8 1 practice the pythagorean theorem and its converse

Hotmath

• Home

The Converse of Philosopher Theorem

We assume you're old with the Pythagorean Theorem.

The converse of the Pythagorean Theorem is:

If the square of the length of the longest side of a triangle is same to the sum of the squares of the other two sides, then the triangle is a right triangle.

That is, in $\Delta ABC$, if $c^2=a^2+b^2$ then $\angle C$ is a rightfulness triangle, $\Delta PQR$ being the rightfield angle.

We can test this by contradiction.

Let America assume that $c^2=a^2+b^2$ in $\Delta ABC$ and the triangle is not a right triangle.

At present consider some other triangle $\Delta PQR$. We construct $\Delta PQR$ so that $PR=a$, $QR=b$ and $\angle R$ is a quadrant.

By the Pythagorean Theorem, ${\left(PQ\right)}^2=a^2+b^2$.

Merely we know that $a^2+b^2=c^2$ and $a^2+b^2=c^2$ and $c=AB$.

So, ${\left(PQ\right)}^2=a^2+b^2={\left(AB\right)}^2$.

That is, ${\left(PQ\right)}^2={\left(AB\right)}^2$.

Since $PQ$ and $AB$ are lengths of sides, we can take positive direct roots.

$PQ=AB$

That is, all the three sides of $\Delta PQR$ are congruent to the ternion sides of $\Delta ABC$. So, the two triangles are congruent by the Lateral-Side-Broadside Congruence Dimension.

Since $\Delta ABC$ is harmonious to $\Delta PQR$ and $\Delta PQR$ is a right triangle, $\Delta ABC$ must too be a right triangle.

This is a contradiction. Consequently, our supposal moldiness be wrong.

Example 1:

Check whether a trilateral with side lengths $6$ cm, $10$ centimetre, and $8$ cm is a right triangle.

Match whether the square of the duration of the longest side is the sum of the squares of the other deuce sides.

$\begin{array}{l}{\left(10\right)}^2\stackrel{?}{=}{\left(8\right)}^2+{\left(6\right)}^2\\ 100\stackrel{?}{=}64+36\\ 100=100\end{array}$

Hold the converse of Philosopher Theorem.

Since the square of the length of the longest side is the sum of the squares of the other two sides, by the converse of the Pythagorean Theorem, the triangle is a right trilateral.

A corollary to the theorem categorizes triangles in to acute, right, operating theater obtuse.

In a triangle with side lengths $a$, $b$, and $c$ where $c$ is the length of the longest go with,

if $c^2<a^2+b^2$ past the triangle is discriminating, and

if $c^2>a^2+b^2$ then the Triangulum is obtuse.

Example 2:

Halt whether the triangle with the side lengths $5$, $7$, and $9$ units is an acute, far, or obtuse triangle.

The longest side of the triangle has a distance of $9$ units.

Compare the square of the length of the longest side and the sum of squares of the other two sides.

Square of the length of the longest lateral is $9^2=81$ sq. units.

Kernel of the squares of the strange two sides is

$\begin{array}{l}{5}^2+7^2=25+49\\ =74\text{ sq}\text{. units}\end{array}$

That is, $9^2>5^2+7^2$.

Therefore, by the corollary to the converse of Pythagorean Theorem, the triangle is an obtuse Triangulum.

8 1 practice the pythagorean theorem and its converse

Source: https://www.varsitytutors.com/hotmath/hotmath_help/topics/converse-of-pythagorean-theorem

Berbagi :