Pythagoras Theorem

A Pythagoras Theorem is a mathematical relation amongest three sides of a right triangle namely the hypotenuse, adjacent and opposite.

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The Pythagoras Theorem has the following mathematical formula

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

${\mathrm{hypotenuse}}^2={\mathrm{adjacent}}^2+{\mathrm{opposite}}^2$

${\mathrm{hypotenuse}}^2={\mathrm{base}}^2+{\mathrm{height}}^2$

How to solve Pythagoras Theorem step by step

In the right angled-triangle ABC below find the value of x in meters(m)

Pythagoras Theorem Formula from the above diagram

${\mathrm{BC}}^2+{\mathrm{AB}}^2={\mathrm{AC}}^2$

BC = 7 + x , AB = 5 + x , AC = 2x

Substitute into the formula

${\mathrm{(7\; +\; x)}}^2+{\mathrm{(5\; +\; x)}}^2={\mathrm{(2x)}}^2$

Expand and simplify

$\mathrm{(7\; +\; x)(7\; +\; x)}+\mathrm{(5\; +\; x)(5\; +\; x)}=\mathrm{(2x)(2x)}$

$\mathrm{}7\times 7+7\times \mathrm{x}+\mathrm{x}\times 7+\mathrm{x}\times \mathrm{x}=5\times 5+5\times \mathrm{x}+\mathrm{x}\times 5+\mathrm{x}\times \mathrm{x}+{\mathrm{4x}}^2$

$\mathrm{49\; +\; 7x\; +\; 7x\; +}{x}^2=\mathrm{25\; +\; 5x\; +\; 5x\; +}{x}^2+{\mathrm{4x}}^2$

Adding like terms on each side we have

$\mathrm{49}+\mathrm{14x}+x^2=\mathrm{25}+\mathrm{10x}+{\mathrm{5x}}^2$

Collecting like terms we have

$\mathrm{49\; -\; 25\; +\; 14x\; -\; 10x\; +}{x}^2-{\mathrm{5x}}^2$

Adding and subtracting

$\mathrm{24}+\mathrm{4x}-{\mathrm{4x}}^2=0$

Rewriting the equation we have

${\mathrm{4x}}^2-\mathrm{4x}-\mathrm{24}=0$

From the above quadratic equation find the value of x by factorising

${\mathrm{4x}}^2-\mathrm{4x}-\mathrm{24}=0$

Simplify the quadratic equation

$\frac{{\mathrm{4x}}^2}{4}-\frac{\mathrm{4x}}{4}-\frac{\mathrm{24}}{4}=\frac{0}{4}$

$x^2-x-6=0$

product = -6

Sum = -1

Factors = -3, 2

$x^2+(\mathrm{2x}-\mathrm{3x})-6=0$

$x^2+\mathrm{2x}-\mathrm{3x}-6=0$

$x\left(\mathrm{x\; +\; 2}\right)-3\left(\mathrm{x\; +\; 2}\right)=0$

$\left(\mathrm{x\; +\; 2}\right)=0$ or $\left(\mathrm{x\; -\; 3}\right)=0$

$x=3$ or $x=\mathrm{-2}$

Pythagoras Example

Let AB and BC be the slant height in a diagram below. Hence find the lenth BC

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

Solutions:

${\mathrm{BC}}^2={\mathrm{BD}}^2+{\mathrm{DC}}^2$

Given that BC = ? , BD = 24mm and DC = 7mm

${\mathrm{BC}}^2={\mathrm{24}}^2+7^2$

${\mathrm{BC}}^2=\mathrm{24}+24+7+7$

${\mathrm{BC}}^2=\mathrm{576}+\mathrm{49}$

${\mathrm{BC}}^2=\mathrm{625}$

$\sqrt{\mathrm{BC}}=\sqrt{\mathrm{625}}$

Answers: $\mathrm{BC}=\mathrm{25mm}$

How to find distance using Pythagoras

Calculate the distance x taken by a boy to move from the cliff

$X^2={\mathrm{10}}^2+{\mathrm{12}}^2$

$X^2=100+144$

$X^2=244$

$\sqrt{X^2}=\sqrt{244}$

$X=15.62$

How to use pythagoras theorem

Two buildings are built next to each other and are joined by a footbrigdge of lenth 20 meters and the distance between than is 15 metres. Find the height h

${\mathrm{hypotenuse}}^2={\mathrm{adjacent}}^2+{\mathrm{opposite}}^2$

${\mathrm{20}}^2={\mathrm{15}}^2+h^2$

$h^2={\mathrm{20}}^2-{\mathrm{15}}^2$

$h^2=\mathrm{400}-225$

$h^2=\mathrm{175}$

$\sqrt{h^2}=\sqrt{\mathrm{175}}$

$h=\mathrm{13.23\; m}$

How to solve Pythagoras Theorem