Question

A circle of radius 5 cm has a chord of length 8 cm. Tangents drawn at the end points of the chord meet at a point. Find the perimeter of the triangle formed by the chord and the tangents.

• A. $\frac{80}{7}$ cm
• B. $\frac{27}{2}$ cm
• C. $\frac{81}{5}$ cm
• D. $\frac{64}{3}$ cm

Answer 1

The correct option is D

$\frac{64}{3}$ cm

Consider triangle ABC. Tangents CA and CB form the sides of triangle ABC.

By Theorem- From an external point, the tangents drawn to a circle are equal in length.

So, CA = CB. Thus triangle ABC is an isosceles triangle.

By Theorem - Centre of the circle lies on the bisector of the angle between two tangents

$\therefore$ CO is the angle bisector of $\angle$ ACB. So, CO is perpendicular to chord AB and therefore, OC bisects AB which gives AP = PB = 4 cm.

OA = OB = 5 cm

Now, as $△$ APO forms right $△$, by Pythagoras theorem,
$A{O}^2=A{P}^2+O{P}^2\Rightarrow 5^2=4^2+O{P}^2$
$\therefore OP=3$ cm

Let $AC=x,OC=y$

In $△$ ACO, apply Pythagoras theorem to get ${OC}^2$ = ${OA}^2$ + ${AC}^2$

$y^2=25+x^2$ ...(1)

In $△$ ACP, apply Pythagoras theorem to get ${AC}^2$ = ${AP}^2$ + ${CP}^2$

$x^2=16+{(y-3)}^2$ ...(2)

Subtracting (1) from (2), we get

$25=6y-7\Rightarrow y=\frac{16}{3}$ cm

Substituting value of y in equation (1),

$x=\frac{20}{3}$ cm

Thus, AC = BC = $\frac{20}{3}$ cm ; AB = 8 cm

$\therefore$ The perimeter of the triangle formed by chord and tangents (i.e. $△$ ABC)
= AB + BC +AC
= $8+\frac{20}{3}+\frac{20}{3}$
= $\frac{64}{3}$ cm.