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In right triangles, there is a fundamental relationship between the lengths of the *legs* and the length of the hypotenuse. This relationship is presented in the Pythagorean Theorem.

A right triangle is a specific type of triangle that contains a right angle. The side that lies opposite the right angle is always the longest and is known as the hypotenuse. The other sides are commonly called legs.

For a right triangle with legs $a$ and $b,$ and hypotenuse $c,$ the following is true.

$a_{2}+b_{2}=c_{2}$

This can be proven using area.

For any right triangle, the Pythagorean Theorem can be used to calculate the length on an unknown side when the lengths of the two other sides are known. Also, **if** the Pythagorean Theorem is true for a triangle in which all sides are known, then it can be concluded that the triangle is a right triangle.

Draw four copies of one right triangle with legs $a$ and $b$ and hypotenuse $c.$ Arrange them so they create a square.

Notice that the outer square has side lengths of $(a+b).$ Thus, it follows that the area of the square is $(a+b)_{2}.$ Alternately, the square can be seen as a composite shape, consisting of the four triangles and the inner square. Thus, its area can be expressed as the sum of the area of the triangles and the inner square. $area of inner squarearea of triangles :c_{2}:4(21 ab)⇔2ab $ The total area can be expressed as follows. $2ab+c_{2}$ Both expressions represent the area of the larger square. Therefore, they can be set equal to each other. $(a+b)_{2}=2ab$ The equation can be simplified using the fact that the square of a binomial is a perfect square trinomial.$(a+b)_{2}=2ab+c_{2}$

ExpandPosPerfectSquare$(a+b)_{2}=a_{2}+2ab+b_{2}$

$a_{2}+2ab+b_{2}=2ab+c_{2}$

SubEqn$LHS−2ab=RHS−2ab$

$a_{2}+b_{2}=c_{2}$

Find the length of $x.$

Show Solution

Since the triangle is a right triangle, we can use the Pythagorean Theorem to find the length of $x.$

The hypotenuse is the side opposite the right angle, for this triangle it measures $20$ units. The two other sides are the legs, $x$ and $12.$ We can now form an equation with the Pythagorean Theorem, that can be solved for $x$.$a_{2}+b_{2}=c_{2}$

SubstituteValuesSubstitute values

$12_{2}+x_{2}=20_{2}$

CalcPowCalculate power

$144+x_{2}=400$

SubEqn$LHS−144=RHS−144$

$x_{2}=400−144$

SubTermSubtract term

$x_{2}=256$

SqrtEqn$LHS =RHS $

$x=±256 $

CalcRootCalculate root

$x=±16$

$x>0$

$x=16$

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