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

Step-by-step explanation:

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 207588

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.

Answer 2

Answer:

AD = 3 cm

Step-by-step explanation:

According to the information provided in the question it is given as

In the given figure, a circle is inscribed in the quadrilateral A B C D. Given AB=6 cm,

BC=7 cm

CD=4 cm.

We need to find  AD.​

Here in the question it is given that a circle inscribe in  the quadrilateral having four sides in which three side are given we need to find the third side

Let us consider that circle touches the side AB,BC ,CD, AD  at p,q,r,s respectively

$\mathrm{AP}=\mathrm{AS}$

$\mathrm{BP}=\mathrm{BQ}$

$\mathrm{DR}=\mathrm{DS}$

$\mathrm{CR}=\mathrm{CQ}$

Lengths of two tangents drawn from an external point of circle, are equal

Hence by Adding all these we get,

$(\mathrm{AP}+\mathrm{BP})+(\mathrm{CR}+\mathrm{RD})=(\mathrm{BQ}+\mathrm{QC})+(\mathrm{DS}+\mathrm{SA})$

$\mathrm{AB}+\mathrm{CD}=\mathrm{BC}+\mathrm{DA}$

$\Rightarrow 6+4=7+\mathrm{AD}$

$\Rightarrow \mathrm{AD}=10-7=3 \mathrm{~cm}$

Hence AD= 3 cm