A TV transmitting antenna is 98 m high and the receiving antenna is at the ground level. If the radius of the earth is 6400 km, the surface area covered by the transmitting antenna is approximately:

The TV transmitting antenna height is $$h = 98$$ m and the Earth's radius is $$R = 6400$$ km. To derive the maximum range, consider the geometry of a right triangle formed by the Earth's center, the antenna top, and the farthest point on the horizon. By the Pythagorean theorem:

$$(R + h)^2 = R^2 + d^2$$

Rearranging gives

$$d^2 = R^2 + 2Rh + h^2 - R^2 = 2Rh + h^2$$

Since $$h \ll R$$, we can neglect $$h^2$$, yielding

$$d = \sqrt{2Rh}$$

Converting $$h = 98$$ m to kilometers ($$h = 0.098$$ km) and substituting into the formula gives

$$d = \sqrt{2 \times 6400 \times 0.098} = \sqrt{1254.4} \approx 35.42\text{ km}$$

The transmitting antenna thus covers a circular disc of radius $$d$$ on the Earth's surface. Since

$$d^2 = 1254.4\text{ km}^2$$

the area is

$$A = \pi d^2 = \pi \times 1254.4 \approx 3.14159 \times 1254.4 \approx 3940\text{ km}^2$$

This is approximately 3942 km². The correct answer is Option B: 3942 km².