Question

In the given figure PQR is right angle triangle. I(PQ)
= 6cm, (QR) = 8cm and I(PR) = 10cm. Seg QM
perpendicular to Seg PR. Find I {QM}​

Answer 1

Step-by-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.