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##### Rectangle

A Rectangle is a four sided-shape where all the angles are right-angled (90o).

### Properties

1. It’s a with four right angles.

2. It’s diagonals bisect each other.

3. The opposite side of  a rectangle is equal.

4. The opposite side of a rectangle is parallel.

### The area and the perimeter of Rectangle

The area of a Rectangle is the plan enclosed by the sides of the rectangle.

The area is found by multiplying the length with the width.

Area = length x width

$\displaystyle A=a\cdot b$

The perimeter of a Rectangle is the total lengths of all sides.

The perimeter is 2 times the (length of one side +the other side)

$\displaystyle P=2\left( a+b \right)$

Example 1: On the figure below the sides of rectangle are $\displaystyle AC=25cm$ and $\displaystyle BC=7cm$. Find the area.

Solution: The diagonal separates the rectangle into two equal right angle triangle.

On triangle $\displaystyle ABC$ we find $\displaystyle AB$ by using the Pythagorean theorem

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

$\displaystyle {{\left( {AB} \right)}^{2}}={{\left( {25} \right)}^{2}}-{{\left( 7 \right)}^{2}}$

$\displaystyle {{\left( {AB} \right)}^{2}}=625-49$

$\displaystyle {{\left( {AB} \right)}^{2}}=567$

$\displaystyle AB=24cm$

Now that we know the width and length of the rectangle we find the area.

$\displaystyle Area=AB\cdot BC=24\cdot 7=168c{{m}^{2}}$

Example 2: On the figure below the sides of the rectangle are$\displaystyle AD=3cm$ and $\displaystyle DC=7cm$. The segment $\displaystyle CE=5cm$. Find the area of the triangle $\displaystyle ACE$?

Solution: Based on the property that opposite sides of rectangle are equal $\displaystyle AD=BC=3cm$

We observe the triangle $\displaystyle EBC$ that is a right angle triangle because of the right angle of the rectangle.

We find $\displaystyle EB$ by using the Pythagorean theorem

$\displaystyle {{\left( {EB} \right)}^{2}}={{\left( {EC} \right)}^{2}}-{{\left( {CB} \right)}^{2}}$

$\displaystyle {{\left( {EB} \right)}^{2}}={{\left( 5 \right)}^{2}}-{{\left( 3 \right)}^{2}}$

$\displaystyle {{\left( {EB} \right)}^{2}}=25-9$

$\displaystyle {{\left( {EB} \right)}^{2}}=16$

$\displaystyle EB=4$

To find the area of the triangle $\displaystyle ACE$ we can find the area of the rectangle and subtract the area of the two right angle triangles $\displaystyle ADC$ and $\displaystyle EBC$

$\displaystyle {{A}_{{ABCD}}}=AD\cdot DC=$$\displaystyle 7\cdot 3=21c{{m}^{2}} \displaystyle {{A}_{{ADC}}}=$$\displaystyle \frac{1}{2}AD\cdot DC=$$\displaystyle \frac{{21}}{2}c{{m}^{2}} \displaystyle {{A}_{{EBC}}}=\frac{1}{2}EB\cdot BC=$$\displaystyle \frac{{3\cdot 4}}{2}=6c{{m}^{2}}$

$\displaystyle {{A}_{{AEC}}}=$$\displaystyle {{A}_{{ABCD}}}-({{A}_{{ADC}}}+{{A}_{{EBC}}}) \displaystyle {{A}_{{AEC}}}=$$\displaystyle 21-(\frac{{21}}{2}+6)=$$\displaystyle \frac{{42}}{2}-\frac{{33}}{2}=\frac{9}{2}c{{m}^{2}}$

So the area of the triangle is $\displaystyle {{A}_{{AEC}}}=\frac{9}{2}=4,5c{{m}^{2}}$

Example 3: The perimeter of a rectangle is $\displaystyle 150cm$. If the widthis $\displaystyle 30cm$, find the length and area?

Solution: We know that the perimeter is found by adding the length of all sides.

Lets note one side with $\displaystyle x$.

From the properties the opposite sides of a rectangle are congruent so:

$\displaystyle P=x+x+30+30=150$

$\displaystyle 2x+60=150$

$\displaystyle 2x=90$

$\displaystyle x=45$

So the length is $\displaystyle 45cm$.

The area  is found by multiplying the length and width.

So $\displaystyle Area=45\cdot 30=1350c{{m}^{2}}$

Example 4: On the figure below we know that the diagonals are $\displaystyle 16cm$. Find x and y.

Solution: From the properties of rectangle we know that diagonals bisect each other and are congruent.

$\displaystyle AC=16$ so $\displaystyle AO=OC=\frac{{16}}{2}=8$

$\displaystyle AO=x+y=8$  and   $\displaystyle OC=3x-y=8$

Now to find $\displaystyle x$ and $\displaystyle y$ we have to solve this Simultaneous Linear Equations with two variables.

Solving simultaneously using Elemination Method

$\displaystyle \left\{ \begin{array}{l}x+y=8\\3x-y=8\end{array} \right.$

$\displaystyle \frac{\begin{array}{l}x+y=8\\3x-y=8\end{array}}{{4x=16}}$

$\displaystyle x=4$

Now we substitute to find the other variable

$\displaystyle 4+y=8$

$\displaystyle y=4$

So $\displaystyle x=4$ and $\displaystyle y=4$

Example 5: Find the length and width of the rectangle if it´s area is $\displaystyle 5c{{m}^{2}}$.

Solution: We know that the area of the rectangle is width x length

$\displaystyle Area=(x+3)(x-1)=5$

$\displaystyle (x+3)(x-1)=5$

$\displaystyle {{x}^{2}}-x+3x-3=5$

$\displaystyle {{x}^{2}}+2x-3-5=0$

$\displaystyle {{x}^{2}}+2x-8=0$

TIP! To solve quadratic equation check Quadratic equation

$\displaystyle {{x}_{1}}=2$ and $\displaystyle {{x}_{2}}=-4$

Since we are talking about width and length only $\displaystyle {{x}_{1}}=2$ is acceptable.

Finding the length: $\displaystyle x+3=2+3=5cm$

Finding the width: $\displaystyle x-1=2-1=1cm$

The proof: $\displaystyle A=5\cdot 1=5c{{m}^{2}}$