## Pythagoras Theorem Maths Notes

**Right angled triangle:**

A triangle in which one of the angles measure 90° is called a right angled triangle.

In ∆ ABC ∠ ABC = 90°.

∴ ∆ ABC is a right angled triangle.

The side opposite to 90° is called the hypotenuse and sides AB and BC are called perpendicular sides or sides containing the right angle. Hypotenuse is the largest side in the right angled triangle.

Remember, in a right angled triangle, the two angles other than right angle are always acute. In the figure, ∠A and ∠C are acute angles.

**Similarity and right angled triangle:**

Theorem : In a right angled triangle, if the altitude is drawn to the hypotenuse, then the two triangles formed are similar to the original triangle and to each other.

Given: In ∆ PQR, ∠PQR = 90°, seg QS ⊥ hypotenuse PR such that P-S-R.

To prove: ∆ PSQ ∆ PQR; ∆ QSR ∆ PQR; ∆ PSQ ∆ QSR.

Proof: In ∆ PSQ and ∆ PQR,

From (1) and (2), re get,

∆PSQ ~ ∆ QSR.

**Theorem of geometric mean:**

Statement: In a right angled triangle, the perpendicular segment to the hypotenuse from the opposite vertex, is the geometric mean of the segments into which the hypotenuse is divided.

Given: In ∆ ABC, ∠ ABC = 90°.

seg BD ⊥ hypotenuse AC such that A-D-C.

To prove: BD^{2} = AD × DC.

Proof: In ∆ ABC, ∠ ABC = 90° ……… (Given)

seg BD ⊥ hypotenuse AC

∴ ∆ ADB ~ ∆ BDC …….. (Similarity of right angled triangles)

∴ \(\frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{BD}}{\mathrm{DC}}\) ……… (Corresponding sides of similar triangles are in proportion)

∴ BD^{2} = AD × DC.

**Pythagoras Theorem:**

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

Given: In ∆ PQR, ∠ PQR = 90°.

To prove : PR^{2} = PQ^{2} + QR^{2}.

Construction: Draw seg QS ⊥ side PR such that P-S-R.

Proof: In ∆ PQR

Adding (2) and (3), we get,

PQ^{2} + QR^{2} = PS × PR + SR × PR

∴ PQ^{2} + QR^{2} = PR(PS + SR)

∴ PQ^{2} + QR^{2} = PR × PR … (P – S -R)

∴ PQ^{2} + QR^{2} = PR^{2} OR PR^{2} = PQ^{2} + QR^{2}.

**Converse of Pythagoras theorem:**

Statement : In a triangle, if the square of one side is equal to the sum of the squares of the remaining two sides, then the triangle is a right angled triangle.

Given: In ∆ ABC, AC^{2} = AB^{2} + BC^{2}.

To prove: ∆ ABC is right angled triangle.

Construction : Draw ∆ XYZ such that ∠XYZ = 90°, XY = AB and YZ = BC.

Proof: In ∆ XYZ,

**Pythagorean Triplet:**

In a triplet of positIve integers, if the square of the largest number is equal to the sum of the squares of the remaining two numbers, then the group of these three numbers is called a Pythagorean triplet.

e.g. if a, b, c are three positive integers and a is the largest integer and a^{2} = b^{2} + c^{2} then (a, b, c) is a Pythagorean triplet.

Remember, numbers in Pythagorean triplet can be written in. any order.

**Formula to find Pythagorean triplet:**

If p, q and r are positive integers and p > q then [(p^{2} + q^{2}), (p^{2} – q^{2}), (2pq)] is a Pythagorean triplet.

This formula can be used to find various Pythagorean triplets by taking different values of p and q.

**Remember this!**

1. Similarity of Right A angled triangles:

In ∆ ABC, if ∠ABC 90° and seg BD ⊥ hypotenuse AC, then ∆ ABC ~ ∆ ADB ~ ∆ BDC.

2. Theorem of geometric mean:

In ∆ PQR,

If ∠ PQR = 90° and seg QS ⊥ hypotenuse PR, then QS^{2} = PS × SR.

3. Pythagoras theorem:

In ∆ PQR, if ∠PQR = 90°, then PR^{2} = pq^{2} + QR^{2}.

4. Converse of Pythagoras theorem:

In ∆ PQR, if PR^{2} = PQ^{2} + QR^{2}, then ∠ PQR = 90° and ∆ PQR is a right angled triangle.

5. 30° – 60° – 90° triangle A theorem:

In ∆ ABC, if ∠C = 30°, ∠A = 60° and ∠B = 90°, B then AB = \(\frac{1}{2}\)AC and BC = \(\frac{\sqrt{3}}{2}\)AC.

