Question

13. Prove that “In a right triangle, the square of the hypotenuse is equal to the sum of the squares of
the other two sides.

Class:- 10

Content Quality Solution Required

❎ Don't Spamming ❎

Answer 1

Figure is attached


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

This theroem is known as Pythagoras theroem...

Answer 2

Answer:

Given :

A right triangle ABC right angled at B.

To prove :

AC² = AB² + BC²

Construction :

Draw BD ⊥ AC

Proof :

In Δ ADB and Δ ABC

∠ A = ∠ A    [ Common angle ]

∠ ADB = ∠ ABC   [ Both are 90° ]

∴  Δ  ADB  Similar to Δ ABC   [ By AA similarity ]

So , AD / AB = AB / AC   [ Sides are proportional ]

= > AB² = AD . AC  ... ( i )

Now in Δ BDC and Δ ABC

∠ C = ∠ C    [ Common angle ]

∠ BDC = ∠ ABC   [ Both are 90° ]

∴  Δ  BDC Similar to Δ ABC   [ By AA similarity ]

So , CD / BC = BC / AC

= > BC² = CD . AC   ... ( ii )

Now adding both equation :

AB² + BC² = CD . AC +  AD . AC

AB² + BC² = AC ( CD + AD )

AB² + BC² = AC² .

AC² = AB² + BC² .

Hence proved .