A field is in the shape of a rhombus. Its perimeter is 584 m and the length of one of its diagonals is 220 m. What is the area (in m
2
) of the field?

Let the side of the rhombus be a, and its diagonals be d
1
​
and d
2
​
.

Given:
Perimeter = 4a=584 m⟹a=
4
584
​
=146 m
One diagonal (d
1
​
) = 220 m

In a rhombus, the diagonals bisect each other at right angles (90
∘
).
Therefore, the relationship between the side and the half-diagonals forms a right-angled triangle:

a
2
=(
2
d
1
​

​
)
2
+(
2
d
2
​

​
)
2

Substituting the values:

146
2
=(
2
220
​
)
2
+(
2
d
2
​

​
)
2

21316=110
2
+(
2
d
2
​

​
)
2

21316=12100+(
2
d
2
​

​
)
2

(
2
d
2
​

​
)
2
=21316−12100=9216
2
d
2
​

​
=
9216

​
=96 m
d
2
​
=96×2=192 m
The area of a rhombus is given by:

Area=
2
1
​
×d
1
​
×d
2
​

Area=
2
1
​
×220×192=110×192=21,120 m
2