Question

A is a point at a distance 13 cm from the centre O of a circle of radius 5 cm. AP and AQ are the tangents to the circle at P and Q. If a tangent BC is drawn at a point R lying on the minor arc PQ to intersect AP at B and AQ at C, find the perimeter of the $\Delta ABC.$

• A. 12 cm
• B. 18 cm
• C. 20 cm
• D. 30 cm

Answer 1

The correct option is C 20 cm

Given-

AQ is a tangent to the circle with centre O at Q.

AO intersects the circle at B at which the tangent BC has been drawn.

BC intersects AQ at C.

$OA=13$cm & $OQ=5$cm.

To find out -

The perimeter of $\Delta ABC.$

Solution-

$\because AQ$ is a tangent to the circle with centre O at Q

$⟹\angle AQO={90}^o$ ...(i)

So, $\Delta AQO$ is a right one with hypotenuse AO.

$AQ=\sqrt{{AO}^2-{OQ}^2}=\sqrt{{13}^2-5^2}cm=12$cm ..........(ii)

Again, BC is a tangent to the circle with centre O at B

$⟹\angle ABC={90}^o$ & $\Delta ABC$ is a right one. ..(iii)

So, between $\Delta AQO$ & $\Delta ABC$ we have

$\angle AQO={90}^o=\angle ABC$ ...(from ii & iii),

$\angle BAC$ is common to both.

$\therefore \Delta AQO$ & $\Delta ABC$ are similar.

i.e $\frac{BC}{OQ}=\frac{AB}{AQ}$

$⟹BC=OQ\times \frac{AB}{AQ}=OQ\times \frac{AO-OB}{AQ}=5\times \frac{13-5}{12}cm=\frac{20}{6}cm..$

Now, from iii $\Delta ABC$ is a right one with hypotenuse AC

$\therefore AC=\sqrt{{AB}^2+{BC}^2}=\sqrt{{13}^{22}+{\frac{20}{6}}^2}cm=\frac{26}{3}cm.$

So, perimeter$=AB+BC+AC=8cm+\frac{20}{6}cm+\frac{26}{3}cm=20$cm ...$[AB=AO-OB=(13-5)cm=8cm]$