Question

Converse of Pythagoras theorem states that in a triangle, if the square of larger side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. What are we proving here? We prove $\angle B=$........ .

• A. ${130}^o$
• B. ${90}^o$
• C. ${170}^o$
• D. ${110}^o$

Answer 1

The correct option is B ${90}^o$

Given, a $△ABC$ with

$BC=a,AB=b$ and $AC=c$

$\Rightarrow a^2+b^2=c^2$

Now construct a right angled triangle $DEF$ with $EF=BC=a$

$AB=DE=b$

Let $DF=d$ and $\angle DEF={90}^o$

Since $△DEF$ is a right angled triangle, we can use Pythagoras Theorem,

$\therefore d^2=a^2+b^2$

$\therefore c=d$

i.e, $AC=DF$

There by construction, by SSS test,

$△ABC\cong △DEF$

$\therefore △ABC$ is a right angled triangle with $\angle B={90}^o$