Pythagoras Theorem

Theorem 6.9 In a triangle, if square of one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle. Proof:-  Given : - In a △ABC in which  AC 2 =AB 2 +B C 2 Prove That : -∠B=90° Construction :- A △PQR right triangle at Q such that PQ=AB and QR=BC (In figure) Now, from △PQR , we have:         PR 2 =PQ 2 +QR 2    (Pythagoras Theorem as ∠Q=90°)         PR 2 =AB 2 +B C 2     (By construction)...(1) But   AC 2 =AB 2 +B C 2      (Given)              ...(2) so,   AC=PR      [from(1) and (2)]...(3)  Now, in △ABC and △PQR,                 AB=PQ     (by construction)                 BC=QR     (by construction)                 AC=PR      [from (3)] so,     △ABC≅ △PQR (SSS congrunce) therefore,∠B =∠C        (CPCT) But          ∠Q =90°       (by construction) So,          ∠B = 90° Hence proved

Pythagoras theorem:- In mathematics, the Pythagorean theorem, also known as Pythagoras' theorem, is a fundamental relation in Euclidean geometry among the three sides of a right triangle. It states that the area of the square whose side is the hypotenuse is equal to the sum of the areas of the squares on the other two sides.  c 2  = b 2  + a 2 b=Perpendicular of right triangle a= Base of right triangle c= hypotenuse of right triangle For Example:- Solution :- Perpendicular of right triangle =b=5 Base of right triangle = a Hypotenuse of right triangle=c= 13 According to formula  a 2  = b 2  - c 2  a  = √  (b 2  - c 2 ) a  = √  (13 2  - 5 2 ) a  = √  (169  - 25 ) a  = √  (144 ) a = 12 Ans. Theorem 6.8 In a right triangle, the square of the hypotenuse is equal to the sum of the square of the other two side. Given :- A right triangle ABC right angled at A.   Prove that: - B C 2  = AB 2  + AC 2 Construct