6. Converse of 30° – 60° – 90° triangle theorem:

In ∆ PQR, if ∠ PQR = 90° and pQ = \(\frac{1}{2}\)PR, then ∠ PRQ = 30°

7. 45° – 45° – 90° triangle theorem:

In ∆ XYZ, ∠ XYZ = 90°, ∠ YXZ = ∠ XZY = 45°, then XY = YZ = \(\frac{1}{\sqrt{2}}\) × XZ.

**Application of Pythagoras theorem:**

Let’s understand the application of Pythagoras theorem to acute and obtuse angled triangles. Consider the following examples:

(A) In acute angled triangle ABC, seg AD ⊥ side BC, B-D-C, ∠B∠90°; then

AC^{2} = AB^{2} + BC^{2} – BC^{2} – 2BC.BD

Proof: In ∆ ADB, ∠ ADB = 90°.

∴ by Pythagoras theorem,

AB^{2} = AD^{2} + BD^{2} . ….. (1)

In ∆ ADC, ∠ ADC = 90°.

∴ by Pythagoras theorem,

AC^{2} = AD^{2} + DC^{2}.

AC^{2} = AD^{2} + (BC – BD)^{2} … (B-D-C)

∴ AC^{2} = AD^{2} + BC^{2} – 2BCBD + BD^{2}

∴ AC^{2} = (AD^{2} + BD^{2}) + BC^{2} – 2BCBD

AC^{2} = AB^{2} + BC^{2} – 2BCBD … [From (1)]

(B) In obtuse angled ∆ ABC, ∠B > 90°, if seg AD ⊥ side BC and D-B-C, then AC^{2} = AB^{2} + BC^{2} + 2BCDB.

Proof: In ∆ ADB, ∠ ADB = 90°.

∴ by Pythagoras theorem,

AB^{2} = D^{2} + DB^{2} …….. (1)

In ∆ ADC, ∠ ADC = 90°.

∴ by Pythagoras theorem,

AC^{2} = AD^{2} + DC^{2}

∴ AC^{2} = AD^{2} + (DB + BC)^{2} ……. (D-B-C)

∴ AC^{2} = AD^{2} + DB^{2} + 2DB∙BC + BC^{2}

∴ AC^{2} = (AD^{2} + DB^{2}) + BC^{2} + 2DB∙BC

∴ AC^{2} = AB^{2} + BC^{2} + 2BCDB. ….. [From (1)]

**Read and understand**

In ∆ ABC, if seg AP is the median then,

AB^{2} + AC^{2} = 2AP^{2} + 2BP^{2}.

This result is known as ‘Apollonius theorem’.

Apollonius theorem establishes the relation between the sides and median of the triangle.

**Apollonius theorem:**

Statement: In ∆ ABC, if M is the midpoint of side BC, then AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}.

Given: In ∆ ABC, M is the midpoint of seg BC.

To prove: AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}.

Construction : Draw seg AD ⊥ side BC such that B-D-C.

Proof: Let’s consider the case, where AM is not perpendicular to seg BC, then out of ∠AMB and ∠AMC, one is acute angle and other is obtuse angle. Let’s consider ∠ AMB as an acute angle and ∠ AMC as an obtuse angle.

∆ ABM is an acute angled triangle,

∴ by application of Pythagoras theorem, we get,

AB^{2} = AM^{2} + BM^{2} – 2BM∙DM … (1)

∆ AMC is an obtuse angled triangle.

By application of Pythagoras theorem, we get,

AC^{2} = AM^{2} + MC^{2} + 2MC∙DM ……. (2)

But MC = BM … (M is the midpoint of side BC) … (3)

Substituting (3) in (2), we get,

AC^{2} = AM^{2} + BM^{2} + 2BM∙DM … (4)

Adding (1) and (4), we get,

AB^{2} + AC^{2} = AM^{2} + BM^{2} – 2BM∙DM + AM^{2} + BM^{2} + 2BM∙DM

∴ AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}

Let’s consider the case, where seg AM ⊥ seg BC.

In ∆ AMB, ∠ AMB = 90°.

∴ by Pythagoras theorem,

AB^{2} = AM^{2} + BM^{2} … (5)

In ∆ AMC, ∠ AMC = 90°.

∴ by Pythagoras theorem,

AC^{2} = AM^{2} + MC^{2} … (6)

But MC = BM …(M is the midpoint of side BC) … (7)

∴ AC^{2} = AM^{2} + BM^{2}…. … [From (6) and (7)] … (8)

Adding (5) and (8), we get,

AB^{2} + AC^{2} = AM^{2} + BM^{2} + AM^{2} + BM^{2}

∴ AB^{2} + AC^{2} = 2AM^{2} + 2BM^{2}.