Question

Construct a right triangle PQR with PQ=6 cm, QR=8cm and angle Q=90° . Draw QS, the perpendicular from Q on PR and draw the circle passing through Q, R and S and construct the tangent from P to the circle .

Answer 1

Step-by-step explanation:

PQ,QR and PR are tangents to the circle at C,A and B.

∴ OC=OB=OA=r [ Radius of the circle ]

⇒ Area of the △PQR=

2

1

×PQ×QR

=

2

1

×3×4

=6cm

2

In right-angled △PQR,

⇒ (PR)

2

=(PQ)

2

+(QR)

2

⇒ (PR)

2

=(3)

2

+(4)

2

⇒ (PR)

2

=9+16

⇒ (PR)

2

=25

∴ PR=5cm

Now,

Area(△PQR)=Area(△OPQ)+Area(△OQR)+Area(△ORP)

⇒ 6=

2

1

r×PQ+

2

1

r×QR×

2

1

r×PR

⇒ 6=

2

1

r(PQ+QR+PR)

⇒ 6=

2

1

r(3+4+5)

⇒ 6=

2

1

r(12)

⇒ 6=6r

∴ r=1cm

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Answer 2

Answer:

the upper one is the correct answer pls mark him brainliest