ICSE Class 9 Maths Chapter 13 Pythagoras Theorem

Chapter 13: Pythagoras Theorem

Proof and Simple Applications with Converse

13.1 Introduction

Buddhayan, an Indian Mathematician (600 B.C.), developed a relationship between the squares described on the hypotenuse of a right-angled triangle and the sum of the squares described on the remaining two sides of the triangle. But the credit of the present form of this relationship goes to a Greek Mathematician Pythagoras and is named as Pythagoras Theorem.

13.2 Pythagoras Theorem

Theorem 9

Area Based Proof

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

Given: A triangle ABC in which \(\angle ABC = 90°\).

To prove: \(AC^2 = AB^2 + BC^2\)

Construction: Draw squares ACDE, ABFG and BCHI on sides AC, AB and BC respectively. Draw BMN perpendicular to AC at point M and DE at point N. Join GC and BE.

Proof:

\(\angle FBC = \angle FBA + \angle ABC = 90° + 90° = 180°\)

\(\Rightarrow\) FBC is a straight line.

Since, GA // FB [Opp. sides of the square]

\(\Rightarrow\) GA is parallel to FC

Since, BMN and AE both are perpendicular to the same line AC

\(\Rightarrow\) AE is parallel to BMN. As each angle of quadrilateral AMNE is 90°, AMNE is a rectangle

Now, let \(\angle BAC = x\)

\(\therefore \angle GAC = \angle BAE\) [Each 90° + x]

\(AG = AB\) [Sides of the same square]

\(AC = AE\) [Sides of the same square]

\(\therefore \triangle GAC \cong \triangle BAE\) [By SAS]

\(\Rightarrow\) area of \(\triangle GAC\) = area of \(\triangle BAE\) .....(i) [Congruent triangles are equal in area]

We know, the area of a triangle is half the area of a parallelogram (rectangle, square, etc.) if both are on the same base and between the same parallels.

Since, \(\triangle GAC\) and square ABFG are on same the base (AG) and between the same parallels (AG // CF)

\(\therefore\) area of \(\triangle GAC = \frac{1}{2} \times\) area of square ABFG .....(ii)

Similarly, \(\triangle BAE\) and rectangle AMNE are on same base (AE) and between the same parallels (AE//BN)

\(\therefore\) area of \(\triangle BAE = \frac{1}{2} \times\) area of rectangle AMNE .....(iii)

From equations (i), (ii) and (iii), we get:

area of square ABFG = area of rectangle AMNE .....(iv)

In the same way, it can be proved that:

area of square BCHI = area of rectangle CMND .....(v)

On adding equations (iv) and (v), we get:

area of square ABFG + area of square BCHI = area of rectangle AMNE + area of rectangle CMND

\(\Rightarrow\) area of square ABFG + area of square BCHI = area of square ACDE

\(\Rightarrow\) square on AB + square on BC = square on AC

\(\Rightarrow AB^2 + BC^2 = AC^2\) i.e., \(AC^2 = AB^2 + BC^2\)

Hence Proved.

Conversely, if in any triangle, the square on the largest side of the triangle is equal to the sum of the squares on remaining two sides, then the triangle is a right-angled triangle and the angle opposite to the largest side is a right-angle.

i.e. if in triangle ABC; BC is the largest side and \(BC^2 = AB^2 + AC^2\), then \(\angle A = 90°\).

Theorem 10

Alternative Proof for Theorem 9

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

Given: A triangle ABC in which \(\angle ABC = 90°\).

To Prove: \(AC^2 = AB^2 + BC^2\)

Construction: Draw BD \(\perp\) AC.

Proof:

Hence Proved.

Important Properties

1. In a right-angled triangle, the hypotenuse is the largest side.

2. If AB is the largest side of a triangle ABC, then:

(i) \(AB^2 = AC^2 + BC^2 \Rightarrow\) ABC is a right-angled triangle with AB as hypotenuse and \(\angle ACB = 90°\).

(ii) \(AB^2 > AC^2 + BC^2 \Rightarrow\) ABC is an obtuse-angled triangle with angle ACB greater than 90°.

(iii) \(AB^2 < AC^2 + BC^2 \Rightarrow\) ABC is an acute-angled triangle.

Teacher's Note

The Pythagorean theorem is fundamental in construction and navigation. When builders lay out the foundation of a house, they use the 3-4-5 right triangle to ensure corners are perfectly square.

Worked Examples

Example 1

A ladder reaches a window which is 15 metres above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 8 metre high. Find the width of the street, if the length of the ladder is 17 metres.

Solution:

According to the given statement, the figure will be as shown alongside. In the figure, PQ is the width of the street, PM and QN are two buildings, A is the foot of the ladder, AB is the first position of the ladder and AB' is its second position.

Clearly, \(AB = AB' = 17\) m, \(PB = 15\) m and, \(QB' = 8\) m

In \(\triangle PAB\), \(PA^2 + PB^2 = AB^2\)

\(\Rightarrow PA^2 + 15^2 = 17^2\)

i.e. \(PA^2 = 17^2 - 15^2 = 289 - 225 = 64 \Rightarrow PA = \sqrt{64}\) m = 8 m

In \(\triangle QAB'\), \(QA^2 + (QB')^2 = (AB')^2\)

\(\Rightarrow QA^2 + 8^2 = 17^2\)

i.e. \(QA^2 = 17^2 - 8^2 = 289 - 64 = 225 \Rightarrow QA = \sqrt{225}\) m = 15 m

\(\therefore\) The width of the street = PQ = PA + QA = 8 m + 15 m = 23 m

Teacher's Note

This practical problem shows how the Pythagorean theorem helps solve real-world situations, such as determining safe distances when moving ladders or positioning equipment.

Example 2

In the given diagram, AB = 3CD = 18 cm and 3BP = 4CP = 36 cm. Show that the measure of angle APD is 90°.

Solution:

The measure of angle APD will be 90°, if \(AP^2 + DP^2 = AD^2\)

\(AB = 3CD = 18\) cm \(\Rightarrow AB = 18\) cm and \(CD = \frac{18}{3}\) cm = 6 cm

\(3BP = 4CP = 36\) cm \(\Rightarrow BP = \frac{36}{3}\) cm = 12 cm and \(CP = \frac{36}{4}\) cm = 9 cm

In \(\triangle DPC\),

\(DP^2 = DC^2 + CP^2 = 6^2 + 9^2 = 36 + 81 = 117\)

In \(\triangle APB\),

\(AP^2 = AB^2 + BP^2 = 18^2 + 12^2 = 324 + 144 = 468\)

\(\therefore DP^2 + AP^2 = 117 + 468 = 585\) ........... I

Draw DE \(\perp\) to AB and join DA

\(DE = CB = 9\) cm + 12 cm = 21 cm

and \(AE = AB - BE = AB - CD = 18\) cm - 6 cm = 12 cm

\(\therefore\) In \(\triangle AED\), \(AD^2 = AE^2 + DE^2 = 12^2 + 21^2 = 144 + 441 = 585\) ........... II

From I and II, we get: \(DP^2 + AP^2 = AD^2\) \(\therefore\) Angle APD = 90°.

Teacher's Note

This example demonstrates how the converse of the Pythagorean theorem can verify right angles in complex geometric configurations without directly measuring the angle.

Pythagorean Triplets

Consider three positive numbers a, b and c with c as the largest of a, b and c. If \(a^2 + b^2 = c^2\); a, b and c are called Pythagorean triplets. For example: 3, 4 and 5 are Pythagorean triplets as \(3^2 + 4^2 = 5^2\).

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