Q16. - Algebra 2

Question

In the figure below, what is the area of the shaded square in square units?

Step-by-Step Solution

Answer

The area of the shaded square is 13 units.

Step 1 ­- Label the given diagram.

The labelled diagram is:

Step 2 ­- Description of step.

From the given diagram it can be noticed that:

$AD=AE+ED$.

Therefore, it can be obtained that:

$\begin{array}{c}AD=AE+ED\\ =2+3\\ =5\end{array}$

Therefore, it can be noticed that:

$AD=AB=BC=CD=5$.

Therefore, $ABCD$ is a square.

Therefore, all the angles in the square $ABCD$ will be of measure $90°$ each.

Step 3 ­- Description of step.

Consider the measure of the angle $\angle DEH$ as x.

In the triangle $\Delta DEH$, it can be noticed that:

$\begin{array}{c}\angle DEH+\angle EHD+\angle EDH=180\\ x+\angle EHD+90=180\\ \angle EHD=180-90-x\\ \angle EHD=90-x\end{array}$

It is given that the shaded figure is a square that implies $EFGH$ is a square.

Therefore, all the angles in the square $EFGH$ will be of measure $90°$ each.

Now, from the given diagram it can also be noticed that the angles $\angle DEH$, $\angle HEF$ and $\angle FEA$ forms the linear pair.

Therefore, it can be obtained that:

$\begin{array}{c}\angle DEH+\angle HEF+\angle FEA=180\\ x+90+\angle FEA=180\\ \angle FEA=180-90-x\\ \angle FEA=90-x\end{array}$

Now as in the triangles $\Delta DEH$ and $\Delta AFE$, it can be noticed that:

$\begin{array}{l}\angle EHD=FEA=90-x\\ \angle EDH=FAE=90\end{array}$

Therefore, by AA similarity, the triangles $\Delta DEH$ and $\Delta AFE$ are similar.

Therefore, $\angle DEH=AFE=x$.

Similarly, it can be obtained that all the triangles $\Delta DEH$, $\Delta AFE$, $\Delta BGF$ and $\Delta CHG$ are similar.

Therefore, the diagram is:

Step 4 ­- Description of step.

Now it can be noticed that one side which is the hypotenuse of each triangle are also equal as they are the sides of the square $EFGH$.

Therefore, the triangles are congruent.

Therefore, it can be said that:

$\Delta DEH\cong \Delta AFE\cong \Delta BGF\cong \Delta CHG$.

Therefore, by corresponding parts of congruent triangles it can be noticed that:

$AF=BG=CH=DE=3$ and $EA=FB=GC=HD=2$

Therefore, the diagram is:

Step 5 ­- Description of step.

Now, in the triangle $\Delta GCH$, it can be noticed that:

$\begin{array}{c}G{C}^2+C{H}^2=G{H}^2\\ 2^2+3^2=G{H}^2\\ \sqrt{2^2+3^2}=GH\\ \sqrt{4+9}=GH\\ \sqrt{13}=GH\end{array}$

Therefore, the measure of the side $GH$ is $\sqrt{13}$.

Step 6 ­- Description of step.

The area of the square is given by:

$area=side\times side$

Therefore, the area of the square $EFGH$ having measure of each side as $\sqrt{13}$ is given by:

$\begin{array}{c}area=side\times side\\ =\sqrt{13}\times \sqrt{13}\\ =13\end{array}$

Therefore, the area of the shaded square is 13 units.