Question

derivation for Pythagoras theorem​

Answer 1

Answer:

Pythagorean Theorem Derivation

Consider a right angled triangle\Delta ABC. It is right angled at B.

Let BD be perpendicular to the side AC.

In   \Delta ABC and \Delta ADB

∠ABC=∠ABD=90∘  

∠DBA=∠BCA

∠A=∠A

Using the AA criterion for the similarity of triangles, ΔABC≅ΔADB

Considering ΔABCandΔBDC

                                                  ∠DAB=∠CBD

∠DBA=∠BCD

 ∠CDB=∠ADB=90∘

Using the AA criterion for the similarity of triangles, ΔBDC≅ΔABC

Thus, it can be concluded ΔADB≅ΔBDC   

So if a perpendicular is drawn from the right-angled vertex of a right triangle to the hypotenuse, then the triangles formed on both sides of the perpendicular are similar to each other and also to the whole triangle.

Now, we are required to prove  AC2=AB2+BC2. We drop a perpendicular BD on the side AC.

We already know that ΔADB≅ΔABC

∴   ADAB=ABBC (Condition for similarity) Or AB2=AC×AD…….. (1)

Also,   ΔBDC≅ΔABC

∴CDBC=BCAC   (Condition for similarity) Or 

Answer 2

Answer:

h=√l^2+p^2

In a right angle triangle the square the largest side or hyptaneous is equal to the sum of square of length and perpendicular

so

,h^2=l^2+p^2

Hence h=√l^2+p^2