Question

state
Prove converse of Pythagoras
theorm.​

Answer 1

Answer:

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Converse of Pythagorean Theorem states that:

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

Given: A ∆PQR in which PR² = PQ² + QR²

To prove: ∠Q = 90°

Construction: Draw a ∆XYZ such that XY = PQ, YZ = QR and ∠Y = 90°

So, by Pythagora’s theorem we get

XZ² = XY² + YZ²

⇒ XZ² = PQ² + QR² ……….. (i), [since XY = PQ and YZ = QR]

But, PR² = PQ² + QR² ………… (ii), [given]

From (i) and (ii) we get,

PR² = XZ² ⇒ PR = XZ

Now, in ∆PQR and ∆XYZ, we get

PQ = XY,

QR = YZ and

PR = XZ

Therefore ∆PQR ≅ ∆XYZ

Hence ∠Q = ∠Y = 90°

Answer 2

Answer:

Step-by-step explanation:

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.

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 △PQR,

From △PQR, ∠Q = 90° (By construction)

PR² = PQ² + QR² (By using Pythagoras theorem)

PR² = AB² + BC²...(1)

AC² = AB² + BC² (Given).....(2)

From Eq (1) and (2), we get

PR = AC

In ∆ABC and ∆PQR,

AB = PQ (By construction)

BC = QR (By construction)

AC = PR (Proved above)

So, △ABC is congruent to △PQR by SSS criteria.

Then, ∠B = ∠Q (By CPCT)

Therefore,∠B = ∠Q = 90°

Hence, ABC is a right angled triangle.