Question

In figure, a circle touches all the four sides of a quadrilateral $ABCD$, whose sides $AB=6cm$, $BC=7cm$, and $CD=4cm$. Length of $AD$ is

• A. $10$ $cm$
• B. $7$ $cm$
• C. $3$ $cm$
• D. $4$ $cm$

Answer 1

Answer: (C)

$3$ $cm$

Given : A circle inscribed in a quadrilateral $ABCD,AB=6cm,BC=7cm,CD=4cm$.
To find : $AD$
Proof : Let the circle touch the sides $AB,BC,CD,DA$, at $P,Q,R$ and $S$, respectively.
$AP=AS$
$BP=BQ$
$DR=DS$
$CR=CQ$ {Lengths of two tangents drawn from an external point of circle, are equal}
Adding all these, we get
$(AP+BP)+(CR+RD)=(BQ+QC)+(DS+SA)$
$AB+CD=BC+DA$
$\Rightarrow 6+4=7+AD$
$\Rightarrow AD=10-7=3cm$.