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  • Brainly User
2014-10-23T19:20:06+05:30
ABCD is a quadrilateral. A circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at P, Q, R and S respectively
.AB = 6 cm, CD = 4 cm and BC = 7 cm
We know that, the length of tangents drawn from an external point to the circle are equal.
AP = AS
BP = BQ
CQ = CR
DR = DS
Now, AB + CD = (AP + PB) + (CR + DR)
= (AS + BQ) + (CQ + DS)
= (AS + DS) + (BQ + CQ)
= AD + BC
∴ AB + CD = AD + BC
⇒ AD = AB + CD – BC
⇒ AD = 6 cm + 4 cm – 7 cm
⇒ AD = 10 cm – 7 cm
⇒ AD = 3 cm
Thus, the length of the side AD is 3 cm.
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thank you sir.
2014-10-23T21:41:03+05:30

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See diagram.

Draw perpendicular (normals) to the sides at the points where the inscribed circle touches the quadrilateral.

Let GC = x cm.  So  DG = 4 -x = DH = 4-x  cm

As CF = x cm, FB = 7-x  so BE = 7 - x  cm

So AE = 6 - (7-x) = x-1  cm

So AH = x-1 and  AD = x-1+4-x = 3 cm

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