Tag Archives: areas

Finding the formulas for areas of triangles

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:

A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }

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

Triangle Diagram

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

A=\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 }

This can be simplified into:

\frac { x\cdot CN }{ 2 } +\frac { \left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { x\cdot CN+\left( c-x \right) \cdot CN }{ 2 } \\ \\ =\frac { CN\left\{ x+\left( c-x \right) \right\} }{ 2 } \\ \\ =\frac { CN\cdot c }{ 2 }

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

\sin { \left( A \right) } =\frac { O }{ H } =\frac { CN }{ b } \\ \\ \therefore \quad b\cdot \sin { \left( A \right) } =CN\\ \\ \sin { \left( B \right) =\frac { O }{ H } } =\frac { CN }{ a } \\ \\ \therefore \quad a\cdot \sin { \left( B \right) } =CN

Now because:

A=\frac { CN\cdot c }{ 2 }

This ultimately means that:

A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } \\ \\ A=\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) } \\ \\ \therefore \quad A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) =\frac { 1 }{ 2 } ac } \cdot \sin { \left( B \right) }

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:

C=\left( 90-A \right) +\left( 90-B \right) \\ \\ =90-A+90-B\\ \\ =180-A-B\\ \\ =180-\left( A+B \right) \\ \\ \therefore \quad \sin { \left( C \right) } =\sin { \left( 180-\left( A+B \right) \right) }

And if we use the trigonometric identity below:

\sin { \left( \alpha -\beta \right) } =\sin { \left( \alpha \right) \cdot \cos { { \left( \beta \right) } } -\cos { \left( \alpha \right) \cdot \sin { \left( \beta \right) } } }

We will reach the conclusion:

\sin { \left( 180-\left( A+B \right) \right) } =\sin { \left( 180 \right) \cdot \cos { \left( A+B \right) -\cos { \left( 180 \right) \cdot \sin { \left( A+B \right) } } } } 

But because:

\sin { \left( 180 \right) =0 } ,\quad \cos { \left( 180 \right) =-1 } \\ \\ \sin { \left( 180-\left( A+B \right) \right) =-\left( -1 \right) \cdot \sin { \left( A+B \right) } } \\ \\ \therefore \quad \sin { \left( C \right) =\sin { \left( A+B \right) } }

Now, sin(A+B) as a trigonometric identity, is:

\sin { \left( A+B \right) =\sin { \left( A \right) \cdot \cos { \left( B \right) +\cos { \left( A \right) \cdot \sin { \left( B \right) } } } } }

And, thanks to SOH CAH TOA…

\sin { \left( A+B \right) =\sin { \left( C \right) } } \\ \\ \sin { \left( A \right) =\frac { CN }{ b } } \\ \\ \cos { \left( B \right) =\frac { A }{ H } } =\frac { \left( c-x \right) }{ a } \\ \\ \cos { \left( A \right) =\frac { A }{ H } =\frac { x }{ b } } \\ \\ \sin { \left( B \right) =\frac { CN }{ a } }

Which means that…

\sin { \left( C \right) =\frac { CN }{ b } \cdot \frac { \left( c-x \right) }{ a } +\frac { x }{ b } \cdot \frac { CN }{ a } } \\ \\ =\frac { CN\left( c-x \right) }{ ab } +\frac { CN\cdot x }{ ab } \\ \\ =\frac { CN\left( c-x \right) +CN\cdot x }{ ab } \\ \\ =\frac { CN\left\{ \left( c-x \right) +x \right\} }{ ab } \\ \\ =\frac { CN\cdot c }{ ab } \\ \\ \therefore \quad ab\cdot \sin { \left( C \right) =CN\cdot c } \\ \\ \therefore \quad \frac { 1 }{ 2 } ab\cdot \sin { \left( C \right) =\frac { CN\cdot c }{ 2 } =A }

As this is the case, we can conclude that:

A=\frac { 1 }{ 2 } bc\cdot \sin { \left( A \right) } =\frac { 1 }{ 2 } ac\cdot \sin { \left( B \right) =\frac { 1 }{ 2 } } ab\cdot \sin { \left( C \right) }

How to derive the formula for the area of an equilateral triangle

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.

Point A

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

Circle...

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

Point B added...

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

Step 4...

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

Another circle...

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

Points C and D added to drawing...

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

Points A, B and C connected.

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

Line through points C and D.

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

Point E added...

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.

*Algebraic skills will be required from this point…

{ AE }^{ 2 }+{ CE }^{ 2 }={ AC }^{ 2 }\\ \\ \Rightarrow \quad { \left( \frac { 1 }{ 2 } r \right)  }^{ 2 }+{ CE }^{ 2 }={ r }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }={ r }^{ 2 }-{ \left( \frac { 1 }{ 2 } r \right)  }^{ 2 }\\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 4r^{ 2 } }{ 4 } -\frac { { r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad { CE }^{ 2 }=\frac { 3{ r }^{ 2 } }{ 4 } \\ \\ \Rightarrow \quad CE=\sqrt { \frac { 3{ r }^{ 2 } }{ 4 }  } \\ \\ \therefore \quad CE=\frac { r\sqrt { 3 }  }{ 2 } 

Length CE found...

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.

A=\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 }  }{ 2 } \cdot \frac { 1 }{ 2 } +\frac { 1 }{ 2 } r\cdot \frac { r\sqrt { 3 }  }{ 2 } \cdot \frac { 1 }{ 2 } \\ \\ =\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } +\frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =2\cdot \frac { 1 }{ 8 } { r }^{ 2 }\sqrt { 3 } \\ \\ =\frac { 1 }{ 4 } { r }^{ 2 }\sqrt { 3 } 

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.

How to quickly double the area of a square (simple geometry lesson)

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”…

Remember that A is the entire area of the square, therefore 1/4 x A is a quarter of the area of the square.

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…

8\times \frac { 1 }{ 4 } A=2A

How about that? 🙂

Latest Mathematics Proofs: August 2016

Recently I discovered a few more proofs, some related to A Level mathematics.  You can access these proofs by clicking on the links below.

Derive the formula of an ellipsehttps://plus.google.com/+mathsvideosforweb/posts/ZF8D3ghSNRD

Discover the distance between point A and B on the edge of a circle: https://plus.google.com/+mathsvideosforweb/posts/9fmuEJ4qeXY

Find the area of a sector within a circle: https://plus.google.com/+mathsvideosforweb/posts/cHt9t9PeSfg

Formula for a torus, derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/TS84TL44BYd

Create the mathematical singularity shown in movie documentaries: https://plus.google.com/+mathsvideosforweb/posts/7p2onSzNYte

Parameterised formula for a torus derived from scratch: https://plus.google.com/+mathsvideosforweb/posts/AZrVMiPVdbv

Derive the formula for an ellipsoid (normal and parameterised) from scratch: https://plus.google.com/+mathsvideosforweb/posts/Wijf94ztn3v

Other news:

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