Find the angle between the diagonal of a rectangle with perimeter 2p and area (3/16)p^2(i.e. 3/116 times ‘p’ square).
Can someone please help me with the solution?
I can get you started, but I’m not sure the question is complete- we’re looking for the angle between and diagonal and what? One of the sides? If no side is specified the only logical solution is that the rectangle is really a square, which means each side is p/2, but that doesn’t give you the right area. Therefore there must be something missing in the question.
The basic solution would be to let x be one side of the rectangle. Then the other side has to be (p – x) since the perimeter adds up to 2p. Also we know that x(p – x) = 3/16(p ^ 2). Therefore (x ^ 2) – px + 3/16(p ^ 2) = 0. You can treat this as a quadratic equation and solve for x in terms of p. The you can find the length of the 2 sides on terms of p, and once you have that you can find the diagonal and the angles using relatively simple trigonometry.
Hope this helps. If anyone else would care to contribute that’d be great.
thanks for the solution oLahav. I have got this problem from one of our lessons. http://gmat.learnhub.com/lesson/5727-circle-and-triangles.
Dear Olahav
But this quadratic equation will have imaginary roots if i am not wrong if the roots are imaginary then u cant form such a rectangle. do guide.
i don’t think the roots are imaginary
the discriminant of the equation is: (p)2 (4)(p2)(3/16) = p2 – (3/4)p2 = (1/4)p^2
which is positive; does this help you understand it better?
Post Comments