Question

In the given figure a triangle ABC is drawn to circumscribe a circle of radius 3 cm such that the segments BD and DC into which BC is divided by the point of contact D are of length 9 cm and 3 cm respectively. Find the length of sides AB and AC.

• A. AB = 9 cm, AC = 15 cm
• B. AB = 15 cm, AC = 9 cm
• C. AB = 10 cm, AC = 5 cm
• D. AB = 5 cm, AC = 10 cm

Answer 1

Answer: (B)

AB = 15 cm, AC = 9 cm

The length of tangents drawn from an external point to a circle are equal.
Mark points F and E as shown in figure.

$\text{So,}BF=BD=9cm,CD=CE=3cm,AF=AE=x\text{(Assume)}$

$\text{Applying Pythagoras theorem in}$
$△ABC,\text{we get,}$

${AB}^2$ = ${BC}^2$+${CA}^2$

${(AF+BF)}^2$ = ${(BD+DC)}^2$ + ${(CE+AE)}^2$

${(x+9)}^2$ = ${(9+3)}^2$ + ${(3+x)}^2$

${(x+9)}^2$ = ${12}^2$ + ${(3+x)}^2$

$x^2+18x+81=144+9+6x+x^2$

$12x=72$

$x=6cm$

$\therefore AB=AF+BF=6+9=15cm$
$\text{and}AC=AE+CE=6+3=9cm$