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Pythagoras Theorem

Triangles II - Concepts

Calculating Perimeter of Triangle

Area of Similar Triangles

Similarity Related To Right Triangle

Pythagoras Theorem

Results Deduced From Pythagoras Theorem

Class - RRB Technician Grade I Signal Subjects

Concept Explanation

Pythagoras Theorem

Pythagoras Theorem:

Pythagoras, the famous Greek philosopher, lived about 572 B.C. to 501 B.C. He proved the relation between the lengths of the sides of a triangle, Although, this theorem was known to the Babylonians 1000 years ealier, but, Pythagoras is believed to have the first to discovered a proof of this theorem. However, long ago ( 800 B.C.), the Indian mathe,atician Baudhaya had stated and proved this property of a right angled triangle.

Let us have a look at the fig. $large Delta ABC$ is a right triangle, right angled at C, so that AB is the hypotenuse and AC and BC are the sides of the right triangle,

Then, $large (AB)^{2}=(BC)^{2}+(CA)^{2}$

i.e., $large (Hypotenuse)^{2}=(Base)^{2}+(perpendicular)^{2}$

Given: A triangle ABC right angled at B

To Prove: $large AB^2+BC^2=AC^2$

Construction: From B draw BD $large perp$ AC

Proof: Since BD $large perp$ AC

$large Delta ADBsimDelta ABC$ [By the above theorem]

$large Rightarrow frac{AD}{AB}=frac{AB}{AC}$

$large Rightarrow AB^2= AD;X;AC$ ............(1)

Now $large Delta BDCsimDelta ABC$ [By the above theorem]

$large Rightarrow frac{BC}{AC}=frac{DC}{BC}$

$large Rightarrow BC^2= AC;X;DC$ ............(2)

Adding Eq (1) and (2)

$large AB^2 +BC^2= AC;X;AD + AC;X;DC$

$large AB^2 + BC^2= AC;(AD + ;DC)$

$large AB^2 +BC^2= AC;X;AC$

$large AB^2 +BC^2= AC^2$

Hence Proved

Example 1: The lengths of two sides of a right triangle are 5 cm and 12 cm. Find the length of the hypotenuse.

Solution: Suppose BC = 5cm and AC = 12 cm.

$large therefore$ By Pythagoras theorem,

$large (AB)^{2}=(BC)^{2}+(AC)^{2}$

$large (AB)^{2}=(5)^{2}+(12)^{2}$

$large =25+144$

$large =169$

$large (AB)^{2}=169$

$large ABtimes AB=13times 13$

$large Rightarrow$ $large or;;AB=13 cm$

$large therefore$ Length of the hypotenuse = 13 cm.

Converse of Pythagoras Theorem:

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

Given: A triangle ABC such that $large AB^2+ BC^2=AC^2$

To Prove : $large angle B =90^0$

Construction : Construct a triangle DEF such that DE= AB, EF= BC and $large angle E =90^0$

Proof: Since $large Delta DEF$ is a right angled triangle with right angle at E

$large therefore DF^2=DE^2+EF^2$

Since DE = AB and EF = BC replacing them in the above equation

$large therefore DF^2=AB^2+BC^2$

But $large AB^2+ BC^2=AC^2$ [Given]

$large therefore DF^2=AC^2$

DF = AC

Now in $large Delta$ ABC and $large Delta$ DEF

AB = DE [By Construction]

BC= EF [By Construction ]

DF = AC [Proved Above]

$large therefore Delta ABCcongDelta DEF$ [By SSS Congruence Criterion]

$large Rightarrow angle B = angle E=90^0$

Hence $large Delta$ ABC is right angled at B

Illustration: ABC is an isosceles triangle with Ac = BC. If $large AB^2=2AC^2$ . Prove that ABC is aright triangle.

In $large Delta$ ABC, we are given that

AC = BC and $large AB^2=2AC^2$

Now $large AC^2= BC^2$

Adding $large AC^2$ To both sides we get

$large AC^2+AC^2=BC^2+AC^2$

$large BC^2+AC^2=2AC^2$

But $large AB^2=2AC^2$

$large therefore BC^2+AC^2=AB^2$

From Converse of Pythagoras Theorem we can say that

Triangle ABC is a right angled at C

Sample Questions

(More Questions for each concept available in Login)

Question : 1

If square of one side of triangle is equal to the sum of the square of other two sides then triangle is ______________________.
A. Acute a triangle	B. Obtuse triangle	C. Right triangle	D. None of these
Right Option : C
View Explanation

Question : 2

A 13-m long ladder reaches a window of building 12 m above the ground. Determine the distance of the foot of the ladder from the building.
A. 7	B. 15	C. 5	D. 9
Right Option : C
View Explanation

Question : 3

The length of the hypotenuse of a right triangle exceeds the length of the base by 2 cm and exceeds twice the length of the altitude by 1 cm. Find the length of each side of the triangle.
A. 5, 13, 17	B. 7, 11, 17	C. 8, 15, 17	D. 11, 15, 17
Right Option : C
View Explanation

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Pythagoras Theorem

Pythagoras Theorem

Students / Parents Reviews [20]

Upmanyu Sharma

Sharandeep Singh

Aman Kumar Shrivastava

Hridey Preet

Shivam Rana

Ayan Ghosh

Archit Segal

Manish Kumar

Yatharthi Sharma

Cheshta

Barkha Arora

Manish Kumar

Barkha Arora

Anika Saxena

Aman Kumar Shrivastava

Aman Kumar Shrivastava

Yogansh Nyasi

Om Umang

Prisha Gupta

Shreya Shrivastava