Step 1: Recall tiling rules.
To tile the plane perfectly, the shape's interior angles at a vertex must exactly sum to \(360^\circ\) when copies meet.
Step 2: Eliminate options.
- (A) Circle → cannot tile the plane, leaves gaps.
- (B) Regular octagon → does not tile by itself; needs squares to fill gaps.
- (C) Regular pentagon → cannot tile due to its \(108^\circ\) angle; multiples do not fit \(360^\circ\).
Step 3: Valid option.
(D) Rhombus → a parallelogram, and parallelograms always tessellate by repetition.
Final Answer:
\[
\boxed{\text{Rhombus}}
\]

In \(\triangle ABC\), \(DE \parallel BC\). If \(AE = (2x+1)\) cm, \(EC = 4\) cm, \(AD = (x+1)\) cm and \(DB = 3\) cm, then the value of \(x\) is

In the adjoining figure, PA and PB are tangents to a circle with centre O such that $\angle P = 90^\circ$. If $AB = 3\sqrt{2}$ cm, then the diameter of the circle is
In the adjoining figure, TS is a tangent to a circle with centre O. The value of $2x^\circ$ is