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Difference between revisions of "Pythagoras Theorem"
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= What is the Pythagorean Theorem?  =
+
== What is the Pythagorean Theorem?  ==
−
== What is the Pythagorean Theorem? ==
+
=== What is the Pythagorean Theorem? ===
 
The Pythagoras Theorem is also referred to as the Pythagorean Theorem<math>.</math> Pythagorean Theorem is used to find a side of any right triangle. It is <math>a^2+b^2=c^2</math>, where <math>a</math> and <math>b</math> are the legs of the triangle, and <math>c</math> is the hypotenuse.
 
The Pythagoras Theorem is also referred to as the Pythagorean Theorem<math>.</math> Pythagorean Theorem is used to find a side of any right triangle. It is <math>a^2+b^2=c^2</math>, where <math>a</math> and <math>b</math> are the legs of the triangle, and <math>c</math> is the hypotenuse.
−
== Why is it useful? ==
+
=== Why is it useful? ===
 +
To find sides and angles of right triangles. Also, Trigonometry is pointless without it. If you know three angles of a triangle you can use the Pythagorean Theorem to find the sides or the area even if the angles are not right. It is probably the most famous Theorem in all of math!
 +
=== Can we prove it? ===
 +
Yes! The are hundreds of proves. I will just show you a few of them. Mathematicians even make a hobby of finding these proves. Even a US president made a published proof! Of, course this was a president in the 1800s because, well, presidents now are not really up to proving something like that. (You know, Trump and the others).
 +
== Proofs ==
 +
=== Proof 1 ===
 +
We use <math>[ABC] </math> to denote the area of triangle <math>ABC </math>.
 +
 
 +
Let <math>H </math> be the perpendicular to side <math>AB </math> from <math>{} C </math>.
 +
 
 +
<center>
 +
<asy>
 +
pair A, B, C, H;
 +
A = (0, 0);
 +
B = (4, 3);
 +
C = (4, 0);
 +
H = foot(C, A, B);
 +
 
 +
draw(A--B--C--cycle);
 +
draw(C--H);
 +
draw(rightanglemark(A, C, B));
 +
draw(rightanglemark(C, H, B));
 +
label("$A$", A, SSW);
 +
label("$B$", B, ENE);
 +
label("$C$", C, SE);
 +
label("$H$", H, NNW);
 +
</asy>
 +
</center>
 +
 
 +
Since <math>ABC, CBH, ACH</math> are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,
 +
<center>
 +
<math> \frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2} </math>.
 +
</center>
 +
But since triangle <math>ABC </math> is composed of triangles <math>CBH </math> and <math>ACH </math>, <math>[ABC] = [CBH] + [ACH] </math>, so <math>AB^2 = CB^2 + AC^2 </math>.  {{Halmos}}
 +
=== Proof 2 ===
 +
 
 +
Consider a circle <math>\omega </math> with center <math>B </math> and radius <math>BC </math>.  Since <math>BC </math> and <math>AC </math> are perpendicular, <math>AC </math> is tangent to <math>\omega </math>.  Let the line <math>AB </math> meet <math>\omega </math> at <math>Y </math> and <math>X </math>, as shown in the diagram:
 +
 
 +
<center>[[Image:Pyth2.png]]</center>
 +
 
 +
Evidently, <math>AY = AB - BC </math> and <math>AX = AB + BC </math>.  By considering the [[Power of a Point | power of point]] <math>A </math> with respect to <math>\omega </math>, we see
 +
 
 +
<center>
 +
<math>AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2 </math>.  {{Halmos}}
 +
</center>
 +
=== Proof 3 ===
 +
 
 +
<math>ABCD</math> and <math>EFGH</math> are squares.
 +
<center>
 +
<asy>
 +
pair A, B,C,D;
 +
A = (-10,10);
 +
B = (10,10);
 +
C = (10,-10);
 +
D = (-10,-10);
 +
 
 +
pair E,F,G,H;
 +
E = (7,10);
 +
F = (10, -7);
 +
G = (-7, -10);
 +
H = (-10, 7);
 +
 
 +
draw(A--B--C--D--cycle);
 +
label("$A$", A, NNW);
 +
label("$B$", B, ENE);
 +
label("$C$", C, ESE);
 +
label("$D$", D, SSW);
 +
 
 +
draw(E--F--G--H--cycle);
 +
label("$E$", E, N);
 +
label("$F$", F,SE);
 +
label("$G$", G, S);
 +
label("$H$", H, W);
 +
 
