Pythagoras Theorem Maths Notes

Pythagoras Theorem Maths Notes

Right angled triangle:

Pythagoras Theorem Notes 1

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.

Pythagoras Theorem Maths Notes

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.
Pythagoras Theorem Notes 2
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,
Pythagoras Theorem Notes 3
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.
Pythagoras Theorem Notes 4
Given: In ∆ ABC, ∠ ABC = 90°.
seg BD ⊥ hypotenuse AC such that A-D-C.
To prove: BD2 = 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)
∴ BD2 = AD × DC.

Pythagoras Theorem Maths Notes

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.
Pythagoras Theorem Notes 5
Given: In ∆ PQR, ∠ PQR = 90°.
To prove : PR2 = PQ2 + QR2.
Construction: Draw seg QS ⊥ side PR such that P-S-R.
Proof: In ∆ PQR
Pythagoras Theorem Notes 6
Adding (2) and (3), we get,
PQ2 + QR2 = PS × PR + SR × PR
∴ PQ2 + QR2 = PR(PS + SR)
∴ PQ2 + QR2 = PR × PR … (P – S -R)
∴ PQ2 + QR2 = PR2 OR PR2 = PQ2 + QR2.

Pythagoras Theorem Maths Notes

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.
Pythagoras Theorem Notes 7
Given: In ∆ ABC, AC2 = AB2 + BC2.
To prove: ∆ ABC is right angled triangle.
Construction : Draw ∆ XYZ such that ∠XYZ = 90°, XY = AB and YZ = BC.
Proof: In ∆ XYZ,
Pythagoras Theorem Notes 8
Pythagoras Theorem Notes 9

Pythagoras Theorem Maths Notes

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 a2 = b2 + c2 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 [(p2 + q2), (p2 – q2), (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.
Pythagoras Theorem Notes 10

2. Theorem of geometric mean:
In ∆ PQR,
If ∠ PQR = 90° and seg QS ⊥ hypotenuse PR, then QS2 = PS × SR.
Pythagoras Theorem Notes 11

3. Pythagoras theorem:
In ∆ PQR, if ∠PQR = 90°, then PR2 = pq2 + QR2.
Pythagoras Theorem Notes 12

4. Converse of Pythagoras theorem:
In ∆ PQR, if PR2 = PQ2 + QR2, then ∠ PQR = 90° and ∆ PQR is a right angled triangle.
Pythagoras Theorem Notes 13

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.
Pythagoras Theorem Notes 14

6. Converse of 30° – 60° – 90° triangle theorem:
In ∆ PQR, if ∠ PQR = 90° and pQ = \(\frac{1}{2}\)PR, then ∠ PRQ = 30°
Pythagoras Theorem Notes 15

7. 45° – 45° – 90° triangle theorem:
In ∆ XYZ, ∠ XYZ = 90°, ∠ YXZ = ∠ XZY = 45°, then XY = YZ = \(\frac{1}{\sqrt{2}}\) × XZ.
Pythagoras Theorem Notes 16

Pythagoras Theorem Maths Notes

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
AC2 = AB2 + BC2 – BC2 – 2BC.BD
Pythagoras Theorem Notes 17
Proof: In ∆ ADB, ∠ ADB = 90°.
∴ by Pythagoras theorem,
AB2 = AD2 + BD2            .            ….. (1)
In ∆ ADC, ∠ ADC = 90°.
∴ by Pythagoras theorem,
AC2 = AD2 + DC2.
AC2 = AD2 + (BC – BD)2              … (B-D-C)
∴ AC2 = AD2 + BC2 – 2BCBD + BD2
∴ AC2 = (AD2 + BD2) + BC2 – 2BCBD
AC2 = AB2 + BC2 – 2BCBD             … [From (1)]

(B) In obtuse angled ∆ ABC, ∠B > 90°, if seg AD ⊥ side BC and D-B-C, then AC2 = AB2 + BC2 + 2BCDB.
Pythagoras Theorem Notes 18
Proof: In ∆ ADB, ∠ ADB = 90°.
∴ by Pythagoras theorem,
AB2 = D2 + DB2                             …….. (1)
In ∆ ADC, ∠ ADC = 90°.
∴ by Pythagoras theorem,
AC2 = AD2 + DC2
∴ AC2 = AD2 + (DB + BC)2             ……. (D-B-C)
∴ AC2 = AD2 + DB2 + 2DB∙BC + BC2
∴ AC2 = (AD2 + DB2) + BC2 + 2DB∙BC
∴ AC2 = AB2 + BC2 + 2BCDB.          ….. [From (1)]

Pythagoras Theorem Maths Notes

Read and understand

Pythagoras Theorem Notes 19

In ∆ ABC, if seg AP is the median then,
AB2 + AC2 = 2AP2 + 2BP2.
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 AB2 + AC2 = 2AM2 + 2BM2.
Pythagoras Theorem Notes 20
Given: In ∆ ABC, M is the midpoint of seg BC.
To prove: AB2 + AC2 = 2AM2 + 2BM2.
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,
AB2 = AM2 + BM2 – 2BM∙DM                … (1)
∆ AMC is an obtuse angled triangle.
By application of Pythagoras theorem, we get,
AC2 = AM2 + MC2 + 2MC∙DM              ……. (2)
But MC = BM            … (M is the midpoint of side BC)         … (3)
Substituting (3) in (2), we get,
AC2 = AM2 + BM2 + 2BM∙DM             … (4)
Adding (1) and (4), we get,
AB2 + AC2 = AM2 + BM2 – 2BM∙DM + AM2 + BM2 + 2BM∙DM
∴ AB2 + AC2 = 2AM2 + 2BM2
Let’s consider the case, where seg AM ⊥ seg BC.
Pythagoras Theorem Notes 21
In ∆ AMB, ∠ AMB = 90°.
∴ by Pythagoras theorem,
AB2 = AM2 + BM2         … (5)
In ∆ AMC, ∠ AMC = 90°.
∴ by Pythagoras theorem,
AC2 = AM2 + MC2              … (6)
But MC = BM …(M is the midpoint of side BC)             … (7)
∴ AC2 = AM2 + BM2….            … [From (6) and (7)] … (8)
Adding (5) and (8), we get,
AB2 + AC2 = AM2 + BM2 + AM2 + BM2
∴ AB2 + AC2 = 2AM2 + 2BM2.

Pythagoras Theorem Maths Notes

Maths Notes

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