Question

prove that in a right angled triangle square of the hypotenuse is equal to sum of the squares of other two sides.

Answer 1

The proof of Pythagorean Theorem in mathematics is very important.

In a right angle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.



States that in a right triangle that, the square of a (a2) plus the square of b (b2) is equal to the square of c (c2).

In short it is written as: a2 + b2 = c2 


Let QR = a, RP = b and PQ = c. Now, draw a square WXYZ of side (b + c).  Take points E, F, G, H on sides WX, XY, YZ and ZW respectively such that WE = XF = YG = ZH = b.

Then, we will get 4 right-angled triangle, hypotenuse of each of them is ‘a’: remaining sides of each of them are band c. Remaining part of the figure is the 

square EFGH, each of whose side is a, so area of the square EFGH is a2.

Now, we are sure that square WXYZ = square EFGH + 4 ∆ GYF

or, (b + c)2 = a2 + 4 ∙ 1/2 b ∙ c

or, b2 + c2 + 2bc = a2 + 2bc

or, b2 + c2 = a2

Answer 2

AC²=AB²+BC²

Given: A right angled ∆ABC, right angled at B

To Prove: AC²=AB²+BC²

Construction: Draw perpendicular BD onto the side AC .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

△ADB∼△ABC. (by AA similarity)

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC

(in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

Hope this helps you!!!