Equilateral and Equiangular Polygons

A polygon is a 2 dimensional geometric figure bound with straight sides. A polygon is called Equilateral if all of its sides are congruent. Common examples of equilateral polygons are a rhombus and regular polygons such as equilateral triangles and squares.  Now, a polygon is equiangular if all of its internal angles are congruent.  Some important facts to consider The only equiangular triangle is the equilateral triangle If P is an equilateral polygon that has more than three sides, it does not have to be equiangular. A rhombus with no right angle is an example of an equilateral but non-equiangular polygon.  Rectangles, including squares, are the only equiangular quadrilaterals Equiangular polygon theorem. Each angle of an equiangular n-gon is  $$\Bigg(\frac{n-2}{n}\Bigg)180^{\circ} = 180^{\circ} -   \frac{360^{\circ}}{n} $$ Viviani's theorem   Vincenzo Viviani (1622 – 1703) was a famous Italian mathematician. With his exceptional intelligence in math...
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Linear Equations - I

Equations are the foundation of mathematics. All forms of math rely on the principle of equality.
An equation states that two expressions have the same value. Expressions are what are on either side of the equation. This may seem obvious, but understanding this is essential.
Variables, or the letters we see in equations, are values that we do not know. We must manipulate the equation to find the value of the variable(s). Coefficients are the numbers located directly left of variables. The coefficient of a variable multiplies the variable’s value. For example, $5x$ means “five times x.”

 Linear equations are the building blocks of algebra. An example of a linear equation is 
$2x + 3 = 6$. To solve this equation, we must obtain the variable alone on one side and a simplified value on the other. This process is called isolating the variable. It uses principles of inverse operations: subtraction cancels addition, division cancels multiplication, etc.

 Example 1 : $2x+3=6$ Solve for $x$.

  • To isolate the variable $x$, we must eliminate the $+3$ and the coefficient $2$.
  • To eliminate $+3$, we subtract $3$ from the left and the right sides of the equation.  
         $ \Rightarrow$ $2x + 3 - 3 = 6 -3 $
         $ \Rightarrow$ $2x  = 3 $

Remember, whatever is done to one side of an equation must also be done to the other. This is the main rule of solving equations. If it is not followed, the two sides of the equation will no longer be equal.
  • The next step is to eliminate the coefficient $2$. Recall that $2x$ means $2 × x$. Reversing multiplication calls for division, so we divide both sides by $2$. 
       $ \Rightarrow$ $ \cfrac {2x}{2} = \cfrac {3}{2} $

  • This leaves us with $x = \cfrac {3}{2}$ . The variable is isolated and the other side is simplified, so we are done.
      $ \Rightarrow$ $ \boxed{x = \frac {3}{2}} $



Example 2 : $ \cfrac{3a+2}{2} = 6a + 3.$ Solve for $a$.
  • Remember, when manipulating one side of an equation, always act on the entire side, not just individual parts.
  • Multiplying both sides by $2$ yields
        $ \Rightarrow$ $3a + 2= 2(6a + 3) $
        $ \Rightarrow$ $3a + 2= 12a + 6 $
  • From here we subtract 3a from both sides to obtain all terms containing $a$ on one side of the equation. Doing so yields
        $ \Rightarrow$ $3a -3a + 2= 12a - 3a +6 $
        $ \Rightarrow$ $2= 9a +6 $
  • Next, we subtract 6 from both sides, obtaining $9a = −4$
  • Finally, we divide both sides by 9, obtaining the solution, 
        $\boxed{a = -\cfrac{4}{9}}$



Example 3 : Solve the equation $ \cfrac{7}{5}y -10 = y.$
  • Subtracting $y$ from both sides gives $ \Rightarrow$ $ \cfrac {2}{5}y - 10 = 0 $
  • Adding $10$ to both sides gives $ \Rightarrow$ $ \cfrac {2}{5}y = 10 $
  • How do we isolate y from here? The reciprocal of $ \cfrac {2}{5}$ is $ \cfrac {5}{2}$, and multiplying both sides by $ \cfrac {5}{2}$ yields $ \cfrac {5}{2} \cdot \cfrac{2}{5}y =  \cfrac {5}{2} \cdot 10$
  • Basic simplification gives us 
        $ \Rightarrow$ $\boxed{y= 25} $

Your turn to try 

$(1)$  $ \cfrac{2x-4}{7} = 4$. Solve for $x$

$(2)$  Solve for $ x + 12 = 2 $ $ \left ( \cfrac {x}{2} + \cfrac {3}{2}  \right ) $

Leave a comment if you need solutions to the above problems or if you are looking for additional practice problems. 

In the next post I will talk about using Cross Multiplication to solve equations and or expressions. Stay safe and stay healthy. 

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Equilateral and Equiangular Polygons

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