TrianglesClass X

Naushaba ZafarK.V. Pragati Vihar

In any right triangle, the square of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of other two sides (the two sides that meet at a right angle).

Pythagoras Theorem

The theorem can be written as an equation relating the lengths of the sides a, b and c, often called the Pythagorean equation:

a² + b² = c²where c represents the length of the hypotenuse, and a

and b represent the lengths of the other two sides

let us take a right triangle ABC, right angled at B. Let BD be theperpendicular to the hypotenuse AC In Δ ADB and Δ ABC,

∠ A = A (common)∠and ADB = ABC (90 deg)∠ ∠So, Δ ADB ~ Δ ABC (AA similarity rule) ……………….(1)Similarly, Δ BDC ~ Δ ABC ……………………………..….(2) So, from (1) and (2), triangles on both sides of the

perpendicular BD are similar to the whole triangle ABC. Also, since Δ ADB ~ Δ ABC and Δ BDC ~ Δ ABC So, Δ ADB ~ Δ BDC

Proof of Pythagoras Theorem using similarity rule

This leads us to a theorem“If a perpendicular is drawn from the vertex of the right

angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.”

(say theorem i)

Applying this theorem to Pythagoras theorem – “In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides”

Proof (Cont.)

Given a right triangle ABC, right angled at B. We need to prove that AC² = AB² + BC² Let us draw BD AC. Now, ⊥ Δ ADB ~ Δ ABC (Theorem i) So, AD/AB = AB/AC (Sides are proportional)or, AD . AC = AB² ...............(1) Also, Δ BDC ~ Δ ABC (Theorem i) So, CD/BC = BC/ACor, CD . AC = BC² ..............(2) Adding (1) and (2), AD . AC + CD . AC = AB² + BC²or, AC (AD + CD) = AB² + BC²or, AC . AC = AB² + BC² or, AC² = AB² + BC² (hence proved)

Proof (cont.)

In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Here, we are given a triangle ABC in which AC² = AB² + BC². We need to prove that B = 90°.∠

In Δ PQR right angled at Q constructed such that PQ = AB and QR = BC

Converse of Pythagoras Theorem

Now, from Δ PQR, we have : PR² = PQ² + QR² (Pythagoras Theorem, as ∠Q = 90°)

or, PR² = AB² + BC² (By construction) ……………………………(1)But AC² = AB² + BC² (Given) ……………………………(2)So, AC = PR [From (1) and (2)] ……………………………(3) Now, in Δ ABC and Δ PQR,AB = PQ and BC = QR (By construction)AC = PR [Proved in (3) above] So, Δ ABC Δ PQR (SSS congruence)≅

◦ Therefore, B = Q (CPCT)∠ ∠But Q = 90° (By construction)∠

So, B = 90° (∠ Hence proved)

Proof of Converse

Given by Bhaskar We start with four copies of the same right triangle. Three of

these have been rotated 90°, 180°, and 270°, respectively.

Each has area ab/2.

Pythagoras Theorem: alternative proof

If we put them together, they form a square with side c. The square has a small square hole with the side (a - b) The area of big square = area of small square + area of four trianglesi.e. c² = (a-b) ² + 4. ab/2 = a² - 2ab + b² + 2ab

= a² + b² (hence proved)

Thus, in right triangle the square of hypotenuse equals the sum of other two sides (Pythagoras Theorem)

Alternative proof (Cont.)

Bhaskar -Proof

Thank YouREFERENCES

- NCERT textbook-http://jwilson.coe.uga.edu/EMT668/EMAT6680.F99/Challen/pythagorean/lesson2/lesson2.html- www.cut-the-knot.org/pythagoras/index.shtml