Question

in a right angle triangle prove that the square of hypotenuse is equal to the sum of the squares of the other two sides

Answer 1

see by pythagorus property if we add base and perpendicular we get the hypotenuse

Answer 2

Given: In ΔABC,  ∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

∠ABC=90°                                                                  

BD is perpendicular to hypotenuse AC              

Therefore, △ADB∼△ABC∼△BDC                            

△ABC∼△ADB

(AB/AC)=(AD/AB)                                              

AB²=AD×AC                         →                                 (1)

△BDC∼△ABC

CD/BC=BC/AC                                                        

BC²=CD×AC                         →                                 (2)

Adding the equations (1) and (2),

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)  

Since, AD + CD = AC

Therefore, AC²=AB²+BC²

Hence Proved!