In Fig., a circle touches all the four sides of a quadrilateral ABCD with AB = 6 cm, BC = 7 cm and CD = 4 cm. Find AD.

Given: AB = 6 cm

BC = 7 cm

CD = 4 cm

To find: The length of AD

Theorem Used:

The length of two tangents drawn from an external point are equal.

Explanation:

Here,

A is the external point, AP and AS are the external points.

By the theorem stated above,
AP=AS
Let, AP=AS=X
Similarly, BP=BQ
CQ=CR
RD=DS
Since, AP=X
⇒ BP = AB - AP=6-X
Now, BP=BQ=6-X
CQ=BC-BQ=7-(6-X)
=1+X
now, CQ=CR=1+X
RD=CD-CR=4-(1+X)
=3-X
RD=DS=3-X
AD=AS+SD
X+3-X=3
AD=3cm

Here,
AP=AS
Let, AP=AS=X
Similarly=BQ
CQ=CR
RD=DS
Since, AP=X
⇒BP=AB-AP=6-X
Now, BP=BQ=6-X
CQ=BC-BQ=7-(6-X)
=1+X
now, CQ=CR=1+X
RD=CD-CR=4-(1+X)
=3-X
RD=DS=3-X
AD=AS+SD
X+3-X=3
AD=3cm