 +
label("a", A--B,N);
 +
label("a", B--F,SE);
 +
label("a", C--G,S);
 +
label("a", H--D,W);
 +
label("b", E--B,N);
 +
label("b", F--C,SE);
 +
label("b", G--D,S);
 +
label("b", A--H,W);
 +
label("c", E--H,NW);
 +
label("c", E--F);
 +
label("c", F--G,SE);
 +
label("c", G--H,SW);
 +
</asy>
 +
</center>
 +
<math>(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2</math>. {{Halmos}}
 +
 
 +
== Pythagorean Triples ==
 +
Pythagorean Triples are a group of integers a,b and c in which <math>a^2+b^2=c^2</math>. These are the first few:
 +
(3,4,5)  (5,12,13) (7,24,25) (8,15,17)
 +
 
 +
(9,40,41) (11,60,61) (12,35,37) (13,84,85)
 +
 
 +
(15,112,113) (16,63,65) (17,144,145) (19,180,181)
 +
 
 +
(20,21,29) (20,99,101) (21,220,221) (23,264,265)
 +
 
 +
(24,143,145) (25,312,313) (27,364,365) (28,45,53)
 +
 
 +
(28,195,197) (29,420,421) (31,480,481) (32,255,257)
 +
 
 +
(33,56,65) (33,544,545) (35,612,613) (36,77,85)
 +
 
 +
(36,323,325) (37,684,685)
 +
 
 +
And on and on...
 +
Remember that if <math>a^2+b^2=c^2</math> then <math>xa^2+xb^2=xc^2</math> so I did not include 6,8,10 or 10,24,26.
 +
 
 +
== Special Right triangles ==
 +
=== 45-90-45 ===
 +
==== Theorem ====
 +
Say you have a right triangle with angles 45, 45 and 90. Then if <math>a</math>(leg) is <math>x</math>, then <math>c</math>(long side) is <math>x\sqrt2</math>
 +
 
 +
==== Proof ====
 +
If this triangle has two equal angles then it has two equal sides. Therefore we can make an equation: <math>x^2+x^2=s^2</math>. <math>s</math> means side and is the hypotenuse. <math>x^2+x^2=s^2 \Rightarrow 2x^2=s^2 \Rightarrow \sqrt{2}x=s</math>. We proved it!
 +
 
 +
=== 30-60-90 ===
 +
==== Theorem ====
 +
If the angles of a right triangle are 30, 60 and 90, and if the short side is <math>x</math> then the long side is <math>2x</math> and the other leg is <math>x\sqrt{3}</math>.
 +
 
 +
==== Proof ====
 +
The long side is clearly <math>2x</math> because of the angles that are twice the size, so we can make an equation: <math>x^2+t^2=(2x)^2</math>. t is the other leg.
 +
 
 +
<math>x^2+t^2=(2x)^2 \Rightarrow t^2=(2x)^2-x^2 \Rightarrow t^2=3x^2 \Rightarrow t=\sqrt{3}x</math> We proved it!
 +
 
 +
== Pythagorean Theorem related word problems ==
 +
=== Problem 1 ===
 +
Hawick is 15 miles south of Abbotsford, and Kelso is 17 miles east of Abbotsford.
 +
What is the distance from Hawick to Kelso?
 +
==== Solution ====
 +
<math>15^2+17^2=x^2 \Rightarrow 514=x^2 \Rightarrow x=\sqrt{514}</math>
 +
=== Problem 2 ===
 +
A zip line starts on a platform that is 40 feet above the ground. The anchor for the zip line is 198 horizontal feet from the base of the platform.
 +
How long is the zip line?
 +
==== Solution ====
 +
<math>40^2+198^2=x^2 \Rightarrow 40804=x^2 \Rightarrow x=202</math>

Latest revision as of 10:12, 23 June 2019

What is the Pythagorean Theorem?

What is the Pythagorean Theorem?

The Pythagoras Theorem is also referred to as the Pythagorean Theorem$.$ Pythagorean Theorem is used to find a side of any right triangle. It is $a^2+b^2=c^2$, where $a$ and $b$ are the legs of the triangle, and $c$ is the hypotenuse.

Why is it useful?

To find sides and angles of right triangles. Also, Trigonometry is pointless without it. If you know three angles of a triangle you can use the Pythagorean Theorem to find the sides or the area even if the angles are not right. It is probably the most famous Theorem in all of math!

Can we prove it?

Yes! The are hundreds of proves. I will just show you a few of them. Mathematicians even make a hobby of finding these proves. Even a US president made a published proof! Of, course this was a president in the 1800s because, well, presidents now are not really up to proving something like that. (You know, Trump and the others).

Proofs

Proof 1

We use $[ABC]$ to denote the area of triangle $ABC$.

Let $H$ be the perpendicular to side $AB$ from ${} C$.

