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Question
PQ is a chord of length 4.8 cm of a circle of radius 3cm. The tangents at P and Q intersect at a point T as shown in the figure. Find the length of TP.
Solution
Let TR = y and TP= x
We know that the perpendicular drawn from the center to me chord bisects It.
∴ PR + RQ
Now, PR + RQ = 4.8
⇒ PR + PR = 4.8
⇒ PR = 2.4
Now, in right triangle POR
By Using Pythagoras theorem, we have
`PO^2 = OR^2 + PR^2`
`⇒3^2 = OR^2+(2.4)^2`
`⇒OR^2 =3.24`
⇒ OR = 1.8
Now, in right triangle TPR
By Using Pythagoras theorem, we have
`TP^2 = TR^2 + PR^2`
` ⇒ x^2 = y^2 +(2.4)^2`
`⇒ x^2 = y^2 + 5.76` ............(1)
Again, In right triangle TPQ
By Using Pythagoras theorem, we have
`TO^2= TP^2 + PO^2`
`⇒ (y +1.8)^2 = x^2 +3^2`
`⇒ y^2 +3.6y + 3.24 = x^2 +9`
`⇒ y^2 + 3.6y = x^2 +5.76` ...... (2)
Solving (1) and (2), we get
x = 4cm and y = 3.2cm
∴TP = 4cm
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