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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Train Problems

The key to solving moving train problems is to understand the distance that the needs to be covered or the relative speeds at which the objects are commuting in the problems. Lets take a look at both of these cases and talk about how to solve these types of questions. 

If the train is passing a Stationary object, think Distance
 - If the problem asks for the time taken by a moving train to pass a pole or standing man or anything similar that basically is a point, you will need to find the time taken by the train to cover the length of the train itself. 
 - If the problem asks for the time taken by a moving train to pass a bridge or a tunnel or anything that has a length of itself, then the time taken by the train to pass that object will be the length of the train $+$ the length of the object.

If the train is passing a Moving object, think Speed
 - If the problem involves $2$ moving objects, you would usually want to calculate the net speed between the two objects. Basically, if the two objects (Object A and Object B) are moving in the same direction, where Object A is moving faster than Object B, you would find the Net speed using the following formula. 
$$\text{Net Speed} = \text{Object A} - \text{Object B}$$
Conversely, if the $2$ objects are moving in opposite directions, then their Net Speed is found by adding the two speeds. In other words, 
$$\text{Net Speed} = \text{Object A} + \text{Object B}$$

You will come across problems that ask to calculate the time it takes for $2$ moving trains to cross each other. The concept we learned earlier about $T = \frac{D}{S}$ still applies here, but our distance is now the sum of the lengths of the $2$ trains and our speed becomes the Net Speed. We can write this as 
$$\text{Time} = \frac{\text{Length of object A + Length of object B}}{\text{Net Speed}}$$

Lets see these ideas in action

Problem 1: A train running at the speed of $60$ mph crosses a pole in $20$ seconds. What is the length of the train in feet?

Solution: We start with the basic $D = S \cdot T$. We know that the speed is $60$ mph, but because our given time to cross the pole is in seconds, we need to convert our speed in miles per second. This brings our speed to $\frac{1}{60}$ miles per second. Plugging it back in, we have $$D = \frac{1\text{ mile}}{60\text{ seconds}} \cdot 20 \text{ seconds} = \frac{1}{3} \text{ mile} = \boxed{1760 \text{ feet}}$$
($1$ mile = $5280$ feet)

Problem 2: $2$ trains of equal length are running on parallel tracks in the same direction at $40$ mph and $30$ mph. The faster train passes the slower train in $30$ seconds. Calculate the lengths of the trains in feet.

Solution: We can start by calling the length of both trains as $x$ feet each because the trains have equal lengths. This means that the total distance must by $2x$ feet. The time for the trains to cross each other is given as $30$ seconds, and the Net Speed would be $40-30 = 10$ mph because we know that if the two trains are moving in the same direction, we have to subtract their speeds for the Net Speed. Thus, using our $D = S \cdot T$ formula, we can say that
$$2x \text{ feet} = \frac{10 \text{ miles}}{\text{hour}} \cdot 30 \text{ sec}$$
We need to convert the miles per hour to miles per second, and doing that gives us
$$2x \text{ feet} = \frac{10 \text{ miles}}{3600 \text{ sec}} \cdot 30 \text{ sec} = \frac{1 \text{ mile}}{12} = 440 \text{ ft} \Rightarrow x = \boxed{220 \text{ ft}}$$

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Equilateral and Equiangular Polygons

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