JAMIA 2026 — Mathematics PYQ

Step 1: Calculate the lengths of the inner sides

The midpoints divide each side of the square into two equal parts of length $\frac{a}{2}$.

This creates four small right-angled triangles at the corners: $△APS,△BPQ,△CQR,$ and $△DRS$.

Using the Pythagorean theorem for the corner triangle $△APS$ (where $\angle A=90^{\circ }$):

$$P{S}^2=A{P}^2+A{S}^2$$

$$P{S}^2={\left(\frac{a}{2}\right)}^2+{\left(\frac{a}{2}\right)}^2$$

$$P{S}^2=\frac{a^2}{4}+\frac{a^2}{4}=\frac{2a^2}{4}=\frac{a^2}{2}$$

$$PS=\frac{a}{\sqrt{2}}$$

Since all four corner triangles are identical (congruent by SAS property), the side lengths of the inner quadrilateral $PQRS$ are all equal:

$$PQ=QR=RS=SP=\frac{a}{\sqrt{2}}$$

Because all four sides are equal, the inner quadrilateral is at least a rhombus.

Step 2: Determine the inner angles

Let's check the angles of the inner figure. Look at the isosceles right-angled triangle $△APS$:

$$\angle APS=\angle ASP=45^{\circ }$$

Similarly, in $△BPQ$:

$$\angle BPQ=\angle BQP=45^{\circ }$$

Since $APB$ is a straight line, the sum of angles on the line is $180^{\circ }$:

$$\angle APS+\angle SPQ+\angle BPQ=180^{\circ }$$

$$45^{\circ }+\angle SPQ+45^{\circ }=180^{\circ }$$

$$90^{\circ }+\angle SPQ=180^{\circ }$$

$$\angle SPQ=90^{\circ }$$

Conclusion:

Since the inner quadrilateral $PQRS$ has all sides equal and all interior angles equal to $90^{\circ }$, it satisfies all the definitive properties of a square.

Correct Answer

The correct option is (c) Square.