Question

3. A circle touches all the sides of the quadrilateral ABCD. If AB = 12.8 cm. BC = 15.2 cm
and CD = 9.3 cm. then find AD.​

Answer 1

Answer:

Since, length of two tangents drawn from an external point of circle are equal.

So, AP=AS

BP = BQ DR = DS

and RC = CQ

Adding all we get

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

(AP + BP) + (DR + RC)

(AS + DC) + (BQ + CQ)

AB + DC = AD + BC

6 + 4 = AD + 7

10 = AD + 7

AD = 10-7 = 3cm

Hence, AD = 3cm.

Step-by-step explanation:

Hope this answer will help you.

Answer 2

Given :-

• A circle touches all the sides of the quadrilateral ABCD.
• AB = 12.8 cm.
• BC = 15.2 cm.
• CD = 9.3 cm.

To Find :-

• Length of AD .

Solution :-

we know that, A circle touches all the sides of the quadrilateral it is called cyclic quadrilateral.

so,

• In cyclic quadrilateral :- AB + CD = BC + AD .

then,

→ 12.8 + 9.3 = 15.2 + AD

→ 22.1 = 15.2 + AD

→ AD = 22.1 - 15.2

→ AD = 6.9 cm . (Ans.)

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