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Re: In quadrilateral ABCD, AB = AD = 5 and BC = 6 and CD = 1. Which of the [#permalink]
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Hello, everyone. I have nothing to add to the excellent analysis provided above by chetan2u. However, for the more visually inclined (or dependent) solver, such as me, I thought I would attach a figure. When you pair the two posts—again, if you need more than a description—the problem should be easier to grasp. Even though it plays on the simple notion of finding a valid side length given two other sides of a triangle, since those triangles are hidden here and we still have to conceptualize what sorts of angles would be possible within one of them, I would probably call this a 700-level question. (Just my opinion.)

Attachment:
Screen Shot 2021-05-07 at 16.19.33.png
Screen Shot 2021-05-07 at 16.19.33.png [ 17.64 KiB | Viewed 7958 times ]


Cheers, and good luck with your studies.

- Andrew
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In quadrilateral ABCD, AB = AD = 5 and BC = 6 and CD = 1. Which of the [#permalink]
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The way I reasoned through it:

If we connect diagonal BD and call that side X, we can create triangle BCD inside the funky quadrilateral (sounds like a rap name from the 80s)

The triangle inequality theorem states that, when you have 1 unknown side:

(Differences between 2 known sides) < X < (SUM of two known sides)

The above inequality must hold for every triangle.

(6 - 1) < X < (6 + 1)

5 < X < 7

X is also the Non Equal side of an Isosceles Triangle ABD, with side lengths:

5
5
5 < X < 7


X can not be 5, so the Angle at the vertex between the 2 equal sides of length 5 can not be 60 degrees (this can’t be an equilateral triangle)

Also, if we assign any value from 5 to 7 (the boundaries) to Diagonal BD, the Pythagorean theorem can not be satisfied.

If Angle A is an Obtuse Angle greater than 90 degrees, then the following relationship will hold true (holds true for any obtuse triangle)

(Longest side)^2 > (5)^2 + (5)^2


Giving diagonal BD the highest boundary value of 7 (which it must be less than) we still can not satisfy this condition:

(7)^2 > (5)^2 + (5)^2
49 is not greater than 50

Therefore, triangle ABD can not be an obtuse triangle with an angle greater than 90 degrees. Each angle must be less than 90 degrees, including the angle at vertex A

Since BD is greater than 5, the angle at vertex A (opposite side BD) can not be 60 degrees. Side BD must be longer than 5, the length that would make it an equilateral triangle.

-if we start with the Angle at vertex A as 60 degrees, angle A will have to “stretch” wider to accommodate Side BD which is longer than 5

60 degrees < angle A < 90 degrees


Answer must be 75 degrees


Perfectly crafted question, answers and all.

Posted from my mobile device
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Re: In quadrilateral ABCD, AB = AD = 5 and BC = 6 and CD = 1. Which of the [#permalink]
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Re: In quadrilateral ABCD, AB = AD = 5 and BC = 6 and CD = 1. Which of the [#permalink]
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