NCERT Exemplar · Class 10 Mathematics Circles

\[TP=\sqrt{OT^2-OP^2}=\sqrt{13^2-5^2}=12\text{ cm} \quad (\triangle OPT,\ \angle OPT=90^\circ) \] \[OE=5\text{ cm},\quad ET=OT-OE=13-5=8\text{ cm} \] \[\angle AET=90^\circ \quad (AB\perp OT \text{ at } E) \] \[\tan(\angle ATE)=\tan(\angle PTO)=\frac{OP}{TP}=\frac{5}{12} \quad (\angle ATE=\angle PTO) \] \[AE=ET\tan(\angle ATE)=8\times\frac{5}{12}=\frac{10}{3}\text{ cm} \] \[BE=AE=\frac{10}{3}\text{ cm} \quad (\text{symmetry about } OT) \] \[AB=AE+BE=\frac{20}{3}\text{ cm} \]Answer: \(\displaystyle AB=\dfrac{20}{3}\) cm

\[\angle OCP=90^\circ \quad (\text{tangent}\perp\text{radius at } C) \] \[\angle OCA=\angle PCA-\angle OCP=110^\circ-90^\circ=20^\circ \] \[\angle OAC=\angle OCA=20^\circ \quad (OA=OC,\ \triangle OCA \text{ isosceles}) \] \[\angle CAB=\angle OAC=20^\circ \] \[\angle ACB=90^\circ \quad (\text{angle in a semicircle}) \] \[\angle CBA=180^\circ-90^\circ-20^\circ=70^\circ \quad (\text{angle sum}, \triangle ABC) \]Answer: \(\displaystyle \angle CBA=70^\circ\)

Let \(\displaystyle M\) be the midpoint of \(\displaystyle BC\); \(\displaystyle A\), \(\displaystyle O\), \(\displaystyle M\) are collinear (\(\displaystyle \triangle ABC\) isosceles). Let \(\displaystyle AM=h\), \(\displaystyle BM=MC=a\). \[AB^2=a^2+h^2=36 \] \[OB^2=a^2+(h-9)^2=81 \quad (OB=R=9) \] \[(h-9)^2-h^2=81-36=45 \Rightarrow -18h+81=45 \Rightarrow h=2 \] \[a^2=36-h^2=32 \Rightarrow a=4\sqrt2 \] \[\text{Area}=\tfrac12\times BC\times AM = a\times h = 8\sqrt2\text{ cm}^2 \]Answer: Area \(\displaystyle =8\sqrt2\) cm\(\displaystyle ^2\)

\[AP=AQ=\sqrt{OA^2-OP^2}=\sqrt{13^2-5^2}=12\text{ cm} \quad (\triangle OPA,\ \angle OPA=90^\circ) \] \[BP=BR,\quad CQ=CR \quad (\text{tangents from } B,\ C \text{ to the circle}) \] \[\text{Perimeter}=AB+BC+CA=(AB+BP)+(CQ+CA) \quad (BC=BR+RC=BP+CQ) \] \[=AP+AQ=24\text{ cm} \]Answer: Perimeter \(\displaystyle =24\) cm