Question

Mr. Graze, a farmer, has a rectangular field as shown. The longest possible length inside the field is the smallest prime number.
Find the ratio of the area of the field to the square of the perimeter of the field.

• A. $\frac{\sqrt{6}}{4{\left(\sqrt{3}+1\right)}^2}$
• B. $\frac{\sqrt{3}}{2{\left(\sqrt{3}+1\right)}^2}$
• C. $\frac{\sqrt{3}}{4{\left(\sqrt{3}+1\right)}^2}$
• D. $\frac{\sqrt{6}}{8{\left(\sqrt{3}+1\right)}^2}$

Answer 1

The correct option is C $\frac{\sqrt{3}}{4{\left(\sqrt{3}+1\right)}^2}$

The longest distance inside a rectangle is its diagonal.
This is equal to the smallest prime number, which is 2.
Hence, the diagonal of the field is equal to 2 units.

Let's use the Pythagoras theorem to calculate the other side of the field.

${\left(\mathrm{AC}\right)}^2={\left(\mathrm{AB}\right)}^2+{\left(\mathrm{BC}\right)}^2{2}^2={\mathrm{AB}}^2+1^{2}4={\mathrm{AB}}^2+1{\mathrm{AB}}^2=3\mathrm{AB}=\sqrt{3}$

Area of the field
$=\mathrm{AB}\times \mathrm{BC}=1\times \sqrt{3}=\sqrt{3}$
Perimeter of the field
$=2(\mathrm{AB}+\mathrm{BC})=2\left(\sqrt{3}+1\right)$
Square of the perimeter
$={\left(2\left(\sqrt{3}+1\right)\right)}^2=4{\left(\sqrt{3}+1\right)}^2$

Hence,

$Ratio=\frac{Area}{Squareofperimeter}=\frac{\sqrt{3}}{4{\left(\sqrt{3}+1\right)}^2}$