Question

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

Answer 1

statement ; in right angled triangle the square of hypotenus is equal to sum of squares of other two sides

given : abc is right triangle ,right angled at b

construction :draw BD ⊥ AC

to prove : AC² = AB² + BC ²

PROOF :by theorem

           WKT,

            triangle ADB is similar to triangle ABC

            AD÷AB=AB÷AC

             AC×AD=AB²---------------------- EQUATION 1

             WKT

              BDC IS SIMILAR TO ABC

                CD×AC=BC×BC

               CD×AC=BC²--------------------------2

                 BY EQUATION  ( 1)  AND (2 )WE GET

          AC²=AB²+BC²  

 HOPE IT MAY HELP U

             

                                           

                                         

Answer 2

AC²=AB²+BC²

Given: A right angled ∆ABC, right angled at B

To Prove: AC²=AB²+BC²

Construction: Draw perpendicular BD onto the side AC .

Proof:

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

Hope this helps you!!!