An equilateral trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if points A and D belong to the x−axis and m∠BAD=30o?
toxic much yiks
An isosceles trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if points A and D belong to the x−axis and m∠BAD=60o? This is what he meant.
An equilateral trapezoid ABCD with height 2 units has all its vertices on the parabola y=a(x+1)(x−5). What is the value of a, if points A and D belong to the x-axis and m∠BAD=30o?
guess can u help me out
An "equilateral" trapezoid is a square. So, I assume you mean an isosceles trapezoid.
Using the 30° angle, we know that since point A is (-1,0) and D is (5,0),
point B is (-1+2√3,2)
point C is (5-2√3,2)
But that cannot be, since the axis of symmetry is at x=2, so Band C lie beyond the axis.
So, assume the worst; check for the slope from A to the vertex. That will be the minimum possible slope such that a trapezoid can be fit under the curve.
A line from (-1,0) to the vertex at (2,9a) has slope 9a/3 = 3a
so, we need
3a = 1/√3
a = 1/3√3
But that means the vertex is at (2,√3) so graph as a max less than 2. The height of ABCD cannot be 2.
That is, there is no parabola with roots at -1 and 5 such that the required trapezoid fits.
Of course, maybe I made a mistake, so I welcome input.