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Given a quadrilateral ABCD, a circle is inscribed in the [#permalink]

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11 May 2004, 03:54

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Given a quadrilateral ABCD, a circle is inscribed in the quadrilateral in such a way that it touches all of the four sides. What is the perimeter of the quadrilateral?

Given a quadrilateral ABCD, a circle is inscribed in the quadrilateral in such a way that it touches all of the four sides. What is the perimeter of the quadrilateral?

(1) AB+BC=10 (2) AB+CD=12

2 alone is sufficient, while 1 alone is not.

The reason behind is that for a every quadrilateral to have a circle inscribed inside it is equivalent to AB + CD = AD + BC.

Why this is so? If you mark points, where circle inscribed touches quadr. by A1, B1, C1, D1, then |AA1| = |AD1|, |BA1| = |BB1|, |CB1| = |CC1|, |DC1| = |DD1| => since AB = AA1 + BA1, BC = BB1 + CB1, CD = CC1 + DC1, DA = DD1 + AD1, AB + CD = BC + AD.

So, opposite sides of quadr. sum to equal values. => P = 2*(AB + CD) = 24.

Tangents to circle are perpendicular to the radius, using this property, we can prove that it will be a rectangle in which circle is inscribed.
Hence answer should 2*(AB+BC) = 2*10 = 20

Tangents to circle are perpendicular to the radius, using this property, we can prove that it will be a rectangle in which circle is inscribed. Hence answer should 2*(AB+BC) = 2*10 = 20

Any comments?

No, a quadrilateral (such that a circle can be inscribed inside it) is not always rectangle. For instance, consider romb (it has all sides equal, but its angles need not be equal to 90).

Emmanuel: I still didn't get your logic
Why this is so? If you mark points, where circle inscribed touches quadr. by A1, B1, C1, D1, then |AA1| = |AD1|, |BA1| = |BB1|, |CB1| = |CC1|, |DC1| = |DD1| => since AB = AA1 + BA1, BC = BB1 + CB1, CD = CC1 + DC1, DA = DD1 + AD1, AB + CD = BC + AD.

How come from above logic you can say (AB+CD=BC+AD)?
(Before joining this group I was under illusion that I am an expert in Maths!!! )

Emmanuel: I still didn't get your logic Why this is so? If you mark points, where circle inscribed touches quadr. by A1, B1, C1, D1, then |AA1| = |AD1|, |BA1| = |BB1|, |CB1| = |CC1|, |DC1| = |DD1| => since AB = AA1 + BA1, BC = BB1 + CB1, CD = CC1 + DC1, DA = DD1 + AD1, AB + CD = BC + AD.

How come from above logic you can say (AB+CD=BC+AD)? (Before joining this group I was under illusion that I am an expert in Maths!!! )