Question

prove Pythagoras theorem

class 10

Answer 1

Hlo mate :-

Solution :-

______________________________________________________________________________________

Proof of Pythagorean Theorem using Algebra:


Given: A ∆ XYZ in which ∠XYZ = 90°.

To prove: XZ2 = XY2 + YZ2

Construction: Draw YO ⊥ XZ

Proof: In ∆XOY and ∆XYZ, we have,

∠X = ∠X                             → common

∠XOY = ∠XYZ                     →  each equal to 90°

Therefore, ∆ XOY ~ ∆ XYZ   → by AA-similarity

⇒ XO/XY = XY/XZ               

⇒ XO × XZ = XY2 ----------------- (i)

In ∆YOZ and ∆XYZ, we have,

∠Z = ∠Z                                     →            common

∠YOZ = ∠XYZ                             →            each equal to 90°

Therefore, ∆ YOZ ~ ∆ XYZ           →            by AA-similarity

⇒ OZ/YZ = YZ/XZ                 

⇒ OZ × XZ = YZ2 ----------------- (ii) 

From (i) and (ii) we get, 

XO × XZ + OZ × XZ = (XY2 + YZ2)

⇒ (XO + OZ) × XZ = (XY2 + YZ2)

⇒ XZ × XZ = (XY2 + YZ2)

⇒ XZ 2 = (XY2 + YZ2)

______________________________________________________________________________________

☆ ☆ ☆ Hop It's helpful ☆ ☆ ☆

:- Regards :- ♡ ♡ 《 Nitishkr1 》 ♡ ♡

Answer 2

here is your answer OK ☺☺☺☺☺☺☺

Given : A right ΔABC right angled at B

To prove : AC2 = AB2 + BC2

Construction : Draw AD ⊥ AC

Proof : ΔABD and ΔABC

∠ADB = ∠ABC = 90°

∠BAD = ∠BAC (common)

∴ ΔADB ∼ ΔABC (by AA similarly criterion)

AD/AB = AB/AC

⇒ AD × AC = AB2 ...... (1)

Now In ΔBDC and ΔABC

∠BDC = ∠ABC = 90°

∠BCD = ∠BCA (common)

∴ ΔBDC ∼ ΔABC (by AA similarly criterion)

CD/BC = BC /AC

⇒ CD × AC = BC2 ........ (2)

Adding (1) and (2) we get

AB2 + BC2 = AD × AC + CD × AC

= AC (AD + CD)

= AC × AC = AC2

∴ AC2 = AB2 + BC2


thanks for asking this question