Zr (Z=40) and Hf (Z=72) have similar atomic and ionic radii because of :

The correct answer is lanthanoid contraction.

Let us understand the reason behind this step-by-step:

1. Position in the Periodic Table:
Zirconium (Zr, $Z=40$) belongs to the 5th period and Group 4 of the d-block.
Hafnium (Hf, $Z=72$) belongs to the 6th period and Group 4 of the d-block, lying directly below zirconium.

2. General Trend in Atomic Radii:
Normally, as we move down a group in the periodic table, the atomic and ionic radii increase because of the addition of new principal energy shells (extra shells of electrons).

3. The Intervention of the Lanthanoids:
Between zirconium ($Z=40$) and hafnium ($Z=72$), the 14 lanthanoid elements (from Cerium, $Z=58$ to Lutetium, $Z=71$) are filled. In these elements, electrons are added to the inner 4f subshell.

4. Shielding Effect and Lanthanoid Contraction:
The f-orbitals have a very diffuse shape, which results in extremely poor shielding of the outer electrons from the nuclear charge.
As the nuclear charge increases by 1 unit for each subsequent element in the lanthanoid series, the poor shielding effect of the 4f electrons cannot counteract the increased nuclear attraction. Consequently, the outer electron cloud is pulled closer to the nucleus. This steady decrease in atomic and ionic size along the lanthanoid series is known as the lanthanoid contraction.

5. Effect on Hafnium:
Due to this lanthanoid contraction, the expected increase in size of Hf (from adding a new shell) is almost exactly compensated for by the contraction in size over the preceding elements.
As a result, Zr and Hf have nearly identical atomic radii (Zr ≈ 160 pm, Hf ≈ 159 pm) and ionic radii (Zr⁴⁺ ≈ 72 pm, Hf⁴⁺ ≈ 71 pm).