Question

In the given figure, a circle is inscribed in the quadrilateral ABCD. Given AB=6cm,

BC=7cm and CD=4cm. Find AD.​

Answer 1

Answer:

A circle with centre O, touches the sides AB,BC,CD,DA, at P,Q,R,Srespectively.

Join OP,OQ,OR,OS

Now two tangents drawn from an external C

point to a circle subtend equal angles at

the centre.

∴(i)AP=AS, (ii)BP=BQ,(iii)CR=CQ, (iv)DR=DS

AB=6 cm,BC=7 cm,CD=4 cm

Now from above solution we have

AB+CD=BC+AD

⇒10=7+AD

⇒AD=3 cm

Answer 2

Answer:

The length of AD is 3 cm.

Step-by-step explanation:

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

⇒6+4=7+AD

⇒AD=10−7=3cm.

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