Question

A rhombus has perimeter 120 m amd one of its diagonal is 50 m . Find the area of the rhombus

Answer 1

Answer:

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Step-by-step explanation:

A≈829.16m²

p Diagonal

50

m

P Perimeter

120

m

Using the formulas

A=pq

2

P=4a

a=p2+q2

2

Solving forA

A=1

4pP2﹣4p2=1

4·50·1202﹣4·502≈829.1562m²

Answer 2

Answer:

The area of rhombus is $829m^2$

Step-by-step explanation:

Given that,

The perimeter of the rhombus is $P=120m$ and

The length of one of its diagonal is $d_{1}=50m$

We are required to find the area of rhombus.

The relation for area of rhombus is

$A=\frac{1}{2}d_{1}d_{2}$

Let the side of the rhombus be s.

Since perimeter is four times the side,we shall write $P=4s= > 120=4s= > s=30m$

In a rhombus from Pythagorean theorem we can write

$s^2=(\frac{d_{1}}{2})^2+(\frac{d_{2}}{2})^2\\= > 30^2=(\frac{50}{2} )^2+\frac{d_{2}^2}{4} \\= > d_{2}^2=275\times4=1100\\= > d_{2}=33.16m$

Hence the area of the rhombus is

$A=\frac{1}{2}d_{1}d_{2}\\=\frac{1}{2}(50)(33.16)=829m^2$

Therefore,

The area of rhombus is $829m^2$

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