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Intern  B
Joined: 11 Jun 2017
Posts: 10
In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75;  [#permalink]

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In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75; Angle CDA = 60; what is the perimeter of the quadrilateral ABCD in terms of x?
Math Expert V
Joined: 02 Aug 2009
Posts: 7684
Re: In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75;  [#permalink]

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harleen.kaur wrote:
In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75; Angle CDA = 60; what is the perimeter of the quadrilateral ABCD in terms of x?

Hi..
Please post the Q properly with all choices etc.

Otherwise for your answer..
Join A with C..
So two triangles ABC... Isosceles RIGHT angle triangle.
So angle ACD = 75-45=30... So ∆ACD becomes 30-60-90 ..
Two sides AB and BC are x, so AC is hypotenuse x√2...
Take ∆ACD, 30-60-90 means 1:√3:2...
Now opposite 60 is x√2... So other two sides are x√(2/3) and 2x√(2/3)...
Perimeter=x+x+x√(2/3)+2x√(2/3)..
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Re: In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75;  [#permalink]

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Hard to explain clearly without drawing a diagram (which is something that would usually be provided in a question like this, if it were a real GMAT question), but: the 75 degree angle is difficult to work with, so we want to divide it up into angles that are familiar. If you draw the line AC, which is a diagonal within the quadrilateral, then because the angle at B is 90 degrees, and because AB = BC, the triangle ABC we've just created is a 45-45-90 triangle. So using the 1 to 1 to √2 ratio, since the short side in this triangle is x, the hypotenuse, which is AC, is √2 x.

Now we have the length of AC in the other triangle ACD. We're told the angle at D is 60 degrees, and the angle ACD is 30 degrees (since it's 75 - 45). So triangle ACD is a 30-60-90 triangle. We found above that the length of the side AC, opposite the 60 degree angle, is √2 x. Using the 30-60-90 triangle ratio, we know the sides are in a 1 to √3 to 2 ratio, so if the middle-length side is √2 x, the short side, AD, is (√2 / √3) x = (√6 / 3) x, and the long side, CD, is (2√2 / √3)x = (2 √6 / 3)x.

We now know the lengths of the four sides of the quadrilateral: they are x, x, (√6/3)x and (2√6/3)x. Adding these we get the perimeter:

2x + (√6/3)x + (2√6/3)x = 2x + (3√6/3)x = 2x + √6 x = x (2 + √6)
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# In a quadrilateral ABCD, AB = BC = x; Angle ABC = 90; Angle BCD = 75;

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