As given that height of antenna $h_T=20 \mathrm{~m}$ Radius of earth $=6.4 \times 10^6 \mathrm{~m}$ Receiving antenna is at ground level : -
(i) Range of transmitting antenna $\left(r_1\right)$ $=\sqrt{2 \mathrm{hR}}=\sqrt{2 \times 20 \times 6.4 \times 10^6 \mathrm{~m}}=16000 \mathrm{~m}=16 \mathrm{~km}$
Area covered by transmitting antenna $=\pi(\text { range })^2$ $=3.14 \times 16 \times 16=803.84 \mathrm{~km}^2$
(ii) Receving antenna at a height of $\mathrm{H}=25 \mathrm{~m}$ from ground level. Range due to both antenna $\left(\mathrm{r}_2\right)$.
$$
\begin{aligned}
&\text { Range }\left(\mathrm{r}_2\right)=\sqrt{2 \mathrm{hR}}+\sqrt{2 \mathrm{HR}} \\
&=\sqrt{2 \times 20 \times 6.4 \times 10^6}+\sqrt{2 \times 25 \times 6.4 \times 10^6} \\
&=16 \times 10^3+17.9 \times 10^3=33.9 \times 10^3 \mathrm{~m}=33.9 \mathrm{~km}
\end{aligned}
$$
Area covered by the transmitted and receiving antenna $=\pi(\text { Range })^2=3.14 \times 33.9 \times 33.9$
$=3608.52 \mathrm{~km}^2$
Percentage increase in area in two transmissions
$$
=\frac{\text { Difference in area }}{\text { Initial area }} \times 100
$$

$$
=\frac{(3608.52-803.84)}{803.84} \times 100=348.9 \%
$$

$$
\text { Thus, the percentage increase in area covered is } 348.9 \% \text {. }
$$