Questions & Answers

Question

Answers

Answer

Verified

128.7k+ views

* Pythagoras theorem states that in a right angled triangle, the sum of square of base and square of height is equal to square of the hypotenuse.

If we have a right angled triangle, \[\vartriangle ABC\] with right angle, \[\angle B = {90^ \circ }\]

Then using the Pythagoras theorem we can write that \[A{C^2} = A{B^2} + B{C^2}\]

Let us assume a triangle given to us in the question as \[\vartriangle ABC\], which has a square of one side equal to the sum of squares of other sides.

Now we are given for \[\vartriangle ABC\], let the side whose square is equal to the sum of squares of the other two sides be \[AC\]. So, we can write

\[A{C^2} = A{B^2} + B{C^2}\] … (i)

Now we assume another triangle which is a right angle triangle, say \[\vartriangle DEF\] with \[\angle E = {90^ \circ }\] and two sides equal to sides of \[\vartriangle ABC\], say \[AB = DE,BC = EF\]

In right angle triangle, \[\vartriangle DEF\], using Pythagoras theorem we can write

Square of hypotenuse is equal to sum of square of base and square of height.

Here, base is \[EF\], height is \[DE\] and hypotenuse is \[DF\].

\[ \Rightarrow D{F^2} = D{E^2} + E{F^2}\] … (ii)

Now we know from equation (i)

\[A{C^2} = A{B^2} + B{C^2}\]

Substitute the values of \[AB = DE,BC = EF\] in equation (i)

\[ \Rightarrow A{C^2} = D{E^2} + E{F^2}\]

But we know from equation (ii) \[ \Rightarrow D{F^2} = D{E^2} + E{F^2}\]

\[ \Rightarrow A{C^2} = D{F^2}\]

Taking square root on both sides of the equation,

\[

\Rightarrow \sqrt {A{C^2}} = \sqrt {D{F^2}} \\

\Rightarrow AC = DF \\

\]

Therefore, now we have all three sides of \[\vartriangle ABC\] equal to three sides of \[\vartriangle DEF\]as

\[

AB = DE \\

BC = EF \\

AC = DF \\

\]

So, by Side Side Side (SSS) congruence rule, we can say that \[\vartriangle ABC \cong \vartriangle DEF\].

Since, we know two triangles which are congruent have their corresponding sides equal and their corresponding angles equal.

Here, corresponding angles of \[\vartriangle ABC \cong \vartriangle DEF\] are

\[

\angle A = \angle D \\

\angle B = \angle E \\

\angle C = \angle F \\

\]

So we can say \[\angle B = \angle E = {90^ \circ }\] because \[\angle E = {90^ \circ }\].

Therefore, \[\vartriangle ABC\] is a right angled triangle with \[\angle B = {90^ \circ }\].

Hence Proved.