In this post I’ll be demonstrating how one can derive the three formulas which can be used to find the areas of triangles.

These formulas are in fact:

To begin with, let’s start by looking at the diagram below:

Now, if you look at the diagram carefully – you will notice that the area of the triangle is:

This can be simplified into:

Because of SOH CAH TOA, what we can also say is that:

Now because:

This ultimately means that:

Alright, so far so good… Now we must put the icing on the cake and attach the final piece of the jigsaw puzzle to the formula above. In order to find the three equations which can be used to find the areas of triangles, we must now discover the expression for sin(C). We can discover its expression by first saying that:

In this post I’ll be showing you how to derive the formula for the area of an equilateral triangle – in easy steps. In order to understand this derivation properly, you need to be familiar with Pythagoras’ theorem and also a few algebraic rules. What you’ll also need is a ruler, pair of compasses, a pencil and a sheet of paper.

Ready? Let me begin…

Step 1: Put a point on a blank sheet of paper and name it A.

Step 2: Put the needle of your compass on the point A and draw a circle around it.

Step 3: Add a point B to this circle, on its edge.

Step 4: Put the needle of your compass on the point B and your pencil on the point A.

Step 5: Draw another circle with a radius the length AB.

Step 6: Now add a few extra points to your drawing. Call these points C and D.

Step 7: Connect the points A, B and C forming a triangle.

Step 8: Draw a line going through the points C and D.

Step 9: Where the line going through C and D intersects the triangle, place the point E.

Step 10: Now look at your latest work very carefully… What you will notice is that the lengths AB, AC and BC are all equal to one another. This is because both the circles you drew – are exactly the same size. They each have radiuses equal in proportion. In simple terms, AB=AC=BC.

What you have to do now is name these lengths (r) for radius. Here’s the thing though, because the line going through C and D splits the triangle (equilateral, as each of its sides has the same length) down its middle, the length AE is equal to 1/2 x r, and similarly the length BE is equal to 1/2 x r. Together, the length AE + BE = AB = r.

Step 11: Remember that I said that the line going through C and D splits the triangle down its middle. Also, notice that this exact line is perpendicular to the length AB. Now, because of this, at the point E, you’ve got two right angles. Name these two right angles big R.

[Knowing that these two angles are equal to 90 degrees is vital – because you’ll be able to use Pythagoras’ theorem to find the length CE.]

Step 12: Find the length CE using Pythagoras’ theorem, Adjacent² + Opposite² = Hypotenuse². You will need this length to find the area of the equilateral triangle you’ve produced.

Step 13: Derive the formula for the area (A) of the equilateral triangle. Remember that the area of a right angled triangle is L x W x 1/2.

Presto!!! Keep in mind that you can transform the variable (r) into any variable you wish. This variable (r) is the length of each side of the equilateral triangle you were working with. The formula you’ve derived can be used to find the area of any equilateral triangle.

In this post, I’ll be demonstrating how you can quickly double the area of a square using a simple geometrical trick.

Let’s say you have an ordinary square, like the one below…

Firstly, what you have to do is name the area of this square “A”…

Then, what you do next is divide this square (diagonally) into 4 equal parts…

After you have done this, you then name each part of this square “1/4 x A”…

Notice now, that to double the area of this square, all you have to do, is double the number of the 1/4 x A right angled triangles which currently exist – then configure them – like this…

As you can see, you’ve now got eight of these 1/4 x A right angled triangles neatly configured…

Not only are you left with a new square, double the size of your original square (follow the lines on the outside of the shape), but a handy equation, which proves that you doubled the area of the square you started off with…

In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…

In this diagram what you will discover is that:

*Please read the following contents carefully…

A lengtha exists, which starts at the origin O and ends at a;

A lengthx exists, which starts at the origin O and ends at x;

A lengthx+𝛿x exists, which starts at the origin O and ends at x+𝛿x;

A length𝛿x exists, which starts at x and ends at x+𝛿x;

A heighty exists, which starts at the origin O and ends at y;

A heighty+𝛿y exists, which starts at the origin O and ends at y+𝛿y;

A height𝛿y exists, which starts at y and ends at y+𝛿y;

There is a curve called y=f(x);

There is an area underneath the curve called A which commences at a and ends at x;

There is an area underneath the curve called 𝛿A which commences at x and ends at x+𝛿x (Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called 𝛿A);

There is a rectangle that exists called QRUT. It has an area which is y𝛿x;

There is a rectangle that exists called PRUS. It has an area which is (y+𝛿y)𝛿x;

𝛿A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.

Producing the formula with the information we’ve discovered…

Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…

Let’s start off by saying that:

Area QRUT < 𝛿A < Area PRUS

Which is something we already discovered.

If this is the case, we can say that:

y𝛿x < 𝛿A < (y+𝛿y)𝛿x

Now, check out what happens when we divide all the elements of this expression by 𝛿x:

What we end up with is…

Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:

This is because as 𝛿x approaches 0, 𝛿y approaches 0 leaving (𝛿A)/(𝛿x) sandwiched between y and y+0.000000000000000001 which is virtually y.

And, also…

As a consequence, this ultimately means that:

This is incredibly significant, because if we then integrate both sides of this equation, we get:

And…

Now, this equation can actually be used to find the areaA underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.

Finalising the formula…

Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:

However, it is not complete. We need to know what the constant C is. So…

If we say that at x=a the areaA underneath the curve is 0, watch what happens… Look at what we get…

Which means that:

Hence, we can conclude that:

And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b…

This is probably the formula you’re most familiar with…

Which is the formula which can be used to find areas underneath curves.

If you are still confused and would like to go through this proof once again, please watch my video below…

I’ve also got a new Facebook page related to stuff about the universe and also mathematics. You can add me as a friend by accessing this link: http://www.facebook.com/tiago.hands

GCSE + A Level Mathematics Proofs, Videos and Tutorials.

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