Mathematics
In △PQR, ∠Q = 90° and QM is perpendicular to PR. Prove that :
(i) PQ2 = PM × PR
(ii) QR2 = PR × MR
(iii) PQ2 + QR2 = PR2
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Answer
△PQR is shown in the figure below:
(i) In △PQR and △PMQ,
⇒ ∠PMQ = ∠PQR [Both = 90°]
⇒ ∠QPM = ∠RPQ [Common]
∴ △PQR ~ △PMQ [By AA]
Since, corresponding sides of similar triangles are proportional we have :
⇒
⇒ PQ2 = PM × PR
Hence, proved that PQ2 = PM × PR.
(ii) In △QRM and △PRQ,
⇒ ∠QMR = ∠PQR [Both = 90°]
⇒ ∠QRM = ∠QRP [Common]
∴ △QRM ~ △PRQ [By AA]
Since, corresponding sides of similar triangles are proportional we have :
⇒
⇒ QR2 = PR × MR
Hence, proved that QR2 = PR × MR.
(iii) Adding equations from (i) and (ii) we get,
⇒ PQ2 + QR2 = PM × PR + PR × MR ………(1)
⇒ PQ2 + QR2 = PR(PM + MR)
From figure,
PM + MR = PR
⇒ PQ2 + QR2 = PR2.
Hence, proved that PQ2 + QR2 = PR2.
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