[Maths Class Notes] on Pythagorean Theorem Formula Pdf for Exam

One of the well-recognized formulas in modern mathematics is the Pythagorean Theorem, which renders us with the association between the sides in a right triangle. A right triangle has two legs and a hypotenuse. The two legs meet at an angle of 90° while the hypotenuse is the longest side of the right triangle and is that side which is opposite to the right angle. Simply, a Pythagoras equation describes the relationship between the three sides of a right-angled triangle.

The Pythagorean Theorem explains the link in every right triangle is:

a² + b² = c²

Formula For Pythagoras Theorem

The formula for Pythagoras Theorem is given by:

Perpendicular² + Base² = Hypotenuse²

Or

a² + b² = c²

Where a, b and c represents the sides of the right-angled triangle with hypotenuse as c.

Use of Pythagorean Theorem Formula

The Pythagoras theorem is used to calculate the sides of a right-angled triangle. If we are given the lengths of two sides of a right-angled triangle, we can simply determine the length of the 3rd side. (Note that it only works for right-angled triangles!)

The theorem is frequently used in Trigonometry, where we apply trigonometric ratios such as sine, cos, tan; to find out the length of the sides of the right triangle.

Derivation of Pythagorean Theorem

Take into account a right-angled triangle ΔMNO. From the figure shown below, it is right-angled at N.

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Pythagorean Theorem Derivation – 1

Let NP be perpendicular to the side MO.

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Pythagoras Theorem Derivation – 2

From the above-given figure, consider the ΔMNO and ΔMPN,

In ΔMNO and ΔMPN,

∠MNO = ∠MPN = 90°

∠M = ∠M → common

Using the MM criterion for the similarity of triangles we have,

Δ MNO ~ Δ MPN

Thus, MP/MN = MN/MO

⇒ MN² = MO x MP…(1)

Considering ΔMNO and ΔNPO from the figure below:

Pythagorean Theorem Derivation -3

∠O = ∠O  → common

 ∠OPN = ∠MNO = 90°

Applying the principle of the Angle Angle(AA) criterion for the similarity of triangles, we come to the conclusion that,

ΔNPO ~ ΔMNO

Thus, OP/NO = NO/MO

⇒ NO² = MO x OP …..(2)

From the similarity of triangles, we come to the conclusion that,

∠MPN = ∠OPN = 90°

That said, if a perpendicular is constructed from the right triangle of a right-angled vertex to the hypotenuse, then the triangles so formed on both sides of the perpendicular are identical to each other and as well the whole triangle.

To Prove: MO² = MN² + NO²

By adding up the equation (1) and equation (2), we obtain:

MN² + NO² = (MO x MP) + (MO x OP)

MN² + NO² = MO (MP + OP)….(3)

Since MP + OP = MO, substituting the value in equation (3).

MN² + NO² = MO (MO)

Now, it becomes

 MN² + NO² = MO²

Therefore, the Pythagorean theorem is proved.

Solved Examples

Example:

Calculate the hypotenuse of a right-angled triangle whose lengths of two sides are 6 cm and 9 cm.

Solution: Given the criteria are:

Perpendicular = 9 cm

Base = 6 cm

Applying the Pythagoras theorem we have

Hypotenuse² = Perpendicular² + Base²

Now, putting the values we have will get:

Hypotenuse² = 9² + 6²

Hypotenuse² = 81 + 36

Hypotenuse =√117

Hypotenuse  = √10.8.

Example:

Solve the right-angled triangle with the two given sides 8, b, 17

Solution:

Begin with: a² + b² = c²

Put in the values we know: 8² + b² = 17² = 353

Calculate squares:  64 + b² = 289

Take 64 from both sides: 64 − 64 + b² = 289 − 64

Calculate: b² = 225

Square root of both sides:  b = √225

Calculate: b = 15

Example:

Determine the distance of diagonal across a square of size 2?

Solution:

Begin with: a² + b² = c²

Put in the values we know: 2² + 2² = c²

Calculating the squares: 2 + 2 = c²

2 + 2 = 4:   4 = c²

Now, let’s swap the sides:  c² = 4

Square root of both sides:  c = √4

This is about: 2.