Question

One circle touches all sides of $\square ABCD$. If $AB=5,BC=8,CD=6$; then $AD=$

• A. $3$
• B. $7$
• C. $4$
• D. $9$

Answer 1

The correct option is A $3$

Given that a circle touches all the four sides of $\square ABCD$.

$\therefore$ $AB,BC,CD$ and $AD$ are chords of circle
$\therefore$ $AB+CD=AD+BC$
$\therefore$ $5+6=AD+8$
$\therefore$ $AD=3$