37
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 $MQ = 13$ cm, $NB = 8$ cm, $BQ = 35$ cm and $TP = 80$ cm.
(i) Name the quadrilateral $MQBN$.
(ii) Is $MN$ parallel to $PA$? Justify your answer.
(iii) Find length $TB$.
(iv) Find length $MN$.
(i) Name the quadrilateral $MQBN$.
(ii) Is $MN$ parallel to $PA$? Justify your answer.
(iii) Find length $TB$.
(iv) Find length $MN$.
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Sol. (i) $BN \parallel QM$ and $QB \parallel MN$
The quadrilateral $MQBN$ is a trapezium. (I) (1 Mark)
(ii) No, as $AN \neq PM$ (II) (1 Mark)
(iii) $TQ = TP = 80$ cm
$\therefore TB = 80 - 35 = 45$ cm (III) (1 Mark)
Note: If $MNT$ is considered a straight line and similarity of triangles is used to find $TB$ then $TB = 56$ cm may be considered as correct answer.
(iv) $MN^2 = NS^2 + MS^2$
$= 35^2 + (13 - 8)^2$ (IV) (1 Mark)
$= 1225 + 25$ (V) ($\frac{1}{2}$ Mark)
$= 1250$
$\therefore MN = 25\sqrt{2}$ cm (VI) ($\frac{1}{2}$ Mark)
The quadrilateral $MQBN$ is a trapezium. (I) (1 Mark)
(ii) No, as $AN \neq PM$ (II) (1 Mark)
(iii) $TQ = TP = 80$ cm
$\therefore TB = 80 - 35 = 45$ cm (III) (1 Mark)
Note: If $MNT$ is considered a straight line and similarity of triangles is used to find $TB$ then $TB = 56$ cm may be considered as correct answer.
(iv) $MN^2 = NS^2 + MS^2$
$= 35^2 + (13 - 8)^2$ (IV) (1 Mark)
$= 1225 + 25$ (V) ($\frac{1}{2}$ Mark)
$= 1250$
$\therefore MN = 25\sqrt{2}$ cm (VI) ($\frac{1}{2}$ Mark)