Question

Given figure shows, a right $△$ circumscribing a circle. If AP = 8 cm and BQ = 4 cm, then the length of AC =

• A. 16 cm
• B. 20 cm
• C. 24 cm

Answer 1

The correct option is B 20 cm

Let us assume $CR$ = $\text{x}cm$
By Theorem- Tangents from an external point to a circle are equal in lengths.
$\therefore AR=AP=8cm$,
$BP=BQ=4cm$, and
$CQ=CR=\text{x}cm$

Applying pythagoras theorem, we get,
$A{B}^2+B{C}^2=A{C}^2$
$\Rightarrow (AP+PB{)}^2+(BQ+CQ{)}^2=(AR+CR{)}^2$
$\Rightarrow (8+4{)}^2+(4+\text{x}{)}^2=(8+\text{x}{)}^2$
$\Rightarrow {12}^2+(4+\text{x}{)}^2=(8+\text{x}{)}^2$
$\Rightarrow 144+16+{\text{x}}^2+8\text{x}=64+{\text{x}}^2+16\text{x}$
$\Rightarrow 8\text{x}=144+16-64$
$\therefore \text{x}=\frac{96}{8}=12cm$

$\because CR=\text{x}=12cm$ and $AC=AR+CR$
$\therefore AC=8+12=20cm$