\(\large{\underline{\sf Attachment :- \: Diagram \: of \: the \: proof}}\)
\(\mathtt{\huge{\underline{\red{Pythagoras\:Theorem}}}}\)
✴ 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,
\(\mathtt{\huge{\fbox{\fbox{\blue{Proof}}}}}\)
Given :-
- A △PQR in which an ∠PQR = 90° .
To Prove :-
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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