Question

Q.5 a) In the adjoining figure, PQ 1 QR and QRI RS also QR = 6 cm, PQ = RS = 4 cm.

Find the length of PS, if PS bisects QR at 0.
​

Answer 1

In ΔPQR

∠PQR=90

∘

[Given]

QS⊥PR [From vertex Q to hypotenuse PR]

∴QS

2

=PS×SR (i) [By theorem]

Now , in ΔPSQ we have

QS

2

=PQ

2

−PS

2

[By Pythagoras theorem]

=6

2

−4

2

=36−16

=QS

2

=20

⇒QS=2

5

cm

QS

2

=PS×SR (i)

⇒(2

5

)

2

=4×SR

⇒

4

20

=SR

⇒SR=5cm

Now , QS⊥PR

∴∠QSR=90

∘

⇒QR

2

=QS

2

+SR

2

[By Pythagoras theorem]

=(2

5

)

2

+5

2

=20+25

⇒QR

2

=45

⇒QR=3

5

cm

Hence , QS=2

5

cm,RS=5cm and QR=3

5

cm