Answer
$2$ square units
Work Step by Step
$(1)$ A rectangle having length $l$ units and breadth $b$ units has area $A=lb$ square units. Let rectangles in the figure $(b)$ be labeled as $R_1, R_2, R_3, \ldots$ so rectangles $R_1, R_2, R_3, \ldots$ will have area $A_1, A_2, A_3, \ldots$ square units respectively. The rectangle $R_1$ has length $1$ unit and breadth $1$ unit. The rectangle $R_2$ has length $1$ unit and breadth $\frac{1}{2}$ unit. The rectangle $R_3$ has length $1$ unit and breadth $\frac{1}{4}$ unit. The rectangle $R_4$ has length $1$ unit and breadth $\frac{1}{8}$ unit and so on. Hence,
Area of rectangle $R_1 = 1 \times 1 = 1$ square unit
Area of rectangle $R_2 = 1 \times \frac{1}{2} = \frac{1}{2}$ square unit
Area of rectangle $R_3 = 1 \times \frac{1}{4} = \frac{1}{4}$ square unit
Area of rectangle $R_4 = 1 \times \frac{1}{8} = \frac{1}{8}$ square unit
and so on.
$(2)$ We find sum of areas of rectangles $R_1, R_2, R_3, \ldots$ It is
\begin{align*}
\sum &= A_1 + A_2 + A_3 + A_4 + \cdots \\
&= 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots \\
&= 1 + \left(\frac{1}{2}\right)^2 + \left(\frac{1}{2}\right)^3 + \left(\frac{1}{2}\right)^4 + \cdots \\
&= \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k
\end{align*}
$(3)$ The series $\sum = \sum_{k=0}^{\infty} \left(\frac{1}{2}\right)^k$ is a geometric series $\sum_{k=0}^{\infty} ar^k$ with $a=1$ and $r=\frac{1}{2}$. Its value is
\begin{align*}
\sum &= \frac{a}{1-r} \\
&= \frac{1}{1-\frac{1}{2}} \\
&= 2
\end{align*}
$(4)$ Thus the sum of areas of rectangles $R_1, R_2, R_3, \ldots$ is $2$ square units.
\begin{align*}
A_1 + A_2 + A_3 + A_4 + \cdots &= 2 \text{ square units.}
\end{align*}
