Question

how to prove Pythagoras theorem​

Answer 1

given,

in Rt.angle ∆PQR,which is right angled at Q.

to prove: PQ^2 + QR^2= PR^2

Draw QS perpendicular to PR .

proof: ∆RSQ ~ ∆RQP.

=> RQ/RP = RS/RQ

=> RQ^2= RS*RP--------->(1)

∆QSP~∆RQP

=> SP/QP= QP/RP

=> QP^2=RP*SP-------->(2)

ADD eq'n (1) and (2),

RP^2+QP^2= (RS*RP)+(RP*SP)

RQ^2+QP^2=RP(RS+SP)

RQ^2+QP^2=RP*RP

RQ^2+QP^2=RP^2

Hence, proved.

hope it helps............

Answer 2

hey buddy...

Gd mrng....

here is your answer------->>>

.....PYTHAGORAS THEOREM....

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)...

#hope it helps you...

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