Pythagoras Theorem: Formula, Theorem, Proof and Examples

Jasmine Grover logo

Jasmine Grover

Content Strategy Manager

Pythagoras theorem is a geometric theorem in which the sum of squares on the foot of the right triangle is equal to the square of its hypotenuse. This theorem is used to find the length of the sides and angles of a right-angled triangle that is not known. A right-angled triangle is a polygon having 3 sides, having one of the angles as right angle (90°). Using this theorem, we can extract the perpendicular, base, and hypotenuse formulas.

Check out: Who is the father of mathematics?


Pythagoras Theorem

[Click Here for Sample Questions]

The theorem states that “The square of the hypotenuse in a right-angled triangle is equal to the sum of squares of the other two sides”. For a right-angled triangle, the Hypotenuse is considered the longest side which is opposite to the 90° angle. The other two sides referred to in the definition are called Base or adjacent and Perpendicular or the height.

The sides of a right-angled triangle (a, b, c) come with a positive integer value when they are squared and put into the equation known as the Pythagorean Triple.

The Pythagoras theorem is only applicable when the triangle is a Right-angled Triangle.

Pythagoras Theorem

Pythagoras Theorem

The video below explains this:

Pythagoras Theorem Detailed Video Explanation:


Pythagoras Theorem Formula 

[Click Here for Sample Questions]

Consider the above right-angled triangle PQR. In this triangle –

PQ= “a” is the perpendicular,

QR= “b” is the base and 

PR= “c” is the hypotenuse.

Angle Q is 90° angle

So, according to the definition given by Pythagoras, the Pythagorean Theorem Formula is given by-

Hypotenuse2 = Perpendicular2 + Base2

i.e. c2=a2+b2 

Consider 3 squares a, b, c on three sides of a triangle as shown in the figure below.

Now, by Pythagoras Theorem-

Pythagoras Theorem Formula

Pythagoras Theorem Formula

Area of square “c” = Area of square “a” + Area of square “b”.

Note: “a” and “b” are the other two sides and “c” is the longest axis.

Also Read:

Related Articles
Basic proportionality Theorem Ceva’s Theorem                   
Locus  Bayes’ Theorem                 

Proof of Pythagoras Theorem

[Click Here for Sample Questions]

Let us consider a Right-angled triangle PQR, right-angled at Q.

To Prove: PR2=PQ2+QR2

Construction: Draw a perpendicular QS meeting at PR at S.

Construction: Draw a perpendicular QS meeting at PR at S.

A perpendicular QS meeting at PR at S

Proof

We know, ΔRSQ~ΔRQP

Therefore, RSRQ=RQRP (Corresponding sides of Similar Triangle)

Or, RQ2=RS * RP…………………..(1)

Also, ΔQSP~ΔRQP

Therefore, PSQP=QPRP (Corresponding sides of Similar Triangle)

Or, QP2=PS * RP…………………..(2)

Adding the equations (1) & (2) we get,

RQ2 + QP2= RS * RP + PS * RP 

RQ2 + QP2= RP( RS+PS)

Since RS+PS=RP

Therefore, RP2=RQ2+QP

Or, c2= b2+a2

Or, c2 =a2+b2

Or, Hypotenuse2 = Perpendicular2 + Base2

Hence, The Pythagoras Theorem is proved.


The Converse of Pythagoras Theorem

[Click Here for Sample Questions]

The converse of Pythagoras theorem can be stated as: A triangle can be proved to be a right-angle triangle when the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides. 

In \(\bigtriangleup\) ABC, if C= a+ b2, then \(\angle\)C is a right angle triangle, \(\bigtriangleup\)PQR being the right angle

The Converse of Pythagoras Theorem

The Converse of Pythagoras Theorem

This can be proved by contraction

Let’s assume c2 = a2 + b2 in \(\bigtriangleup\)ABC and the triangle is not right triangle.

Then let's assume another \(\bigtriangleup\)PQR. Through the construction of \(\bigtriangleup\)PQR so that PR = a, QR = b and \(\angle\)R is a right angle.

The Converse of Pythagoras Theorem

The Converse of Pythagoras Theorem

By the Pythagoras theorem, (PQ)2 = a2 + b2

But we are aware that a2 + b2 = c2 and c = AB

Hence (PQ)2 = a2 + b2 = (AB)2

That is (PQ)2 = (AB)2

Since PQ and AB are the lengths of the sides, we can consider them to be positive square roots

PQ = AB

That is all the three given sides of \(\bigtriangleup\)PQR is congruent to all the three sides of \(\bigtriangleup\)ABC. Hence the two given triangles are congruent by the Side-Side-Side property of congruency.

