In a right-angled triangle the square on the hypotenuse is equal to the sum of the squares on the two other sides.


Let in the triangle ABC, B = 90°, hypotenuse AC = b, AB = c and BC = a. It is required to prove that, AC2 = AB2 + BC 2, i.e. b2 = c2 + a2.


Produce BC up to D such that CD = AB = c. Also draw perpendicular DE at D on BC produced, so that DE = BC = a. Join C, E and A, E.


Steps-1: In ΔABC and ΔCDE, AB = CD = c, BC = DE = a and included ∠ABC = included ∠CDE. [each right angle]
Hence, ΔABC ≅ ΔCDE.
∴ AC = CE = b and ∠BAC = ∠ECD.

Steps-2: Again, since AB⊥BD and ED⊥BD, AB ll ED.
Therefore, ABDE is a trapezium.

Steps-3: Moreover, ∠ACB + ∠BAC = ∠ACB + ∠ECD = 1 right angle.
∴ ∠ACE = 1 right angle.
Now, area of the trapezium ABDE = area of (Δ region ABC + Δ region CDE + Δ region ACE)


1. Area of trapezium =1/2 Χ (sum of parallel sides Χ distance between parallel sides).

2. Pythagoras Theorem: In 6th century B.C. Greek philosopher Pythagoras discovered an important property of right-angled triangle. This property of right-angled triangle is known as Pythagorean property. It is believed that before the birth of Pythagoras, in Egyptian and Greek era, this special property of right-angled triangle was in use.

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