91Ó°ÊÓ

The formula for the area of a square is $A=s^2$, where s is the side length and the formula for the area of a triangle is $A=\frac{1}{2}×b×h$, where b is the base and h is the height of the triangle.

The figure has been labeled as follows:

Note that ABCD is a square since two of its adjacent sides, AB and AD are equal. Then, ABCD is a square of side length 4.1 cm, as $BC=4.1\mathrm{cm}$. Then, $CD=4.1\mathrm{cm}$as CD is also a side of the square ABCD.

Also, note that CDE is an isosceles triangle with $CE=DE$, such that its height is $EF=2.6\mathrm{cm}$and the base is $CD=4.1\mathrm{cm}$.

It is clear from the figure that $A=A_1+A_2$, where $A_1$ is the area of the square ABCD and $A_2$ is the area of the triangle CDE.

Put $s=4.1$in $A=s^2$to get,

$$\begin{gathered} A_1={\left(4.1\right)}^2 \\ =16.81 \end{gathered}$$

So, $A_1=16.81$.

Put $b=4.1$and $h=2.6$in $A=\frac{1}{2}×b×h$to get,

$$\begin{gathered} A_2=\frac{1}{2}×\left(4.1\right)×\left(2.6\right) \\ =4.1×1.3 \\ =5.33 \end{gathered}$$

So, role="math" localid="1642491226282" $A_2=5.33$.

Put $A_1=16.81$and $A_2=5.33$in $A=A_1+A_2$to get,

$$\begin{gathered} A=16.81+5.33 \\ =22.14 \end{gathered}$$

So, $A=22.14$.

Since the units of all measurements of lengths are given as centimeters, the units of both $A_1$and $A_2$are square centimeters. Thus, the unit of the required area is also square centimeters. So, the required area is 22.14 square centimeters.

Rounding off to the nearest tenth, the area of the given figure is 22.1 square centimeters.