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A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is $ 30 \mathrm{~m} $ and its longer diagonal is $ 48 \mathrm{~m} $, how much area of grass field will each cow be getting?
Given:
A rhombus shaped field has green grass for 18 cows to graze.
Each side of the rhombus is \( 30 \mathrm{~m} \) and its longer diagonal is \( 48 \mathrm{~m} \).
To do:
We have to find the area of grass field will each cow be getting.
Solution:
We know that,
The diagonal of a rhombus divides it into two pairs of congruent triangles.
Join AC which divides the rhombus ABCD into two pairs of congruent triangles.
In triangle $ABC$,
$a=30\ m, b=30\ m$ and $c=48\ m$
The semi-perimeter of the triangle $s=\frac{a+b+c}{2}$
$=\frac{30+30+48}{2}$
$=\frac{108}{2}$
$=54\ m$
Therefore, by Heron's formula,
Area $=\sqrt{s(s-a)(s-b)(s-c)}$
$=\sqrt{54(54-30)(54-30)(54-48)}$
$=\sqrt{54(24)(24)(6)}$
$=\sqrt{186624}$
$=432\ m^2$
The area of the quadrilateral $=2 \times 432\ m^2$
$=864\ m^2$
The area grazed by each cow $=\frac{\text { Total area }}{\text { Number of cows }}$
$=\frac{864}{18}$
$=48 \mathrm{~m}^{2}$
Therefore, each cow will be getting $48\ m^2$ of area.
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