Since the \(\bigtriangleup\)ABC is congruent to \(\bigtriangleup\)PQR and it is a right angle triangle, therefore \(\bigtriangleup\)ABC is also a right-angle triangle.

This is a contradiction, hence the assumptions we made must be wrong. 


Application of Pythagoras Theorem

[Click Here for Sample Questions]

  • This theorem helps us to find the diagonal of a square.
  • Pythagoras Theorem helps us to find whether a triangle is a right-angled triangle or not.
  • For a Right-angled triangle using this Theorem, we can find the length of the unknown side of the other two sides are known
  • Pythagoras theorem enables us to calculate the diagonal length of a roof, the distance between the foot of a slanted ladder, the perpendicular height of the wall or bean on which the ladder is slanted.
  • Pythagoras theorem is very useful for civil engineering projects
  • This theorem is also used for navigation purposes as it is difficult to measure the distance across water bodies.
  • Cartographers also use the Pythagoras theorem to find the distance between two points on a 2D sheet or map
  • Pythagoras theorem can also be used to find the steepness of the hill slope or mountain slope

Things to Remember

  • Pythagoras theorem:  \(c = \sqrt{a^2+b^2}\)
  • Pythagoras theorem states the relation between all the three sides of a right triangle.
  • It helps in the calculation of the sides and angles of a triangle.
  • Pythagoras theorem helps in understanding whether a triangle is a right-angled triangle or not.
  • It helps in the trigonometry ratio calculation along with distance and height measurement.

Also Read:


Sample Questions

Ques: Find the value of x. (2 marks)
Find the value of x

Ans.- Since x is on the opposite side of the right-angle so it’s the hypotenuse

Now, From the Theorem-

Hypotenuse2 = Perpendicular2 + Base2

x2 = 152 + 102

⇒ x= 152+102

⇒ x = 18

Therefore, the value of unknown side “x” is 18.

Ques: The sides of a triangle are 10,8 and 6 units. Check whether it is a right-angled triangle or not. (3 marks)

Ans.- Using Pythagoras theorem, 

Hypotenuse2 = Perpendicular2 + Base2

Given,

Perpendicular=8

Base=6

Hypotenuse=10 (since it is the longest measuring length)

From the formula,

82+62=102

⇒ 64+36=100

⇒ 100=100

L.H.S = R.H.S.

Therefore, the angle opposite to 10 is a right-angle.

Ques: A bamboo is placed against the door such that its foot is at the distance of 2.5m from the door and its top reaches a Wallpaper that is 6m above the ground and the wallpaper is fixed above the door. Find the length of the bamboo? (2 marks)

Ans.- Let AB be a ladder and CA be the Bamboo with the wallpaper at A.

Also, BC=2.5m and CA=6m

A bamboo is placed against the door such that its foot is at the distance of 2.5m from the door and its top reaches a Wallpaper that is 6m above the ground and the wallpaper is fixed above the door. Find the length of the bamboo

From Pythagoras theorem, 

Hypotenuse2 = Perpendicular2 + Base2

AB2=BC2 + CA2

⇒AB2=2.52 + 62

⇒AB= (42.25)1/2

⇒AB= 6.5 m

The length of the bamboo is 6.5 m

Ques: A fenced rectangular plot of length 15cm and diagonal 17cm. Find the perimeter of the rectangular plot? (3 marks)

Ans.- The angle between the two sides of a rectangular plot is 90 degrees. Thus, the diagonal divides the rectangle into two right-angled triangles where the diagonal forms the hypotenuse. The length of perpendicular of that triangle is given by the length of that rectangular plot and similarly, the breadth corresponds to the base of that triangle.

A fenced rectangular plot of length 15cm and diagonal 17cm. Find the perimeter of the rectangular plot?

For the right angled triangle ABD,

AB2+BD2=AD2

⇒172=152+BD2

⇒289-225=BD2

⇒ BD= 64

⇒ BD= 8

The base of the right-angled triangle is 8cm which is the breadth of the rectangle.

Therefore, Perimeter of the rectangle= 2* (Length + Breadth)

= 2(15+8)

= 46

The perimeter of the given rectangular plot is equal to 46cm.

Ques: Aniket, Roshni, and Karanveer were having a birthday party at Deepshika’s house. After the party gets over, Roshni and Karanveer went back to their respective house in a rented car hired through Uber while Aniket stayed back. Roshni’s house was 12 miles straight towards the east, from Deepshika’s house. Karanveer’s house was 5 miles straight south to Deepshika’s house. How far was their house (Roshni’s and Karanveer’s)? (2 marks)

Ans.- Let us visualize the whole content in the question with reference to a right-angled triangle. That means Roshni and Karanveer are hypotenuses apart from each other. So the distance between both of their house can be extracted by: d2=122+52=169, where d is the distance of the house. On solving we get d = 13. Therefore, the houses are positioned 13 miles away from one another.

