Question

PROVE THAT IN A RIGHT ANGLED TRIANGLE SQUARE OF THE HYPOTENUSE IS EQUAL TO SUM OF THE SQUARES OF OTHER TWO SIDES

Answer 1

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SOLUTION
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Given:

A right angled ∆ABC, right angled at B



To Prove- AC²=AB²+BC²


⚫Construction: draw perpendicular BD onto the side AC .


Proof:

$<b>$We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.


We have

△ADB∼△ABC. (by AA similarity)


Therefore, AD/ AB=AB/AC


(In similar Triangles corresponding sides are proportional)



AB²=AD×AC……..(1)


Also, △BDC∼△ABC


Therefore, CD/BC=BC/AC


(in similar Triangles corresponding sides are proportional)


Or, BC²=CD×AC……..(2)



Adding the equations (1) and (2) we get,



AB²+BC²=AD×AC+CD×AC


AB²+BC²=AC(AD+CD)


( From the figure AD + CD = AC)


AB²+BC²=AC . AC



Therefore, AC²=AB²+BC²

This theroem is known as Pythagoras theroem...

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Answer 2

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