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(A) The area covered for broadcasting is given by $A = \pi d_T^2$,where $d_T$ is the range of the transmitting antenna.Given $A = 6.4\pi 	imes 10^9 \

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$31.25$

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(A) The area covered for broadcasting is given by $A = \pi d_T^2$,where $d_T$ is the range of the transmitting antenna.Given $A = 6.4\pi 	imes 10^9 \ m^2$,we have $\pi d_T^2 = 6.4\pi 	imes 10^9$.$d_T^2 = 64 	imes 10^8 \ m^2$,so $d_T = 8 	imes 10^4 \ m = 80 \ km$.The total line-of-sight distance $d_M$ between two antennas of heights $h_T$ and $h_r$ is $d_M = d_T + d_r = \sqrt{2Rh_T} + \sqrt{2Rh_r}$.Given $d_M = 100 \ km$ and $d_T = 80 \ km$,we have $d_r = d_M - d_T = 100 \ km - 80 \ km = 20 \ km$.The range of the receiving antenna is $d_r = \sqrt{2Rh_r}$.$20 	imes 10^3 \ m = \sqrt{2 	imes (6400 	imes 10^3 \ m) 	imes h_r}$.Squaring both sides: $(20 	imes 10^3)^2 = 2 	imes 6400 	imes 10^3 	imes h_r$.$400 	imes 10^6 = 12800 	imes 10^3 	imes h_r$.$h_r = \frac{400 	imes 10^6}{12.8 	imes 10^6} = \frac{400}{12.8} = 31.25 \ m$.

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