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##### Square

Square is a four sided shape which have all the sides of equal length and also the angles are equal, all 90°.

### The area and the perimeter of a square

The area of a square is the plane surrounded by the sides.

To find the area of the square we multiply side times side or we just square the side since all sides are equal.

$\displaystyle A=a\cdot a={{a}^{2}}$

The perimeter of a square is the total lengths of all sides. Since all sides are equal, we can say the perimeter of a square is 4 times of its side.

$\displaystyle P=4\cdot a$

### Properties

1. All the interior angles are right angles (90°)

2. All the sides of the square are equal.

3. The opposite sides of the square are parallel to each other.

4. The diagonal bisects each other at 90°

5. The two diagonals of the square are equal to each other.

6. The diagonal of the square divides it into two congruent isosceles triangles.

Example 1: On the figure below the perimeter of the square is $\displaystyle 28cm$. Find the diagonal  $\displaystyle AC$.

Solution: From the properties we know that all sides are equal $\displaystyle AB=BC=CD=DA=x$

The perimeter is: $\displaystyle P=x+x+x+x=28$

$\displaystyle 4x=28$

$\displaystyle x=7$

From the properties we also know that the diagonal of the square divides it into two congruent isosceles triangles.
So the triangle $\displaystyle ABC$ is an isosceles right angle triangle.

To find the diagonal we use the Pythagorean theorem

$\displaystyle {{\left( {AC} \right)}^{2}}={{\left( {AB} \right)}^{2}}+{{\left( {BC} \right)}^{2}}$

$\displaystyle {{\left( {AC} \right)}^{2}}={{(7)}^{2}}+{{(7)}^{2}}$

$\displaystyle {{\left( {AC} \right)}^{2}}=49+49$

$\displaystyle {{\left( {AC} \right)}^{2}}=98$

$\displaystyle AC=\sqrt{{98}}=7\sqrt{2}$

So the diagonal is $\displaystyle AC=7\sqrt{2}$

Example 2: The perimeter of the square is $\displaystyle \mathbf{a}$ and the area is $latex \displaystyle \mathbf{a}$.We know the sum of the perimeter and area is $\displaystyle a+b=45$. Find the length of the sides.

Solution: Firstly lets suppose that the side of our square is an unknown variable $\displaystyle x$

The perimeter: $\displaystyle P=x+x+x+x=a$

$\displaystyle 4x=a$

The area: $\displaystyle A=x\cdot x=b$

$\displaystyle {{x}^{2}}=b$

The sum of the perimeter and area is $\displaystyle a+b=45$

Substituting $\displaystyle a$ and $\displaystyle b$ we get

$\displaystyle 4x+{{x}^{2}}=45$

$\displaystyle {{x}^{2}}+4x-45=0$

$\displaystyle {{x}_{1}}=5$ and $\displaystyle {{x}_{2}}=-9$

Since we are talking about length of the side only $\displaystyle {{x}_{1}}=5$ is acceptable.

From the properties, all sides of a square are equal, so we found all the sides are 5 units.

TIP! To solve quadratic equations check Quadratic equation

Example 3: The length of a diagonal of a square is $\displaystyle 4\sqrt{2}$. Find its area and perimeter.

Solution: From the properties we know that the diagonal of the square divides it into two congruent  isosceles triangles.

So the triangle $\displaystyle ABC$ is an  isosceles right angle triangle.

Since the sides are equal we labbel them with the variable $\displaystyle x$.

To find the sides we use the Pythagorean theorem

$\displaystyle {{\left( {AB} \right)}^{2}}+{{\left( {BC} \right)}^{2}}={{\left( {AC} \right)}^{2}}$

$\displaystyle {{x}^{2}}+{{x}^{2}}={{\left( {4\sqrt{2}} \right)}^{2}}$

$\displaystyle 2{{x}^{2}}=32$

$\displaystyle {{x}^{2}}=16$

$\displaystyle x=4$ or ($\displaystyle x=-4$)

The perimeter: $\displaystyle P=4+4+4+4=16$

The area: $\displaystyle A=4\cdot 4=16$

We see that the area and the perimeter are the same, since we aren’t given units to measure we say:
The area is $\displaystyle 16\text{ }square\text{ }units$ and the perimeter is $\displaystyle 16\text{ }units$.

Be careful! When we talk about length of sides we are going to accept only the positive value of the variable.

Example 4: $\displaystyle ABCD$ is a square and $\displaystyle RST$ is  an equilateral triangle.The area of the lined figure is 3 time bigger then the area of the triangle. We have $\displaystyle RT=2$. Find $\displaystyle AB$.

Solution: Firstly lets label the area of the lined figure with $\displaystyle {{A}_{f}}$

The area of the triangle $\displaystyle {{A}_{{RTS}}}$

The area of the square $\displaystyle {{A}_{{ABCD}}}$

We know that the area of the lined figure is 3 times bigger then the area of the triangle $\displaystyle {{A}_{f}}=3{{A}_{{RTS}}}$

The area of the equilateral triangle is: $\displaystyle {{A}_{{RTS}}}=\frac{{{{a}^{2}}\sqrt{3}}}{4}$

Check out the area of an equilateral triangle

Where with “a” is labeled the side of the triangle that on our case is 2.

Substituting we get: $\displaystyle {{A}_{{RTS}}}=\frac{{{{{\left( 2 \right)}}^{2}}\sqrt{3}}}{4}=\sqrt{3}$

The area of the lined figure is $\displaystyle {{A}_{f}}=3{{A}_{{RTS}}}=3\sqrt{3}$

The area of the square is found by adding the area of the lined figure with the equilateral triangle:

$\displaystyle {{A}_{{ABCD}}}={{A}_{f}}+{{A}_{{RTS}}}$

Also the area of the square is found by squaring one side: $\displaystyle {{A}_{{ABCD}}}={{x}^{2}}$

We labeled the sides with the variable x since is unknown. $\displaystyle AB=x$

So we get:

$\displaystyle {{A}_{{ABCD}}}=3\sqrt{3}+\sqrt{3}=4\sqrt{3}$

$\displaystyle {{x}^{2}}=4\sqrt{3}$

$\displaystyle x=2\sqrt[4]{3}$

So the side is $\displaystyle AB=2\sqrt[4]{3}$

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