Question

prove that in a right angle triangle the square on the hypotenuse is equal to the sum of the square on the other two sides

Answer 1

Hello Mate!

Since by Pythagoras theorum

H² = B² + P²

In ∆ABC, H is BC, AB is B and AC is P

BC² = AB² + AC²

BC × BC = ( AB × AB ) + ( AC × AC )

ar(sq. BCDE) = ar(sq. ABMN) + ar(sq. ACFG)

Hence, in a right angle triangle the square on the hypotenuse is equal to the sum of the square on the other two sides.

Have great future ahead!

Answer 2

Proof => We are given a right triangle ABC right angled at B

we need to prove that AC ² = AB ² + BC ²

Let's draw-

BD ⊥ AC

so, ΔADB ~ ΔABC. (Theorem 6.7)

Now, AD / AB = AB / AC

=> AD. AC = AB ².............................(1)

Also, ΔBDC ~ ΔABC. (theorem 6.7)

so, CD / BC = BC / AC

=> CD. AC = BC ² .........................(2)

Adding (1) and (2) ,

AD. AC + CD. AC = AB² + BC²

AC(AD+CD) = AB² + BC²

AC. AC = AB² + BC²

AC = AB² + BC²