If ABCD is a square, what are the coordinates of C?

In this example, all the right triangles formed would be 30 - 60 - 90 as per diagram below;

If Side =2 then the diagonal equals \(2\sqrt{2}\)

The diagonal equals the distance between the 2 points. So I assume we can use the distance between 2 points formula.

D=\(\sqrt{(x2-x1)^2+(y2-y1)^2}\)
\(D^2\)=\((x2-x1)^2+(y2-y1)^2\)

\(2\sqrt{2}^2\)=\((x2-0)^2+(y2-1)^2\)


\(2\sqrt{2}^2\)=\((1+\sqrt{3})^2+(\sqrt{3}-1)^2\)

8=\((1+\sqrt{3})^2+(\sqrt{3}-1)^2\)

8=4+2

???

Can anyone help and tell me where I went wrong.

I thought I could apply the distance between 2 points formula because we knew the distance, and points X1 and Y1.

If anyone can help please do so. I did some practice problems using the formula and it works, however, it does not work for this problem. I emailed Magoosh, where the Q comes from, but they never respond.

You can. Everything is correct except the last calculation.

8=\((1+\sqrt{3})^2+(\sqrt{3}-1)^2\)

\(8= 1^2 + \sqrt{3}^2 + 2*1*\sqrt{3} + \sqrt{3}^2 + 1^2 - 2*\sqrt{3}*1\)
\(8 = 1 + 3 + 3 + 1 = 8\)

There are many other ways of solving this question:

One method is to note that the sides of 30-60-90 triangle are in the ratio 1:\(\sqrt{3}\):2. Since OA has length 1, OD has length \(\sqrt{3}\). Because the square is tilted, x coordinate of C will be more than x coordinate of D. So options (A), (B) and (E) are out. Also, the square is slightly titled so the x coordinate of C will not be twice of x coordinate of D. Hence option (C) is not possible either. Answer must be (D).


We see that coordinates of D are (\(\sqrt{3}\), 0) because of the 30-60-90 triangle. Now imagine that the x axis is turned 90 degrees at D. So the y coordinate of C will be \(\sqrt{3}\). Also since OA is of length 1, when you rotate it by 90 degrees, the x coordinate of C is 1 more than the x coordinate of C i.e. it is \(\sqrt{3} + 1\).

Karishma,

Thank you so much. This is huge help. I can not believe I missed that.

Is this a special pythagorean triangle like the ones with sides 3,4,5 and 5,12,13?

Pythagorean triplets give you the ratio of sides where all three lengths of sides are integers - the most common one is 3, 4, 5.

This is the ratio of sides in case the right triangles are special triangles:

Triangle with angles 30-60-90 (occur often) - Here ratio of sides is 1: root 3: 2 (Not a pythagorean triplet since all sides are not integers)
Isosceles right triangle i.e. angles are 45-45-90 - Here ratio of sides is 1:1:root 2 (Not a pythagorean triplet since all sides are not integers)

Can anyone Pls. explain , how did we get 1 for one of the sides of the blue triangle. I solved till that point and could not proceed.

As mentioned above, after you get the angles as 30-60-90 in the blue triangle, you need to remember that the sides for a 30-60-90 triangle are in the ratio \(1:\sqrt{3}:2\) with '1' side opposite to 30 degree angle. Thus you get 1 for that length.

Another way is by trigonometry: in the blue triangle, you know that the hypotenuse =2

Thus in this triangle, cos (60) = base / hypotenuse ---> 1/2 = base / 2 ---> base = 1 (as cos(60) = 0.5)

An explanation using property of slopes!

Product of the slopes of Perpendicular lines = -1

\(m_1*m_2 = -1\)
Since AO = 1 therefore OD = √3 [30-60-90 Triangle ratio]
Let, Slope of Line AD, \(m_1 = -1/√3\)
i.e. Slope of Line DC, \(m_2 = √3\)
in Triangle i.e. Ratio of Y-Co-ordinate of C to Horizontal Distance of CD on X-Axis = √3/1

i.e. x Co-ordinate of C = OD+1= √3+1

i.e. Co-ordinates of C = ((√3+1),√3)

i.e. Option D

since we have 30-60-90 triangle, we can deduct that the sides of the small triangle with the base on the x axis is 1-sqrt(3)-2
which makes the length of a side of the square = 2.
since all the sides are equal, we can draw a line from point c to the x-axis to form a 30-60-90 triangle.
since we have a similar triangle with the first one, we see that the hypotenuse is the same = 2. which makes that the point on y axis is sqrt(3)

now, only points A, C, and D have y coordinate of point c sqrt(3). B and E are thus eliminated.
now, from the origin to the point D, we have a distance of sqrt(3). The distance from point D to the line perpendicular to the x-axis which form a 30-60-90 triangle is 1. Thus, the x-coordinate of the point C must be 1+sqrt(3)

Eliminate A, and C.

D is the answer.

For Red Triangle
cos 60 =adjancent side/hyp
1/2 =1/h
h=2

cos30=\sqrt{3}/2=x/2
x axis length = \sqrt{3}

now the blue side
cos 60=1/2=x/2
x=1

so total lenght of the x axis is 1+\sqrt{3}

only D is the suitable option

VeritasKarishma

How distance formula helped find the coordinate?? I did not get this method.
Kindly help.

VeritasKarishma

I believe it's just trying out options and getting 8 as distance??
Will that not take time? Or am I missing something here??

Yes, it is taking options and plugging them in to find the one that yields a diagonal length of \(2\sqrt{2}\). Personally, I don't favour this method. Check out my reply above to see how I would solve this question.

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