[asy] pair A, B, C, H; A = (0, 0); B = (4, 3); C = (4, 0); H = foot(C, A, B); draw(A--B--C--cycle); draw(C--H); draw(rightanglemark(A, C, B)); draw(rightanglemark(C, H, B)); label("$A$", A, SSW); label("$B$", B, ENE); label("$C$", C, SE); label("$H$", H, NNW); [/asy]

Since $ABC, CBH, ACH$ are similar right triangles, and the areas of similar triangles are proportional to the squares of corresponding side lengths,

$\frac{[ABC]}{AB^2} = \frac{[CBH]}{CB^2} = \frac{[ACH]}{AC^2}$.

But since triangle $ABC$ is composed of triangles $CBH$ and $ACH$, $[ABC] = [CBH] + [ACH]$, so $AB^2 = CB^2 + AC^2$. ∎

Proof 2

Consider a circle $\omega$ with center $B$ and radius $BC$. Since $BC$ and $AC$ are perpendicular, $AC$ is tangent to $\omega$. Let the line $AB$ meet $\omega$ at $Y$ and $X$, as shown in the diagram:

Pyth2.png

Evidently, $AY = AB - BC$ and $AX = AB + BC$. By considering the power of point $A$ with respect to $\omega$, we see

$AC^2 = AY \cdot AX = (AB-BC)(AB+BC) = AB^2 - BC^2$. ∎

Proof 3

$ABCD$ and $EFGH$ are squares.

[asy] pair A, B,C,D; A = (-10,10); B = (10,10); C = (10,-10); D = (-10,-10); pair E,F,G,H; E = (7,10); F = (10, -7); G = (-7, -10); H = (-10, 7); draw(A--B--C--D--cycle); label("$A$", A, NNW); label("$B$", B, ENE); label("$C$", C, ESE); label("$D$", D, SSW); draw(E--F--G--H--cycle); label("$E$", E, N); label("$F$", F,SE); label("$G$", G, S); label("$H$", H, W); label("a", A--B,N); label("a", B--F,SE); label("a", C--G,S); label("a", H--D,W); label("b", E--B,N); label("b", F--C,SE); label("b", G--D,S); label("b", A--H,W); label("c", E--H,NW); label("c", E--F); label("c", F--G,SE); label("c", G--H,SW); [/asy]

$(a+b)^2=c^2+4\left(\frac{1}{2}ab\right)\implies a^2+2ab+b^2=c^2+2ab\implies a^2 + b^2=c^2$. ∎

Pythagorean Triples

Pythagorean Triples are a group of integers a,b and c in which $a^2+b^2=c^2$. These are the first few: (3,4,5) (5,12,13) (7,24,25) (8,15,17)

(9,40,41) (11,60,61) (12,35,37) (13,84,85)

(15,112,113) (16,63,65) (17,144,145) (19,180,181)

(20,21,29) (20,99,101) (21,220,221) (23,264,265)

(24,143,145) (25,312,313) (27,364,365) (28,45,53)

(28,195,197) (29,420,421) (31,480,481) (32,255,257)

(33,56,65) (33,544,545) (35,612,613) (36,77,85)

(36,323,325) (37,684,685)

And on and on... Remember that if $a^2+b^2=c^2$ then $xa^2+xb^2=xc^2$ so I did not include 6,8,10 or 10,24,26.

Special Right triangles

45-90-45

Theorem

Say you have a right triangle with angles 45, 45 and 90. Then if $a$(leg) is $x$, then $c$(long side) is $x\sqrt2$

Proof

If this triangle has two equal angles then it has two equal sides. Therefore we can make an equation: $x^2+x^2=s^2$. $s$ means side and is the hypotenuse. $x^2+x^2=s^2 \Rightarrow 2x^2=s^2 \Rightarrow \sqrt{2}x=s$. We proved it!

30-60-90

Theorem

If the angles of a right triangle are 30, 60 and 90, and if the short side is $x$ then the long side is $2x$ and the other leg is $x\sqrt{3}$.

Proof

The long side is clearly $2x$ because of the angles that are twice the size, so we can make an equation: $x^2+t^2=(2x)^2$. t is the other leg.

$x^2+t^2=(2x)^2 \Rightarrow t^2=(2x)^2-x^2 \Rightarrow t^2=3x^2 \Rightarrow t=\sqrt{3}x$ We proved it!

Pythagorean Theorem related word problems

Problem 1

Hawick is 15 miles south of Abbotsford, and Kelso is 17 miles east of Abbotsford. What is the distance from Hawick to Kelso?

Solution

$15^2+17^2=x^2 \Rightarrow 514=x^2 \Rightarrow x=\sqrt{514}$

Problem 2

A zip line starts on a platform that is 40 feet above the ground. The anchor for the zip line is 198 horizontal feet from the base of the platform. How long is the zip line?

Solution

$40^2+198^2=x^2 \Rightarrow 40804=x^2 \Rightarrow x=202$

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