PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
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PQ is a chord of length 4.8 cm of a circle of radius 3 cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
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Answer:
Radius of the circle is 3cm and PQ=4.8cm.
PQ is a chord of the circle with centre O.
The tangents at P and Q intersect at point T
So, TP and TQ are tangents
OP and OT are joined.
Join OQ
We have two right triangles: ΔOPT and ΔOQT
Here,
OT=OT (Common)
PT=QT (tangents of the circle)
OP=OQ (radius of the same circle)
By Side - Side - Side Criterion
ΔOPT≅ΔOQT
Therefore, ∠POT=∠OQT
Again, from triangles ΔOPR and ΔOQR
OR=OR (Common)
OP=OQ (radius of the same circle)
∠POR=∠OQR (from above result)
By Side - Angle - Side Criterion
ΔOPR≅ΔOQT
Therefore, ∠ORP=∠ORQ
Now,
∠ORP+∠ORQ=180
∘
(Sum of linear angles =180 degrees)
∠ORP+∠ORP=180
∘
∠ORP=90
∘
This implies, OR⊥PQ and RT⊥PQ
Also OR perpendicular from center to a chord bisects the chord,
PR=QR=PQ/2=4.8/2=2.4cm
Applying Pythagoras Theorem on right triangle ΔOPR,
(OP)
2
=(OR)
2
+(PR)
2
(3)
2
=(OR)
2
+(2.4)
2
OR=1.8cm
Applying Pythagoras Theorem on right angled ΔTPR,
(PT)
2
=(PR)
2
+(TR)
2
___(1)
Also, OP⊥OT
Applying Pythagoras Theorem on right ΔOPT,
(PT)
2
+(OP)
2
=(OT)
2
[(PR)
2
+(TR)
2
]+(OP)
2
=(TR+OR)
2
(Using equation (1) and from figure)
(2.4)
2
+(TR)
2
+(3)
2
=(TR+1.8)
2
4.76+(TR)
2
+9=(TR)
2
+2(1.8)TR+(1.8)
2
13.76=3.6TR+3.24
TR=2.9cm [approx.]
From (1) =PT
2
=(2.4)
2
+(2.9)
2
PT
2
=4.76+8.41
or PT=3.63cm [approx.]
Answer: Length of TP is 3.63
Answer:
TP = 4 c.m
Step-by-step explanation:
OP = 3
PR = 2.4
[tex]or = \sqrt{ {3}^{2} - {2.4}^{2} } [/tex]
OR = 1.8
take triangle OPR
[tex] \tan(o) = \frac{2.4}{1.8} [/tex]
[tex] \tan(o) = \frac{4}{3} [/tex]
take triangle POT
[tex] \tan(o) = \frac{tp}{po} [/tex]
[tex] \frac{4}{3} = \frac{tp}{3} [/tex]
So
TP = 4 c.m