If a parallelogram has vertices A(3,3), B(8,0), C(1,-6), and D(-4,-3), what are the lengths of the diagonals?

Question

This is trigonometry homework. I'd like to understand it for a test, so help would be appreciated. Thanks.

I forgot to add:
IT MUST BE SOLVED BY USING THE FOLLOWING: cosine law, sine law, Pythagorean, or a combination of them. This isn't a very advanced math class and it needs to be done using these. Thanks.

AND PLEASE, NOT THAT FORMULA WITH SQUARE ROOTS! IT'S NOT IN THIS CURRICULUM!

FOR THE 99TH TIME PLEASE. PLEASE. IF YOU WANT TO BE CHOSEN AS BEST ANSWER, DO NOT GIVE ME THE SOLUTION IN GEOMETRY. IT IS NOT THE ANSWER. THIS HAS BEEN DONE MULTIPLE TIMES AFTER I HAVE CLEARLY SPECIFIED NOT TO DO THIS. I WILL REPORT ABUSE AS POINT-GAMING TO ANYONE WHO DECIDES TO DO THIS.

Answer 1

Using Pythagoras

AC^2 = (3-1)^2 + (3 - -6)^2
AC^2 = 2^2 + 9^2
AC^2 = 4 + 81
AC^2 = 85
AC = 9.2195 (4 DP)

BD^2 = (8 - -4)^2 + (0 - -3)^2
BD^2 = 12^2 + 3^2
BD^2 = 144 + 9
BD^2 = 153
BD = 12.369

Answer 2

If a parallelogram has vertices A(3,3), B(8,0), C(1,-6), and D(-4,-3), what are the lengths of the diagonals?

Diagonal AC = sqrt[(3-1)^2 + (3 -(-6))^2]
= sqrt(4 +81)=sqrt(85)

Diagonal BD = sqrt[(0-(-3))^2 + (8 -(-4))^2
sqrt( 9 +144) = sqrt (153)=sqrt(9*17) =3* sqrt (17)

BTW, this is a Geometry problem, not a trigonometry problem.

Answer 3

Well the diagonals are going to go from (3,3) to (1,-6) and (8,0) to (-4,-3). The distance from 3,3, to 1,-6 is square root of( (3-1)squared + (3- -6)) so you end up with the square root of (2 sqaured +9 squared), which is the sqaure root of 85. To find the distance of the other diagonal, set it up the same, but with the other coordinates in there. You should end up getting the square root of 153 as the answer.

Edit: it would not let me reply to your message: This is what I came up with
I found this link with the formula:
http://mathforum.org/dr.math/faq/formulas/faq.quad.html#parallelogram


p = sqrt[a2+b2-2ab cos(A)]
q = sqrt[a2+b2-2ab cos(B)]

In this situation. p= AC and q= BD a=CD and b=CA.
In the equation A (as in cos A) is the angel ACD and B is the angle BAC. It helps to see the picture at the link though to understand, and then label your graph accordingly. You can find the angles by finding the tangent (opposite/adjacent) and then taking the reverse tangent to find the angle. I hope this is what you are looking for, and I hope it helps. Let me know if it is confusing or you have further questions.

Answer 4

If you graph the parallelogram, you'll see that the diagonals are AC and BD. So, all you have to do is find the distance between A and C and the distance between B and D.

AC = sqrt((3-1)^2+(3-(-6))^2) = sqrt(4+81) = sqrt(85)
BD = sqrt((8-(-4))^2+(0-(-3))^2) = sqrt(144+9) = sqrt(153)

Answer 5

This problem can be solve with analytic geometry, specifically using the distance formula:

√[(x2 -x1)^2 + (y2 - y1)^2]

In the given problem, the diagonals are AC and BD, therefore:
AC = √[(1-3)^2 + (-6 - 3)^2]
AC = √[(-2)^2 + (-9)^2]
AC = √[4 + 81]
*AC = √(85)

BD = √[(-4 - 8)^2 + (-3 - 0)^2]
BD = √[(-12)^2 + (-3)^2]
BD = √[144 + 9]
*BD = √(153)