In the given figure, \(TP\) and \(TQ\) are tangents to a circle with centre \(M\), touching another circle with centre \(N\) at \(A\) and \(B\) respectively. It is given that \(MQ = 13 \text{ cm}\), \(NB = 8 \text{ cm}\), \(BQ = 35 \text{ cm}\) and \(TP = 80 \text{ cm}\).
(i) Name the quadrilateral MQBN. (1)
(ii) Is MN parallel to PA? Justify your answer. (1)
(iii) Find length TB. (1)
(iv) Find length MN. (2)
Show Hint
When finding the distance between centers in a configuration with parallel radii, always construct a right triangle by drawing a perpendicular from the smaller radius to the larger one. This allows the use of the Pythagoras theorem.
Step 1: Understanding the Concept:
Tangents from an external point to a circle are equal in length.
Radius is perpendicular to the tangent at the point of contact. Step 3: Detailed Explanation: (i) In quadrilateral \(MQBN\), \(MQ \perp TQ\) and \(NB \perp TQ\) (Radius \(\perp\) Tangent).
Since both are perpendicular to the same line \(TQ\), \(MQ \parallel NB\).
A quadrilateral with one pair of opposite sides parallel is a Trapezium. (ii)No, \(MN\) is not parallel to \(PA\). \(PA\) is a chord/segment on the tangents, while \(MN\) is the line joining the centers. There is no geometric condition satisfyng parallelism here. (iii) Since \(TP\) and \(TQ\) are tangents from \(T\) to the circle with center \(M\), \(TP = TQ = 80 \text{ cm}\).
Now, \(TQ = TB + BQ\).
\(80 = TB + 35 \Rightarrow TB = 45 \text{ cm}\). (iv) We know \(MQ \parallel NB\). To find the distance between centers \(MN\) in trapezium \(MQBN\):
Draw a line from \(N\) perpendicular to \(MQ\), say at point \(X\).
\(QX = NB = 8 \text{ cm}\).
\(MX = MQ - QX = 13 - 8 = 5 \text{ cm}\).
In right \(\triangle MXN\), \(NX = BQ = 35 \text{ cm}\).
\[ MN^2 = MX^2 + NX^2 = 5^2 + 35^2 = 25 + 1225 = 1250 \]
\[ MN = \sqrt{1250} = 25\sqrt{2} \approx 35.36 \text{ cm} \] Step 4: Final Answer:
(i) Trapezium, (iii) \(45 \text{ cm}\), (iv) \(35.36 \text{ cm}\).