Ques. A triangle has sides 8 cm, 11 cm and 15 cm. Determine if it is a right triangle. (2 marks)

Ans. Let’s assume hypotenuse = 15 cm (it is the longest side) = c

c2 = 15= 225

Other sides a = 8cm, b = 11cm

a2 + b2 = 82 + 112 = 64 + 121 = 185

185 \(\neq\) 225

a+ b\(\neq\) c2

Hence it is not a right angle triangle. 

Ques. ABC is a right triangle. AC is its hypotenuse. Length of side AB is \(2\sqrt{5}\). Side BC is twice of side AB. Find the length of AC. (3 marks)

Ans. Let AB = a, BC = b, AC = c

AB = a = \(2\sqrt{5}\)

BC is two times of AB, b = 2a = \(2\sqrt{5}\)

AC = Hypotenuse = c

On applying Pythagoras theorem a+ b2 = c2

(\(2\sqrt{5}\))2 + (\( 4\sqrt{5}\))= c2

4(5) + 16(5) = c2

c2 = 20 + 80 = 100

c = 10

AC = 10

Ques. The hypotenuse of a right triangle is 6 cm. Its area is 9 cm2. Find the sides. (5 marks)
The hypotenuse of a right triangle is 6 cm. Its area is 9 cm2. Find the sides.

Ans. Let AB = a and BC = b

In \(\bigtriangleup\)ABC, base = b and altitude = a

Area of the triangle = \(\frac{1}{2} \times (base \times altitude)\)

Hence \(\frac{1}{2} \times (ab)\) = 9

ab = 18 (Equation 1)

Using the Pythagoras formula, \(a^2 + b^2 = 6^2 = 36\)

We add an substract 2ab to complete the square:

\(a^2 + b^2 - 2ab + 2ab = 36\)

\((a - b)^2 + 2ab = 36\)

\((a-b)^2 + 36 = 36 \) (Using ab = 18 from Equation 1)

\((a - b)^2 = 0\)

a = b

By substituting a by b in Equation 1 we get:

\(b^2 = 18 \)

\(b = 3\sqrt{2}\)

Side AB = BC = \(3\sqrt{2}\) cm

Ques. In a right-angled triangle, the longest side is 8 cm. One of the remaining sides is \(4\sqrt{3}\) cm long. Find the length of the other side. (3 marks)

Ans. Let the lengths of sides be a, b, and c (hypotenuse)

The hypotenuse is the longest side, hence c = 8

Let b = \(4\sqrt{3}\) 

From the Pythagoras theorem, \(a^2 + b^2 = c^2\)

\(a^2 + (4\sqrt{3})^2 = 8^2\)

\(a^2 + 16(3) = 64\)

\(a^2 + 48 = 64\)

\(a^2 = 16\)

a = 4

Therefore the third side = 4 cm

For Latest Updates on Upcoming Board Exams, Click Here: https://t.me/class_10_12_board_updates


Also Check:

CBSE X Related Questions

1.
An umbrella has 8 ribs which are equally spaced (see Fig. 11.10). Assuming umbrella to be a flat circle of radius 45 cm, find the area between the two consecutive ribs of the umbrella.
An umbrella has 8 ribs which are equally spaced

      2.

      A vertical pole of length 6 m casts a shadow 4 m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.

          3.

          The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :

          Length (in mm)

          Number of leaves

          118 - 126

          3

          127 - 135 

          5

          136 - 144

          9

          145 - 153

          12

          154 - 162

          5

          163 - 171

          4

          172 - 180

          2

          Find the median length of the leaves. 
          (Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)

              4.
              Check whether \(6n\) can end with the digit \(0\) for any natural number \(n\).

                  5.
                  Which of the following pairs of linear equations are consistent/inconsistent? If consistent, obtain the solution graphically: (i) \(x + y = 5\),\( 2x + 2y = 10\) (ii)\( x – y = 8 , 3x – 3y = 16\) (iii) \(2x + y – 6 = 0\) , \(4x – 2y – 4 = 0\) (iv) \(2x – 2y – 2 = 0,\) \( 4x – 4y – 5 = 0\)

                      6.
                      Find the sums given below :
                      1. \(7 + 10\frac 12+ 14 + ....... + 84\)
                      2. \(34 + 32 + 30 + ....... + 10\)
                      3. \(–5 + (–8) + (–11) + ....... + (–230)\)

                          Comments



                          No Comments To Show