question_answer

Match the columns.

Column-l	Column-ll
(a) Radius of the given  circle is	(p) 20 cm
(b) In the given figure, value of x is	(q) 6 cm
(c) Perimeter of APST with  \[PQ=10\text{ }cm\]is	(r) 5 cm

 

A)                (a) \[\to \] (p), (b) \[\to \] (r), (c) \[\to \] (q)

B)        (a) \[\to \] (r), (b) \[\to \] (q), (c) \[\to \] (p)

C)        (a) \[\to \] (q), (b) \[\to \] (r), (c) \[\to \] (p)

D)        (a) \[\to \] (r), (b) \[\to \] (p), (c) \[\to \] (q)

Correct Answer: B

Solution :

(a) Draw OC perpendicular on AB,              In   \[\Delta \text{ }OBC\]& \[\Delta \text{ }OAC\]                         \[\angle OAC=\angle OCA\]   (each \[{{90}^{o}}\])       \[OB=OA\]  (radii of same circle) OC is common side in both triangles \[\therefore \] \[\Delta \text{ }OBC\text{ }\cong \,\,\Delta \text{ }OAC\]       (RHS congruence) \[\Rightarrow \]            \[BC=AC=\frac{1}{2}AB=\frac{7}{2}\] Now, in \[\Delta \,OCP,\] \[{{(OC)}^{2}}={{(OP)}^{2}}-{{(CP)}^{2}}\] \[\Rightarrow \]            \[{{(OC)}^{2}}=169-{{\left( \frac{7}{2}+9 \right)}^{2}}=\frac{51}{4}\] Now, in \[\Delta \,\,OCA\] \[{{(OA)}^{2}}={{(OC)}^{2}}+{{(CA)}^{2}}\] \[=\frac{51}{4}+{{\left( \frac{7}{2} \right)}^{2}}=\frac{51}{4}+\frac{49}{4}=\frac{100}{4}=25\Rightarrow OA=5\]Hence, radius of circle = 5 cm (b) In   \[\Delta OPT~\]             \[{{(OP)}^{2}}={{(OT)}^{2}}+{{(PT)}^{2}}\] \[\Rightarrow \]            \[{{(x+4)}^{2}}={{x}^{2}}+{{(8)}^{2}}\] \[\Rightarrow \]            \[{{x}^{2}}+16+8x={{x}^{2}}+64\] \[\Rightarrow \]       \[8x=48\,\,\,\,\Rightarrow \,\,\,\,x=6\,\,cm\] (c)  \[PO=10\text{ }cm\] We know, length of tangents drawn from an external point to a circle are equal. \[\Rightarrow \]            PQ = PR                     .....(i)    Also,    \[SQ=SU\]                  ....(ii)    and      \[TU=TR\]                   ....(iii)   Now, perimeter of \[PST\] \[=PS+ST+PT=PS+SU+UT+PT\] \[=PS+SQ+TR+PT\]  (Using (ii) & (iii)) \[=PQ+PR=PQ+PQ\]     (Using (i)) \[=2PQ=2\times 10=20\,cm.\]