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Pythagoras Theorem

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Jetzt kostenlos anmeldenThe Pythagoras theorem is attributed to a Greek mathematician Pythagoras and his group, the Brotherhood of the Pythagoras over 2000 years ago. Their contribution in mathematics developed an algebraic method applied in geometry.

The Pythagoras theorem tells us that if the sides of a right angled-triangle or right triangle are squares, the area of the biggest square is the same as the sum of the area of the two smaller squares.

So, the area of the biggest square equals the sum of the areas of both smaller squares;

${a}^{2}={b}^{2}+{c}^{2}$

This is what the Pythagoras theorem explains.

Therefore, the Pythagoras theorem states that when a triangle has one of it angles equal to 90 degrees, then the square of the longest side equals the sum of the squares of the two other sides.

The longest side is called the** ****hypotenuse**, the vertical side is called the **opposite** and the horizontal side is called the **adjacent**.

So the formula of the Pythagoras theorem is;

$hypotenus{e}^{2}=opposit{e}^{2}+adjacen{t}^{2}$

Find the value of x in the figure below;

Using Pythagoras theorem, we can see that our opposite and adjacent is given but out hypotenuse is given as x. Thus;

${\mathrm{hypotenuse}}^{2}={\mathrm{opposite}}^{2}+{\mathrm{adjacent}}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}={5}^{2}+{12}^{2}\phantom{\rule{0ex}{0ex}}{x}^{2}=25+144\phantom{\rule{0ex}{0ex}}{x}^{2}=169\phantom{\rule{0ex}{0ex}}$

Find the square root of both sides

$x=13cm$

If a right angled triangle has equal dimension in two of it sides and the longest side measures 8cm. Find the other sides.

Solution.

From the question our hypotenuse is given as 8cm. However, the opposite and adjacent are not given. Also, we are told opposite = adjacent.

let the adjacent = y; that means opposite = y. Therefore:

Using pythagoras theorem, ${\mathrm{hypotenuse}}^{2}={\mathrm{opposite}}^{2}+{\mathrm{adjacent}}^{2}\phantom{\rule{0ex}{0ex}}{8}^{2}={y}^{2}+{y}^{2}\phantom{\rule{0ex}{0ex}}64=2{y}^{2}\phantom{\rule{0ex}{0ex}}$

Divide both sides by 2

$\frac{64}{2}=\frac{2{y}^{2}}{2}\phantom{\rule{0ex}{0ex}}32={y}^{2}\phantom{\rule{0ex}{0ex}}$

Find the square root of both sides of the equation

$\sqrt{32}=\sqrt{{y}^{2}}\phantom{\rule{0ex}{0ex}}y=4\sqrt{2}cm\phantom{\rule{0ex}{0ex}}$

So the opposite is 4√2cm and the adjacent is 4√2cm.

If sinØ = 2/5, Find cosØ and tanØ.

Solution

$Sin\xd8=\frac{opposite}{hypotenuse}=\frac{2}{5}\phantom{\rule{0ex}{0ex}}$

This means that the opposite is 2 and the adjacent is 5. Meanwhile, we need to find the adjacent:${\mathrm{hypotenuse}}^{2}={\mathrm{opposite}}^{2}+{\mathrm{adjacent}}^{2}\phantom{\rule{0ex}{0ex}}{5}^{2}={2}^{2}+adjacen{t}^{2}\phantom{\rule{0ex}{0ex}}9=4+adjacen{t}^{2}\phantom{\rule{0ex}{0ex}}$

Subtract 4 from both sides of the equation.

$9-4=4+adjacen{t}^{2}-4\phantom{\rule{0ex}{0ex}}5=adjacen{t}^{2}\phantom{\rule{0ex}{0ex}}$

Take the square roots.

$adjacent=\sqrt{5}\phantom{\rule{0ex}{0ex}}$

Now, we have values for all sides.

$\mathrm{cos}\varnothing =\frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}\phantom{\rule{0ex}{0ex}}\mathrm{cos}\varnothing =\frac{\sqrt{5}}{5}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Tan}\varnothing =\frac{\mathrm{opposite}}{\mathrm{adjacent}}\phantom{\rule{0ex}{0ex}}tan\varnothing =\frac{2}{\sqrt{5}}\phantom{\rule{0ex}{0ex}}$

Rationalize by multiplying the denominator and numerator by √5.

$tan\varnothing =\frac{2\times \sqrt{5}}{\sqrt{5}\times \sqrt{5}}\phantom{\rule{0ex}{0ex}}tan\varnothing =\frac{2\sqrt{5}}{5}$

A Pythagorean triple consists of 3 sets of numbers which prove correctly the Pythagoras theorem. This means that the square of the highest number among this numbers must be equal to the sum of the squares of the other two numbers in the set.

Determine if the following is a Pythagorean triple.

1. 7, 12 and 5

2. 8, 15 and 17

Solution

1. To confirm if the series 7, 12 and 5 are Pythagorean triples, take the square of the largest number.

The largest number is 12 and its square is 144.

You should sum the squares of the other two numbers in the series.

the square of 7 is 49

the square of 5 is 25

49+25 = 74

$144\ne 74\phantom{\rule{0ex}{0ex}}{12}^{2}\ne {7}^{2}+{5}^{2}$

This means that the series 7, 12 and 5 is not a Pythagorean triple.

2. To confirm if the series 8, 15 and 17 are Pythagorean triples, take the square of the largest number.

The largest number is 17 and its square is 289.

You should sum the squares of the other two numbers in the series.

the square of 8 is 64

the square of 15 is 225

64+225 = 289

$289=289\phantom{\rule{0ex}{0ex}}{17}^{2}={8}^{2}+{15}^{2}$

This proves that the set 8, 15 and 17 is a Pythagorean triple.

- The Pythagoras theorem tells us that if the sides of a right angled-triangle or right triangle are squares, the area of the biggest square is the same as the sum of the area of the two smaller square.
- The longest side is called the
**hypotenuse**, the vertical side is called the**opposite**and the horizontal side is called the**adjacent**. The formula of the Pythagoras theorem is; $hypotenus{e}^{2}=opposit{e}^{2}+adjacen{t}^{2}$

A Pythagorean triple consists of 3 sets of numbers which prove correctly the Pythagoras theorem.

The hypothenuse in Pythagorean theorem is the longest side of the right triangle.

The formula of the Pythagoras theorem is; hypothenuse square = opposite square + adjacent square

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