Question

What does Pythagoras’ famous theorem involve?

Answer 1

Pythagoras' theorem involves:
-a^2+b^2=c^2 is the theorem.
-a and b and the sides of the triangle
-c is the base of the triangle

Answer 2

$\large{\underline{\sf Attachment :- \: Diagram \: of \: the \: proof}}$

$\mathtt{\huge{\underline{\red{Pythagoras\:Theorem}}}}$

✴ It states that the Hypotenuse² is equals to Perpendicular² + Base².

In other words, The square of Hypotenuse or side opposite to 90° is equals to the sum of its Perpendicular & Base sq, where Hypotaneous is largest side.

${ \sf{Hypotaneus²\:=\:Perpendicular²\:+\:Base²}}$

${\sf{C²\:=\:A²\:+\:B²}}$

Where,

• C = Hypotenuse

• A = Perpendicular

• B = Base

$\mathtt{\huge{\fbox{\fbox{\blue{Proof}}}}}$

Given :-

• A △PQR in which an ∠PQR = 90° .

To Prove :-

• PR² = PQ² + QR²

Proof :-

We have to draw QD ⊥ PR ,

In △PDQ and △PQR , we have

∠P = ∠P ( common ) .

∠PDQ = ∠PQR [ each equal to 90° ] .

∴ △PDQ ∼ △PQR [ By AA-similarity ] .

PD/PQ = PQ/PR .

PQ² = PD × PR ----------(1) .

In △QDR and △PQR , we have

∠R = ∠R ( common ) .

∠QDR = ∠PQR [ each equal to 90° ] .

∴ △QDR ∼ △PQR [ By AA-similarity ] .

DR/QR = QR/PR .

QR² = DR × PR ----------(2) .

Taking sum of 1 & 2 equation,

PQ² + QR² = PD × PR + DR × PR .

PQ² + QR² = PR( PD + DR ) .

PQ² + QR² = PR × PR

PQ² + QR² = PR²

PR² = PQ² + QR²

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