Question

In figure, a circle touches all the four sides of a quadrilateral ABCD whose sides are AB = 6 cm, BC = 9 cm and CD = 8cm, find the length of side AD.​

Answer 1

Answer:

5 cm

Step-by-step explanation:

Hey Simi, here's your answer:

Let the circle touches the sides AB, BC, CD and DA of the quadrilateral ABCD at points P, Q, R and S respectively.

Since the lengths of the tangents drawn from an external point are equal. So

DR = DS (tangents on circle from point D)

CR = CQ (tangents on circle from point C)

BP = BQ (tangents on circle from point B)

AP = AS (tangents on circle from point A)

Adding all these equations

DR + CR + BP + AP = DS + CQ + BQ + AS

(DR + CR) + (BP + AP) = (DS + AS) + (CQ + BQ)

CD + AB = AD + BC

AD = 8 cm + 6 cm - 9 cm = 5 cm

Pls mark it as the brainliest :)

Answer 2

Hey sis

Your answer

We know that quadrilateral circumscribing the circle where tangents are equal

Let PQRS be a circle and ABCD be quadrilateral then

We know that tangent drawn from external point are equal so, AQ=AP ,BQ=BR,CS=CR,DS=DP add all these so, AB+CD= AD+BC. Let AD =X so from question 6+8=9+X so 14-9=X and X=5cm