2><u>GIVEN:-</u></h2>

  • \rm{Area\:of\:Rhombus = 80m}
  • \rm{Diagonal\:of\:Rhombus = 24m}

<h2><u>TO FIND :-</u></h2>

  • The area of Rhombus.

<h2><u>CONSTRUCTION:-</u></h2>

  • Draw another Rhombus from B to D.

<h2><u>CONCEPT USED</u></h2>

  • <u>All Sides</u> of Rhombus are equal.

  • Diagonal Bisect each other at right angled.

<h2><u>FORMULAE USED:-</u></h2>

  • {\boxed{\rm{\blue{Area\:of\:Rhombus = \dfrac{Product\:of\:Diagonal}{2}}}}}

  • {\boxed{\rm{\red{Pythogoras\:Theorem}}}}

<h3>Now,</h3>

\rm \implies{Perimeter\:of\:Rhombus = 4\times{sides}}

\rm \implies{80 = 4\times{sides}}⟹80=4×sides

\implies\rm{Side = \dfrac{80}{4}}

\implies\rm{Side = 20m}

Now, In right angled triangle ∆OBC,

\implies\rm{OC = 12m}

\implies\rm{BC = 20m}

\implies\rm{OB = ?}

Using Pythogoras Theorem,

\implies\rm{(OB)^2+(OC)^2 = (AB)^2}

\implies\rm{(OB)^2 = (20)^2-(12)^2}

\implies\rm{(OB)^2 = \sqrt{256}}

\implies\rm{OB = 16m}

Therefore,

\implies\rm{2OB = BD}

\implies\rm{2\times{16} = BD}

\implies\rm{32 = BD}

Now,

\rm{Area\:of\:Rhombus = \dfrac{Product\:of\:Diagonal}{2}}

\rm{Area\:of\:Rhombus = \dfrac{32\times{24}}{2}}

\rm{Area\:of\:Rhombus = \dfrac{768}{2}}

\rm{Area\:of\:Rhombus = 384m^2}

\sf  \therefore   Hence,  \:  The \:  Area  \: of \:  Rhombus  \: is \:  384m².

Triss [41]3 years ago
4 0

Answer:

The area of the rhombus is A=384\ m^2

Step-by-step explanation:

we know that

A rhombus have all four sides congruent.

The diagonals bisect each other

The diagonals are perpendicular

see the attached figure to better understand the problem

step 1

Find the length side of the rhombus

Divide the perimeter by 4

Let

b ---> the length side of rhombus

b=\frac{P}{4}

b=\frac{80}{4}=20\ m

step 2

Find the length of the other diagonal BD

In the right triangle AOB

Let

AC =24\ m ----> the length of one diagonal

BD ---> the length of the other diagonal

we have

AO=24/2=12\ m\\AB=20\ m

AB^2=AO^2+OB^2

substitute the values

20^2=12^2+OB^2

OB^2=400-144\\OB=16\ m

BD=2(OB)=2(16)=32\ m

step 3

Find the area of the rhombus

The area of the rhombus is equal to

A=\frac{1}{2}(AC)(BD)

substitute

A=\frac{1}{2}(24)(32)=384\ m^2

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To make comparison and statistical analysis easier, numerical data is systematically and logically represented in tabulations as rows and columns. Grouping relevant data together makes comparison easier and aids in statistical analysis and interpretation.

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