Question

Show that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides​

Answer 1

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Answer 2

Let us consider that :

∠ABC is a right angle.

AC is the hypotenuse.

AB is the perpendicular.

BC is the base.

Given: In ΔABC,  ∠ABC=90°

Construction: BD is a perpendicular on side AC

To Prove: (AC)²=(AB)²+(BC)²

Proof:

In △ABC,

∠ABC=90°                                                                  (Given)

BD is perpendicular to hypotenuse AC              (Construction)  

Therefore, △ADB∼△ABC∼△BDC                             (Similarity of right-angled triangle)

△ABC∼△ADB

(AB/AC)=(AD/AB)      (congruent sides of similar triangles)

 AB²=AD×AC                      ___(1)

△BDC∼△ABC

CD/BC=BC/AC    (congruent sides of similar triangles)

BC²=CD×AC                       ___(2)

Adding the equations (1) and (2),

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)  

Since, AD + CD = AC

Therefore, AC²=AB²+BC²

Hence Proved.

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