A rhombus is formed by joining the mid points of the sides of a unit square. What is the diameter of the largest circle that can be inscribed within the rhombus?

The correct option is 1/√2.

Step-by-step Explanation:

Let the unit square be represented by $ABCD$ as shown in the diagram, where the side length of the square is 1 unit. The midpoints of the sides $AB,BC,CD,$ and $DA$ are denoted by $P,Q,R,$ and $S$ respectively. Joining these midpoints forms the rhombus $PQRS$.

1. Finding the side length of the rhombus:
Since $P$ and $S$ are the midpoints of the sides of a unit square, we have:
$$AP=AS=\frac{1}{2}\text{units}$$
Using the Pythagorean theorem in the right-angled triangle $△SAP$ (where $\angle A={90}^{\circ }$):
$$SP=\sqrt{{\left(\frac{1}{2}\right)}^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{1}{4}+\frac{1}{4}}=\frac{1}{\sqrt{2}}\text{units}$$
Since all sides of a rhombus are equal:
$$SP=PQ=RQ=PS=\frac{1}{\sqrt{2}}\text{units}$$

2. Finding the radius of the inscribed circle:
Let $O$ be the center of the square and the inscribed circle, and let $r$ be the radius of the circle. The diagonals of the rhombus, $PR$ and $SQ$, intersect perpendicularly at the center $O$ and have lengths equal to the side length of the square, which is 1 unit. Thus, the semi-diagonals are:
$$OR=OQ=\frac{1}{2}\text{units}$$
In the right-angled triangle $△ORQ$ (where $\angle ROQ={90}^{\circ }$), the line segment $OT$ is perpendicular to the side $RQ$ and represents the radius $r$ of the inscribed circle. We can express the area of $△ORQ$ in two different ways:
$$\text{Area}=\frac{1}{2}\times \text{base}\times \text{height}$$
Using the perpendicular sides $OR$ and $OQ$:
$$\text{Area}=\frac{1}{2}\times OR\times OQ=\frac{1}{2}\times \frac{1}{2}\times \frac{1}{2}=\frac{1}{8}$$
Using the hypotenuse $RQ$ and height $r$:
$$\text{Area}=\frac{1}{2}\times RQ\times r=\frac{1}{2}\times \frac{1}{\sqrt{2}}\times r$$
Equating both expressions for the area:
$$\frac{1}{8}=\frac{r}{2\sqrt{2}}$$
Solving for $r$:
$$r=\frac{2\sqrt{2}}{8}=\frac{1}{2\sqrt{2}}\text{units}$$

3. Calculating the diameter of the inscribed circle:
The diameter ($d$) of the inscribed circle is twice its radius:
$$d=2r=2\times \frac{1}{2\sqrt{2}}=\frac{1}{\sqrt{2}}\text{units}$$