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.
Answers
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.
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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