Question

Prove the converse of Pythagoras theorem with statement!

Answer 1

$\huge\underline\mathfrak\red{Statement}$

In a triangle, if the square of one side is equal to the sum of square of other two sides then prove that the triangle is right angled triangle.

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$\huge\underline\mathfrak\red{Solution}$

Given : AC² = AB² + BC²

To prove : ABC is a right angled triangle.

Construction : Draw a right angled triangle PQR such that, angle Q = 90°, AB = PQ, BC = QR.

Proof : In triangle PQR,

Angle Q = 90° ( by construction )

Also,

PR² = PQ² + QR² ( By using Pythagoras theorem )...(1)

But,

AC² = AB² + BC² ( Given )

Also, AB = PQ and BC = QR ( by construction )

Therefore,

AC² = PQ²+ QR²....(2)

From eq (1) and (2),

PR² = AC²

So, PR = AC

Now,

In ∆ABC and ∆PQR,

AB = PQ ( By construction )

BC = QR ( By construction )

AC = PR ( Proved above )

Hence,

∆ABC is congruent to ∆PQR by SSS criteria.

Therefore, Angle B = Angle Q ( By CPCT )

But,

Angle Q = 90° ( By construction )

Therefore,

Angle B = 90°

Thus, ABC is a right angled triangle with Angle B = 90°

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Hence proved!

Answer 2

CONVERSE OF THE THEOREM:-

ACCORDING TO IT IF THE SQUARE OF THE ONE SIDE IS EQUAL TO THE SUM OF THE SQUARES OF THE OTHER TWO SIDES OF THEN THE ANGLE OPPOSITE TO THE FIRST SIDE IS A RIGHT ANGLE (90°)