Question

prove the conversation of Pythagoras Theorm

Answer 1

Here is your solution

Converse of Pythagoras Theorem

In a △ ,if the square of one side is equal to the sum of the square of the other two sides then the angle opposite to the first side is a right angle .

GIVEN :-

A △ABC in which

$AC {}^{2} = AB {}^{2} + BC {}^{2}$

TO PROVE :-

$∠B= 90 \:°$

CONSTRUCTION :-

Draw a △ DEF such that DE = AB , EF = BC and ∠E = 90°

PROOF :-

In △DEF , we have ∠E = 90°
now using pythagoream theorem
we have

$DF {}^{2} = DE {}^{2} + EF{}^{2} \\ DF = AB {}^{2} + BC {}^{2}........(1) \\ but \: \: AC = AB {}^{2} + \: BC {}^{2} ........(2)$
From equation (1) and (2) we get

$AC {}^{2} = df {}^{2} \\ AC= DF$
hence

△ABC ≅ △DEF

so

$∠B =∠E = 90° \: proved$

Answer 2

Step-by-step explanation:

Statement:

In a Triangle the square of longer side is equal to the sum of squares of the other two sides, then the triangle is a right angled triangle.

Given -

A Triangle ABC such that

BC² = AB² + AC²

To Prove -

Angle A = 90°

Construction -

Draw a ∆DEF such that AB = DE and AC = DF and Angle D = 90°

Proof -

In ∆ABC,

BC² = AB² + AC² - Given

In ∆ DEF

EF² = DE² + DF²

Therefore,

EF² = AB² + AC²

(Since AB = DE, AC = DF)

Therefore,

BC² = EF² ie - BC = EF

Now, In ∆ABC and ∆DEF

AB = DE - By Construction

AC = DF - By Construction

BC = EF

Therefore

∆ABC ≅ ∆DEF by SSS test.

Thus,

Angle A = Angle D - CPCT

But, Angle D = 90° ( As per construction)

Therefore

Angle A = 90°

Hence Proved!