Pythagoras Theorem and Its Applications

You are at one end of the amusement park and your friend is at the other end. There are two ways you can reach your friend, either you go west for 3 miles and then walk north for another 4 miles which makes it a total of 7 miles, or you go right through the centre of the amusement park walking diagonally. You need to find the distance you need to travel for the second option. How do you do it? Well, you take the help of the Pythagoras Theorem…

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The Pythagoras Theorem

Let’s have a look at what Mr Pythagoras stated when he came up with the Theorem,

Statement: In a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the remaining two sides.

To understand it better we break down the statement.

1. A right-angled triangle is a triangle with a 90-degree angle.
2. The hypotenuse is the longest side of the right-angled triangle.
3. The remaining sides of the triangle are called the base and the perpendicular.

In the diagram above,

∠ABC is a right angle.

AC is the hypotenuse.

AB is known as the perpendicular.

BC is the base.

So according to the Pythagoras Theorem,

(AC)²=(AB)²+(BC)²

But then should we merely trust a single statement? I don’t think so. We need proof!

Browse more Topics under Triangles

• Properties of Triangles
• Congruent Triangles
• Similarity of Triangles
• Inequalities of Triangles
• Basic Proportionality Theorem and Equal Intercept Theorem

Download Triangles Cheat Sheet PDF

Proof of the Pythagoras Theorem using Similarity of Triangles:

Given: In ΔABC,  m∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

$In\; △ABC,$

m∠ABC=90°                                                                  (Given)

seg BD is perpendicular to hypotenuse AC              (Construction)

$Therefore,\; △ADB\sim △AB\mathrm{C\sim △BDC \; \; \; \; \; \; \; \; \; \; \; \; \; \; (Similarity\; of\; right-angled\; triangle)}$

$△ABC\sim △ADB$

$\mathrm{(AB/AC)=(AD/AB) \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; (congruent\; sides\; of\; similar\; triangles)}$

$A{B}^2=AD\times AC$(1)

$△BDC\sim △ABC$

$\frac{C\mathrm{D/}}{BC}=\frac{B\mathrm{C/}}{AC}$                                                        (congruent sides of similar triangles)

$B{C}^2=CD\times AC$(2)

Adding the equations (1) and (2),

$A{B}^2+B{C}^2=AD\times AC+CD\times AC$

$AB2+BC2=AC(AD+CD)$

Since, AD + CD = AC

Therefore, $A{C}^2=A{B}^2+B{C}^2$

Hence Proved. 

There also exists a Converse of the Pythagoras theorem that states, “If the square of one side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle”.

Solved Examples

Now that we proved the theorem lets work with some examples to understand it better.

Q: In a right-angled triangle $△LMN,\; LM\; is\; 3,\; MN\; is\; 4\; find\; the\; hypotenuse\; LN.$

Solution:

Given:  LM= 3, MN=4

To find: Hypotenuse LN=?

Since $△LMN\; is\; a\; right-angled\; triangle,\; we\; use\; the\; Pythagoras\; theorem\; to\; find\; the\; hypotenuse.$

$By\; Pythagoras\; Theorem,$

$LN2=LM2+MN2$

= 3² + 4² = 9 + 16

LN²= 25

LN= √25 =5

Therefore  LM is 5.

Q: In a right-angled triangle $△PQR,\; PQ\; is\; 5,\; hypotenuse\; PR\; is\; 13\; find\; the\; base\; QR.$

Solution:

Given:  PQ= 5, PR=13

To find: QR=?

Since $△PQR\; is\; a\; right-angled\; triangle,\; we\; use\; the\; Pythagoras\; theorem\; to\; find\; the\; SIDE\; QR.$

$By\; Pythagoras\; Theorem,$

$PR²=PQ²+QR²$

13²= 5²+QR²

169= 25 + QR²

QR²= 169-25

QR²=144

QR= √144 =12

Therefore QR is 12.

The Pythagorean Triples

Pythagorean Triples are a set of 3 numbers (with each number representing a side of the triangle) that are most commonly used for the Pythagoras theorem.

Let us assume a to be the perpendicular, b to be the base and c to be the hypotenuse of any given right angle triangle.

The simplest triple is 1, 0, 1

The following is a list of some of the commonly used Pythagorean Triples.

a	b	c
1	0	1
3	4	5
5	12	13
7	24	25
9	40	41
11	60	61
13	84	85
15	112	113

Let us try solving them.

For example, we take the triplet 5,12,13.

Since the longest side of the triangle is the hypotenuse, c=13, a=5 and b=12

The square of 12 and 13 is 144 and 169 respectively while the square of 5 is 25.

c²=169

a²+b²= 25+144 = 169

Therefore c²=a²+b²

Let us take another example of the triplet 2 3, 4, 5

c=5, a=3 and b=4

The square of 4 and 5 is 16 and 25 respectively while the square of 3 is 9.

c²=25

a²+b²= 9+16 = 25

Therefore c²=a²+b²

Scaling up the Triples

The primitive Pythagorean Triples can be scaled up to create further sets of triples.

For example, multiplying (3,4,5) by 2 will give us the triple(6,8,10)

 

Question- What is Pythagoras theorem used for?

Answer- We use the Pythagoras theorem for two-dimensional navigation. It is useful in finding out the shortest distance with the help of two lengths. Thus, you see that distances north and west are the two legs of the triangle so the shortest line which connects them is diagonal.

Question- What does Pythagoras theorem mean?

Answer– Pythagoras theorem in geometry refers to the square of the length of the hypotenuse of a right triangle which equals the sum of the squares of the lengths of the other two sides.

Question- How do I find the length of a triangle?

Answer- You can make use of Pythagoras’ Theorem to find the length of a triangle. We see that the hypotenuse is the longest side of a right triangle. It is situated opposite the right angle. Thus, if you are aware of the lengths of the two sides, you just need to square these two lengths. Then, add the result and take the square root of the sum to get the length of the hypotenuse.

Question- Define the Pythagorean Theorem.

Answer- The Pythagorean Theorem refers to a formula that provides a relationship between the sides of a right triangle. The Pythagorean Theorem is only applicable to Right triangles. A Right triangle is a triangle having a 90-degree angle.