Question

converse theorem of pythagoras

Answer 1

The converse of the Pythagorean Theorem is:

If the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.

That is, in ΔABC, if c2=a2+b2 then ∠C is a right triangle, ΔPQR being the right angle.

We can prove this by contradiction.

Let us assume that c2=a2+b2 in ΔABC and the triangle is not a right triangle.

Now consider another triangle ΔPQR. We construct ΔPQR so that PR=a, QR=b and ∠R is a right angle.

By the Pythagorean Theorem, (PQ)2=a2+b2.

But we know that a2+b2=c2 and a2+b2=c2 and c=AB.

So, (PQ)2=a2+b2=(AB)2.

That is, (PQ)2=(AB)2.

Since PQ and AB are lengths of sides, we can take positive square roots.

PQ=AB

That is, all the three sides of ΔPQR are congruent to the three sides of ΔABC. So, the two triangles are congruent by the Side-Side-Side Congruence Property.

Since ΔABC is congruent to ΔPQR and ΔPQR is a right triangle, ΔABC must also be a right triangle.

This is a contradiction. Therefore, our assumption must be wrong.



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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!