$A$ family uses $8 \; kW$ of power.
$(a)$ Direct solar energy is incident on the horizontal surface at an average rate of $200 \; W$ per square meter. If $20 \%$ of this energy can be converted to useful electrical energy, how large an area is needed to supply $8 \; kW$?
$(b)$ Compare this area to that of the roof of a typical house.

(N/A) Power used by the family, $P = 8 \; kW = 8 \times 10^{3} \; W$.
Solar energy received per square meter $= 200 \; W/m^{2}$.
Efficiency of conversion from solar to electrical energy $= 20 \% = 0.2$.
Let the area required to generate the desired electricity be $A$.
The useful power generated is given by: $P = \text{Efficiency} \times \text{Area} \times \text{Incident solar energy rate}$.
$8 \times 10^{3} = 0.2 \times A \times 200$.
$8000 = 40 \times A$.
$A = \frac{8000}{40} = 200 \; m^{2}$.
$(b)$ The area of a solar panel required to generate $8 \; kW$ of electricity is $200 \; m^{2}$. $A$ typical house roof has dimensions roughly $14 \; m \times 14 \; m = 196 \; m^{2}$. Thus, the required area is approximately equivalent to the roof area of a typical house.