Question

Please ansr this wid explaination!!
I will mark u brainliest if u explain properly!!

Answer 1

Answer:

In Δ POR

angle PRO=90° and OT bisects PQ (theorem of class (theorem of class 9th)

Therefore,PR=QR=x

PR+QR=8(given)

x+x=8

2x=8

x=8/2

x=4cm

PR=4cm

now, by pythagoras theorem

PR^2+OR^2=PO^2

4^2+OR^2=5^2

OR^2=25-16

OR^2=9

OR=3cm

Let TP be y and TR be x

Now in right Δ TPR

PR^2+TR^2=TP^2

4^2+x^2=y^2

16+x^2=y^2                - (1)

NOW  ΔTPO

∠TPO=90°  (BY THEOREM)

THEREFORE BY PYTHAGORAS THEOREM

OT^2=PT^2+PO^2

(OR+RT)^2= y^2+5^2

(3+x)^2=16+x^2+25      (FROM EQ. 1)

9++6x=16+x^2+25

9+6x-16-25=x^2-x^2

6x-32=0

6x=32

x=32/6

x=16/3           -(2)

Putting value of x from eq 2 in 1

16+x^2=y^2

16+(16/3)^2=y^2

16+256/9=y^2

144+256/9=y^2

400/9=y^2

√400/9=y

20/3=y=TP

∴The length of TP is 20/3cm