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In the given figure PQR is right angle triangle. I(PQ) = 6cm, (QR) = 8cm and I(PR) = 10cm. Seg QMperpendicular to Seg PR. Find I {QM} |
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Answer» -step explanation:In ΔPQR ∠PQR=90∘ [Given] QS⊥PR [From vertex Q to hypotenuse PR] ∴QS2=PS×SR (i) [By THEOREM] Now , in ΔPSQ we have QS2=PQ2−PS2 [By PYTHAGORAS theorem] =62−42 =36−16 =QS2=20 ⇒QS=25cm QS2=PS×SR (i) ⇒(25)2=4×SR ⇒420=SR ⇒SR=5cm Now , QS⊥PR ∴∠QSR=90∘ ⇒QR2=QS2+SR2 [By Pythagoras theorem] =(25)2+52 =20+25 ⇒QR2=45 ⇒QR=35cm Hence , QS=25cm,RS=5cm and QR=35cm. |
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