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In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an i

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In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an i  [#permalink]

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New post 28 Mar 2019, 05:59
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A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

72% (01:55) correct 28% (02:17) wrong based on 25 sessions

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Re: In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an i  [#permalink]

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New post 28 Mar 2019, 06:27
1
1
1
using property, sum of two sides of triangle is always greater than the 3rd side
and difference of two sides of a triangle is always less than the 3rd side

in traingle ADB
AD+AB>BD
9+5>bd
14>BD
means BD is less than 14 (I)


in trauingle BCD
BC-CD<BD
17-5<BD
12<BD
means BD is greaer than 12 (II)

from (I) and (II)
BD lies between 12 and 14 and is a integer. so it should be 13
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Re: In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an i  [#permalink]

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New post 28 Mar 2019, 13:00
Bunuel wrote:
Image
In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an integer. What is BD?

(A) 11
(B) 12
(C) 13
(D) 14
(E) 15


Attachment:
570b698fea7c6d66beab04dd6aee44dd24f2b6c1.png


solve the question using ∆ 3rd side rule
∆ ABD and ∆CBD we see range 12-14
so IMO C ;13 possible
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Re: In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an i   [#permalink] 28 Mar 2019, 13:00
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In quadrilateral ABCD, AB = 5, BC = 17, CD = 5, DA = 9, and BD is an i

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