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# Given a quadrilateral ABCD, a circle is inscribed in the

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11 May 2004, 02:54
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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?

(1) AB+BC=10
(2) AB+CD=12
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11 May 2004, 03:08
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hallelujah1234 wrote:
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.
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11 May 2004, 03:42
In my opinion answer should be (A),

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

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11 May 2004, 03:47
mba wrote:
In my opinion answer should be (A),

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

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).

.../\.....
../..\....
..\../....
...\/.....
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11 May 2004, 12:24
at least two sides of the quadrilateral will be equal.
A is enough.
B is not enough
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Chetan

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12 May 2004, 00:04
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!!! )
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12 May 2004, 00:35
mba wrote:
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!!! )

OK, see this:
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4.GIF [ 8.99 KiB | Viewed 5204 times ]

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12 May 2004, 00:49
nice job emmauel..thanks for making it more clear...
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12 May 2004, 23:31
Thanks, emmanuel.

(Now I wonder how come I didn't see this thing earlier, )
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13 May 2004, 06:20
Wow! I really liked Emmanuel's solution.
13 May 2004, 06:20
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