Question

prove Pythagoras theorm​

Answer 1

Hope it helps you ?............

Answer 2

$\bold{Statement}$ : In any right angled triangles, the square of the hypotenuse of that triangle will always be equal to the sum of the squares of other two sides.

$\bold{Given :}$ XYZ is a rt. angled triangle with <Y a right angle and its hypotenuse (XZ) is "a", height (XY) is "b" and base (YZ) is "c"

$\bold{To\:\:prove : }$ a² = b² + c²

$\bold{Construction : }$ A square PQRS is constructed with all its sides equal to (b+c). Inside it, a tilted square EFGH is drawn whose all sides are equal to 'a'.

$\bold{Proof :}$

Area of square EFGH = a²

Area of all four right angled triangles formed = 4 × 1/2 × b × c = 2 bc

Area of square PQRS = (b+c)²

Area of square PQRS = Area of square EFGH + Area of four rt. angled triangles

= a² + 2 bc

By condition,

(b+c)² = a² + 2bc

Or, b²+2bc+c² = a²+2bc

Or, a² = b²+c² [PROVED]

THIS IS THE MOST EASY WAY TO PROVE PYTHAGORAS THEOREM.