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Hypotenuse of an Isosceles Right Triangle
Last updated at February 20, 2026 by Teachoo
Hypotenuse of an Isosceles Right Triangle
Class 8 (Ganita Prakash - 1, 2 & Old NCERT)
Chapter 2 Class 8 - The Baudhayana-Pythagoras Theorem (Ganita Part 2)
- Doubling a square
- Halving a square
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Hypotenuse of an Isosceles Right Triangle
- Decimal Representation of √2
- Figure it out - Page 39, 40
- Formula for Hypotenuse of an Isosceles Right Triangle
- Combining Two squares
- Baudhāyana’s Theorem on Right-angled triangles
- Pythagorean triples
- Figure it out - Page 50
- A Long-Standing Open Problem
- Further Applications of the Baudhāyana - Pythagoras Theorem
- Figure it out - Page 52, 53, 54
Transcript
Hypotenuse of an Isosceles Right TriangleLet’s consider a square PEAR of side 1 unit Drawing Diagonal RE, We see that ∆ PER has PE = PR = 1 ∠ P = 90° Thus, it is an isosceles right angled triangle And, RE is the hypotenuse We have to find length RE Making Square of Double area of PEAR Thus, area of square REST is double the area of square PEAR Since PEAR had side = 1 unit Area of PEAR = 1 × 1 = 1 square unit Thus, Area of REST = 2 × Area of Pear = 2 × 1 = 2 square units Since we need to find hypotenuse RE, let RE = c Square REST is made of side = RE = c Thus, we can write Area of REST = c × c Area of REST = c × c 2 = c2 c2 = 2 c = √𝟐 Thus, hypotenuse is of length √𝟐
