Question

converse of Pythagoras class 10th mathematics theorem​

Answer 1

Theorem - ln a triangle if the square of one side is equal to sum of square of remaining two sides,then the triangle is right angled triangle.

given - in triangle ABC, AC^2= AB ^2 + BC^2

to prove - angle ABC = 90°

construction - draw triangle PQR such that AB = PQ,BC = QR, angle PQR = 90°

proof - In triangle PQR angle Q=90°

PR ^2=PQ^2+ QR^2........(pythagoras theorem )

= AB^2+ BC ^2......(construction )..(1)

=AC^2.......( given )....(2)

PR^2= AC^2

PR=AC..... (3)

triangle ABC ~ triangle PQR......Sss test

angle ABC = angle PQR =90°

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!